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Texas Placement Paper : Technical - Other

TEXAS Instuments
1.which code executes faster?
code1:- for(i=0;i for(j=0;j large_array[i][j]=0;

code2:- for(j=0;j for(i=0;i large_array[i][j]=0;
ans:code1
2.void fn (int *ptr)
{
static int val=100;
ptr=&val;
}
main()
{
int i=10;
printf("%d", i);
fn(&i);
printf("%d", i);
}
ans:prints 10,10
3. C supports - call by value only.
4.main()
{
int i=1;
fork();
fork();
printf("%d",i+1);
}
ans :- prints 2 four times
12.#define max(a,b) (a>b)?a:b
main()
{
int m,n;
m=3+max(2,3);
n=2*max(3,2);
printf("%d,%d",m,n);
}
ans:-m=2,n=3
13.int fn()
{
int x=2;
return x;
}
char * g()
{
char x[4];
strcpy(x,"hello");
return x;
}
main()
{
char *s;
s=g();
fn();
printf("%s\n",s);
}
14.representation of -0.375 --- 111.101
15.1-way set assosiated memory ----- mapped
ans:-direct
16. translation look aside buffers used in______
ans:- cache memory probably
17. DSP
18. faster IPC mechanism
ans:- shared memory
19.some question on gates(figure)
ans:- c or d
20. A+(barA)B= A+B
21. time page
1 1
2 2
3 1
4 3
5 4
6 stop
what are the pages in memory. C memory page table size=21
22. write swap methods with out using temp variables.
23.count no. of 1,s in a word with out counting individaul bits.
24.complexity of binary search tree .



Texas Placement Paper : Technical - C & C++ - 8 September 2005

Texas Instruments Date : 8/9/2005

given an expression tree and asked us to write the in fix of that expression
four choices : 2

global variables in different files are
a:at compiletime
b) loading time
c) linking time
d)execution time

size of(int)
a) always 2 bytes
b) depends on compiler that is being used
c) always 32 bits
d) can,t tell

which one will over flow given two programs
2 prog 1: prog2:
main() main()
{ {
int fact; int fact=0
long int x; for(i=1;i<=n;i++)
fact=factoral(x); fact=fact*i;
} }
int factorial(long int x)
{
if(x>1) return(x*factorial(x-1);
}
a) program 1;
b) program 2;
c) both 1 &2
d) none
}

variables of fuction call are allocated in
a) registers and stack
b) registers and heap
c) stack and heap
d)

avg and worst case time of sorted binary tree

data structure used for proority queue
a) linked list b) double linkedd list c)array d) tree

main(){
char str[5]="hello";
if(str==NULL) printf("string null");
else printf("string not null");
}
what is out put of the program?
a) string is null b) string is not null c) error in program d) it executes but print nothing

There are 0ne 5 pipe line and another 12 pipe line sates are there and flushed time taken to execute five instructions a) 10,17
b) 9,16
c)25,144
d)

for hashing which is best on terms of buckets
a)100 b)50 c)21 d)32 ans 32

void f(int value){
for (i=0;i<16;i++){
if(value &0x8000>>1) printf("1")
else printf("0");
}
}
what is printed?
a) bineray value of argument b)bcd value c) hex value d) octal value

void f(int *p){
static val=100;
val=&p;
}
main(){
int a=10;
printf("%d ",a);
f(&a);
printf("%d ",a);
}
what will be out put?
a)10,10

struck a{
int x;
float y;
char c[10];
}
union b{
int x;
float y;
char c[10];
}
which is true?
a) size of(a)!=sizeof(b);
b)
c)
d)

# define f(a,b) a+b
#defiune g(c,d) c*d
find valueof f(4,g(5,6))
a)26 b)51 c) d)

find avg access time of cache
a)tc*h+(1-h)*tm b)tcH+tmH
c) d) tc is time to access cache tm is time to access when miss occure

main()
{
char a[10]="hello";
strcpy(a,,\0,);
printf("%s",a);
}
out put of the program?
a) string is null b) string is not null c) program error d)

simplyfy k map
1 x x 0
1 x 0 1

int f(int a)
{
a=+b;
//some stuff
}
main()
{
x=fn(a);
y=&fn;
what are x & y types
a) x is int y is pointer to afunction which takes integer value

char a[5][15];
int b[5][15];
address of a 0x1000 and b is 0x2000 find address of a[3][4] and b[3][4]
assume char is 8 bits and int is 32 bits
a) b) c) d)
there are 20 questions all in techinical paper and 36 questions in appititude test in appititude thay have given all diagrams and asked to find what comes next thay are quite easy and i hope if u practice r.s aggraval u can do it easily for tecnical thay have given 1 hr for 20 questions and for not technical thay have given only 40 min and 36 questions
this is the paper i have right now for TI aptitude test consist of all pictorial questions. ie in each question he will give 8 diagrams and ask to find th 9,th diagram in that sequence.You go through RS Agarwal. These aptitude questins are
very easy. Just pratice them. In RS Agarwal gothrough SERIES chapter. It is suffient. There are 35 aptitude
questions. First 25 are very easy. Do these questions in just 15 or 20 minutes. Because last questions are
very touch.

TECHNICAL TEST:
3 flipflops are connected so that after 0 to 5 count occured next number is zero. So what is the counter?
Ans: mod 6 counter

simplication of some boolean expression which is simple. Boolean Expression is A+A,B.
Ans:A+B

Given inorder sequence and preorder sequence and asked to find out postorder sequence.

Some question on value of a static variable.

Given an interger in binary form,find the number of ones in that number without counting each bit.(This questin is not multiple choice question. This question carries more marks. So please take care for this question.)

1-way set associative memory is called-----
a)direct b)something c)1-way set associative 4)something
Ans: c

Fastest IPC mechanism is
a)shared memory b)pipes c)named pipes d)semaphores
Ans:c

Some page references are given. You are asked to implement it with Least Frequently Used algorithm.

Some diagram is given. Iam describinmg the diagram. A 2*1 MUX is given. The inputs are A,B. Output is C. C and A are tied together. What is the diagram.?
Ans:Latch.

This paper is for Electrical & Electronics students. There is separate test for computer Science Students. There are 20 questions.


1)Some circuit is given. Iam describing the circuit.A resistor R & a capacitor C are connected in parallel.
To this circuit another circuit which is having a capacitorof capacity 2C & an impedence Z, is connected in series.
You are asked to find out the value of Z? Note that 2C & Zare connected in series.
a)Z=2C
b)Z=2L
c)Z=L/2
d)Z=2R

Some circuit which consist of only resistors R is given. This is a repetative circuit. U have to find the effctive
resistance of the entire circuit.
A)Rin=R
B)Rin=(5+sqrt(3))/7
C)Rin=(19+sqrt(3))/8
D)None.


Two wave forms are given. You are asked to write the cirsuit to get B(second wave form) from A(first wave form).


#define SUM(a,b) a+b
main()
{
a=2;
b=3;
x=SUM(a,b)*2;
printf("x=%d\n",x);
}
Ans:8.

number(int i)
{
number++;
printf("%d\n",number);
}
main()
{
static int i=0;
number(i);
}
Ans: I don,t know.

Some circuit is given. I can,t describe the circuit. There are
3 resistors,3 capacitors & one inverter.. The question is
What is the value of the frequency such that the circuit oscillates.
A)f=RC
B)f=sqrt(3)/(Pi*R*C)
C)f=1/(Pi*R*C)
D)something
Ans:I don,t know the answer.

7)Question on flipflop. So gothrough all flipflops.

8)There are 5 questions on Nmos & Pmos circuits.


This Paper is for Computer Science Students. THis paper is very easy. You can definitely do it in one hour.
The fastest memory is
(i) DRAM, (ii) ROM, (iii) SRAM, (iv) Main memory
Ans : SRAM

Programing exceptions are
(i) Asynchronous, (ii) Synchronous, (iii) None
Ans : Asynchronous

DSP which architecture is used
(i) MIMD, (ii) SIMD, (iii) Nueman, (iv) Harvard Architecture
Ans : Harvard Architecture

C prog. for searching for an element in linked list

main()
{
unsigned char i;
int sum;
for(i=0; i<300; i++)
sum+ = i;
printf("\nSum = %d\n", sum);
}
Ans : infinite loop

void fn(int *p)
{
static int val = 100;
p = &val;
}
main()
{
int i=10;
printf("i=%d\n", i);
fn(&i);
printf("i=%d\n", i);
}
Ans : i=10 i=10

int a[10[15];
char b[10[15];
(a) location g a[3][4], if base location g a[0][0] is ox1000
(b) location g b[3][4], if base location g b[0][0] is ox2000
int taken 32 bits and char taken 8 bits.
Ans : (a) ox10C4 (b) ox2031


Implement OR gate function with 2*1 MUX
Ans : A ___________
--------|2*1 MUX |
B | |--------o/p
--------| |
| -----------
|_______|C
B=C

Implement 4*1 MUX with 2*1 MUXES

Swapping without using a temporary variables. (2 methods)
(i) x = x+y;
y = x-y;
x = x-y;
(ii) x = x^y;
y = x^y;
x = x^y;

Count no of 1,s in a word without using bit by bit. (This question carries more marks. It is not a multiple choice
question.

Code 1 :
for(i=0; i<1000; i++)
for(j=0; j<100; j++)
x = y;
Code 2 :
for(i=0; i<100; i++)
for(j=0; j<1000; j++)
x = y;
Which code will execute faster
(i) Code 1 and Code 2 are of same speed,
(ii) Code 1,
(iii) Code 2,
(iv) None.
Ans : Code 2

main()
{
int a[10] = {1, 2, 3, ...., 10}, i, x=10, temp;
for(i=0; itemp = a[i];
a[i] = a[x-i-1];
a[x-i-1] = temp;
}
(i) All contents of array a are reversed
(ii) Only some portions are altered
(iii) Remains same
(iv) None
Ans : (iii)

An array is stored in row major order. The memory capacity is 30 MB. And in unix system demand paging is used. Which one will give more page faults?
#define V_L_I 10000
int i, j, array[V_L_I][V_L_I];
Code 1 :
array[i][j] = 1;
Code 1 :
for(j=0; jfor(i=0; iarray[i][j] = 1;
Ans : Code 2

In C which parameter passing technique is used?
(i) call by value,
(ii) call by reference,
(iii) both
Ans : call by value

A circuit is given with 2 exclusive OR gates whose boolean expression will be y = ,(AB) + AB (, indicates bar)

(17) main()
{
int i = 1;
fork();
fork();
printf("\ni = %d\n", i+1);
}
Ans : 4 printfs will occur and i = 2

Compute the complexity of Binary search.
Ans : O(lg n) ( Answer in detail. This is not a multiple choice question. It carries more marks.)

Write expression for the tree graph :
Ans : ((a-b) + c*d)/x

# define MAX(a, b) a>b ? a:b
main()
{
int m, n;
m = 3 + MAX(2, 3);
n = 2 * MAX(3, 2);
printf("m = %d, n = %d\n", m, n)
}
Ans : m=2, n=3

Texas Placement Paper : Technical - Java - 8 September 2005

Date : 8/9/2005
Personal Interview
******************
Some Tips
**************


Something about uIt should about your family,childhood,schooling,B.E.,Future

Tell u,r Strength & Weakness

U should clearly explain u,r projects if they asked.

Area of interest

Have u attended any interviews before? If yes, then why haven,t you get selected?

If u are not selected in my company then how do u feel?

Why are u choose this field?

What is u,r future plan?

If they tell "ask some questions regarding the company" u should ask questions about the company

see the website of the particular company before attend the interview

why should we select u?

Here is some tips that will help u to prepare for hr interviews.Discuss it with friends,"sorry" in case of any mistakes in this.

Prepare in advance --at least 2 days before ,so that u would be more fluent and confident in interview.

When the interviewer asks "Tell about u r self"
start with"WITH PLEASURE SIR" -it is just to say "Thank u 4 giving me an oppourtunity to tell about my self"

.Start with a difference
---------usually students start with "My name is......,My father,s name is.....
Don,t do the same ,Start some thing different like
"People call me as ....."
"My parents named me as ......."
"................is my name"

4.Parents and family
After telling u r name ,tell u r parents name only.
A better approach would be "I,M PROUD TO BE THE SON OF MR.......AND MRS......"

5.Don,t organize as conventional essay
------u can swap the facts just to give the impression that u r spontaneous and had not prepared for this!!!!
For example,
After ur and ur parents details. u need not tell about u r school edu,
but u can start to tell about strengths.Later u can tell about u r school edu( this is just to give an illusion that ur spontaneous)

6.Prepare a speech for atleast 10 min.
--------------some times they expect even more...

7.Tell atleast 5 strengths and justify
Don,t use single line statements like
"I,m a confident person"
"I,m a hard worker"
Rather justify it.
For example if r a hardworker u can tell like this...
"sir ,i firmly belive in the power of hardworking.According to me the formula for success is hardwork only.In the past if i had succeded in any thing it is only because of my hardworking only....Because of my hard working only........................
Because of my hard working only......................
Because of my hard working only I had cleared the aptitude test conducted by u and in front of u.
List ur acheivements and tell hardwork as the key for it.....

Simillarly,What ever u say justify it.Tell hoew u implemented it.How it helped u.What u acheived from it.How it will be usefull for the company.
Some of the strengths are.........

Hard working

Positive attitude

optimistic

Ability to learn from mistakes

Passion to learn new things

Helping tendency

time conciousness

Good memory

Extrovert person

Team worker

Leadership quality ........

Don,t forget to justify,no single line sentences!!!!!
Give practical examples.....
Don,t tell the weakness unless they ask
Don,t tell more than one weakness.
If u tell any weakness ,Don,t forget to tell how u rectified it or how u r working to overcome it.
Don,t be toooo honest in telling u r weakness.


Texas Placement Paper : Technical - C & C++

What will be the output of the following code?

void main ()
{ int i = 0 , a[3] ;
a[i] = i++;
printf (?%d",a[i]) ;
}
Ans: The output for the above code would be a garbage value. In the statement a[i] = i++; the value of the variable i would get assigned first to a[i] i.e. a[0] and then the value of i would get incremented by 1. Since a[i] i.e. a[1] has not been initialized, a[i] will have a garbage value


Why doesn,t the following code give the desired result?
int x = 3000, y = 2000 ;
long int z = x * y ;
Ans: Here the multiplication is carried out between two ints x and y, and the result that would overflow would be truncated before being assigned to the variable z of type long int. However, to get the correct output, we should use an explicit cast to force long arithmetic as shown below:
long int z = ( long int ) x * y ;
Note that ( long int )( x * y ) would not give the desired effect.


Why doesn,t the following statement work?
char str[ ] = "Hello" ;
strcat ( str, ,!, ) ;
Ans: The string function strcat( ) concatenates strings and not a character. The basic difference between a string and a character is that a string is a collection of characters, represented by an array of characters whereas a character is a single character. To make the above statement work writes the statement as shown below:
strcat ( str, "!" ) ;


How do I know how many elements an array can hold?
Ans: The amount of memory an array can consume depends on the data type of an array. In DOS environment, the amount of memory an array can consume depends on the current memory model (i.e. Tiny, Small, Large, Huge, etc.). In general an array cannot consume more than 64 kb. Consider following program, which shows the maximum number of elements an array of type int, float and char can have in case of Small memory model.
main( )
{
int i[32767] ;
float f[16383] ;
char s[65535] ;
}


How do I write code that reads data at memory location specified by segment and offset?
Ans: Use peekb( ) function. This function returns byte(s) read from specific segment and offset locations in memory. The following program illustrates use of this function. In this program from VDU memory we have read characters and its attributes of the first row. The information stored in file is then further read and displayed using peek( ) function.
#include
#include
main( )
{
char far *scr = 0xB8000000 ;
FILE *fp ;
int offset ;
char ch ;
if ( ( fp = fopen ( "scr.dat", "wb" ) ) == NULL )
{
printf ( "\nUnable to open file" ) ;
exit( ) ;
}
// reads and writes to file
for ( offset = 0 ; offset < 160 ; offset++ )
fprintf ( fp, "%c", peekb ( scr, offset ) ) ;
fclose ( fp ) ;
if ( ( fp = fopen ( "scr.dat", "rb" ) ) == NULL )
{
printf ( "\nUnable to open file" ) ;
exit( ) ;
}
// reads and writes to file
for ( offset = 0 ; offset < 160 ; offset++ )
{
fscanf ( fp, "%c", &ch ) ;
printf ( "%c", ch ) ;
}
fclose ( fp ) ;
}


How do I compare character data stored at two different memory locations?
Ans: Sometimes in a program we require to compare memory ranges containing strings. In such a situation we can use functions like memcmp( ) or memicmp( ). The basic difference between two functions is that memcmp( ) does a case-sensitive comparison whereas memicmp( ) ignores case of characters. Following program illustrates the use of both the functions.
#include
main( )
{
char *arr1 = "Kicit" ;
char *arr2 = "kicitNagpur" ;
int c ;
c = memcmp ( arr1, arr2, sizeof ( arr1 ) ) ;
if ( c == 0 )
printf ( "\nStrings arr1 and arr2 compared using memcmp are identical" ) ;
else
printf ( "\nStrings arr1 and arr2 compared using memcmp are not identical"
) ;
c = memicmp ( arr1, arr2, sizeof ( arr1 ) ) ;
if ( c == 0 )
printf ( "\nStrings arr1 and arr2 compared using memicmp are identical" )
;
else
printf ( "\nStrings arr1 and arr2 compared using memicmp are not
identical" ) ;
}


Fixed-size objects are more appropriate as compared to variable size data objects. Using variable-size data objects saves very little space. Variable size data objects usually have some overhead. Manipulation of fixed-size data objects is usually faster and easier. Use fixed size when maximum size is clearly bounded and close to average. And use variable-size data objects when a few of the data items are bigger than the average size. For example,
char *num[10] = { "One", "Two", "Three", "Four",
"Five", "Six", "Seven", "Eight", "Nine", "Ten" } ;
Instead of using the above, use
char num[10][6] = { "One", "Two", "Three", "Four",
"Five", "Six", "Seven", "Eight", "Nine", "Ten" } ;
The first form uses variable-size data objects. It allocates 10 pointers, which are pointing to 10 string constants of variable size. Assuming each pointer is of 4 bytes, it requires 90 bytes. On the other hand, the second form uses fixed size data objects. It allocates 10 arrays of 6 characters each. It requires only 60 bytes of space. So, the variable-size in this case does not offer any advantage over fixed size.


The Spawnl( ) function...
DOS is a single tasking operating system, thus only one program runs at a time. The Spawnl( ) function provides us with the capability of starting the execution of one program from within another program. The first program is called the parent process and the second program that gets called from within the first program is called a child process. Once the second program starts execution, the first is put on hold until the second program completes execution. The first program is then restarted. The following program demonstrates use of spawnl( ) function.
/* Mult.c */
int main ( int argc, char* argv[ ] )
{
int a[3], i, ret ;
if ( argc < 3 || argc > 3 )
{
printf ( "Too many or Too few arguments..." ) ;
exit ( 0 ) ;
}
for ( i = 1 ; i < argc ; i++ )
a[i] = atoi ( argv[i] ) ;
ret = a[1] * a[2] ;
return ret ;
}
/* Spawn.c */
#include
#include
main( )
{
int val ;
val = spawnl ( P_WAIT, "C:\\Mult.exe", "3", "10",
"20", NULL ) ;
printf ( "\nReturned value is: %d", val ) ;
}
Here, there are two programs. The program ,Mult.exe, works as a child process whereas ,Spawn.exe, works as a parent process. On execution of ,Spawn.exe, it invokes ,Mult.exe, and passes the command-line arguments to it. ,Mult.exe, in turn on execution, calculates the product of 10 and 20 and returns the value to val in ,Spawn.exe,. In our call to spawnl( ) function, we have passed 6 parameters, P_WAIT as the mode of execution, path of ,.exe, file to run as child process, total number of arguments to be passed to the child process, list of command line arguments and NULL. P_WAIT will cause our application to freeze execution until the child process has completed its execution. This parameter needs to be passed as the default parameter if you are working under DOS. under other operating systems that support multitasking, this parameter can be P_NOWAIT or P_OVERLAY. P_NOWAIT will cause the parent process to execute along with the child process, P_OVERLAY will load the child process on top of the parent process in the memory.


Are the following two statements identical?
char str[6] = "Kicit" ;
char *str = "Kicit" ;
Ans: No! Arrays are not pointers. An array is a single, pre-allocated chunk of contiguous elements (all of the same type), fixed in size and location. A pointer on the other hand, is a reference to any data element (of a particular type) located anywhere. A pointer must be assigned to point to space allocated elsewhere, but it can be reassigned any time. The array declaration char str[6] ; requests that space for 6 characters be set aside, to be known
by name str. In other words there is a location named str at which six characters are stored. The pointer declaration char *str ; on the other hand, requests a place that holds a pointer, to be known by the name str. This pointer can point almost anywhere to any char, to any contiguous array of chars, or nowhere.


Is the following code fragment correct?
const int x = 10 ;
int arr[x] ;
Ans: No! Here, the variable x is first declared as an int so memory is reserved for it. Then it is qualified by a const qualifier. Hence, const qualified object is not a constant fully. It is an object with read only attribute, and in C, an object associated with memory cannot be used in array dimensions.


How do I write code to retrieve current date and time from the system and display it as a string?
Ans: Use time( ) function to get current date and time and then ctime( ) function to display it as a string. This is shown in following code snippet.
#include
void main( )
{
time_t curtime ;
char ctm[50] ;
time ( &curtime ) ; //retrieves current time &
stores in curtime
printf ( "\nCurrent Date & Time: %s", ctime (
&curtime ) ) ;
}

How do I change the type of cursor and hide a cursor?
Ans: We can change the cursor type by using function _setcursortype( ). This function can change the cursor type to solid cursor and can even hide a cursor. Following code shows how to change the cursor type and hide cursor.
#include
main( )
{
/* Hide cursor */
_setcursortype ( _NOCURSOR ) ;
/* Change cursor to a solid cursor */
_setcursortype ( _SOLIDCURSOR ) ;

/* Change back to the normal cursor */
_setcursortype ( _NORMALCURSOR ) ;
}


How do I write code that would get error number and display error message if any standard error occurs?
Ans: Following code demonstrates this.
#include
#include
#include
main( )
{
char *errmsg ;
FILE *fp ;
fp = fopen ( "C:\file.txt", "r" ) ;
if ( fp == NULL )
{
errmsg = strerror ( errno ) ;
printf ( "\n%s", errmsg ) ;
}
}
Here, we are trying to open ,file.txt, file. However, if the file does not exist, then it would cause an error. As a result, a value (in this case 2) related to the error generated would get set in errno. errno is an external int variable declared in ,stdlib.h, and also in ,errno.h,. Next, we have called sterror( ) function which takes an error number and returns a pointer to standard error message related to the given error number.


How do I write code to get the current drive as well as set the current drive?
Ans: The function getdisk( ) returns the drive number of current drive. The drive number 0 indicates ,A, as the current drive, 1 as ,B, and so on. The Setdisk( ) function sets the current drive. This function takes one argument which is an integer indicating the drive to be set. Following program demonstrates use of both the functions.
#include
main( )
{
int dno, maxdr ;
dno = getdisk( ) ;
printf ( "\nThe current drive is: %c\n", 65 + dno
) ;
maxdr = setdisk ( 3 ) ;
dno = getdisk( ) ;
printf ( "\nNow the current drive is: %c\n", 65 +
dno ) ;
}

The functions memcmp( ) and memicmp( )
The functions memcmp( ) and memicmp( ) compares first n bytes of given two blocks of memory or strings. However, memcmp( ) performs comparison as unsigned chars whereas memicmp( ) performs comparison as chars but ignores case (i.e. upper or lower case). Both the functions return an integer value where 0 indicates that two memory buffers compared are identical. If the value returned is greater than 0 then it indicates that the first buffer is bigger than the second one. The value less than 0 indicate that the first buffer is less than the second buffer. The following code snippet demonstrates use of both
#include
#include
main( )
{
char str1[] = "This string contains some
characters" ;
char str2[] = "this string contains" ;
int result ;
result = memcmp ( str1, str2, strlen ( str2 ) ) ;
printf ( "\nResult after comapring buffer using
memcmp( )" ) ;
show ( result ) ;
result = memicmp ( str1, str2, strlen ( str2 ) ) ;
printf ( "\nResult after comapring buffer using
memicmp( )" ) ;
show ( result ) ;
}
show ( int r )
{
if ( r == 0 )
printf ( "\nThe buffer str1 and str2 hold
identical data" ) ;
if ( r > 0 )
printf ( "\nThe buffer str1 is bigger than buffer
str2" ) ;
if ( r < 0 )
printf ( "\nThe buffer str1 is less than buffer
str2" ) ;
}

How do I write code to find an amount of free disk space available on current drive?
Ans: Use getdfree( ) function as shown in follow code.
#include
#include
#include
#include
main( )
{
int dr ; struct dfree disk ;
long freesp ;
dr = getdisk( ) ;
getdfree ( dr + 1 , &disk ) ;
if ( disk.df_sclus == 0xFFFF )
{
printf ( "\ngetdfree( ) function failed\n");
exit ( 1 ) ;
}
freesp = ( long ) disk.df_avail
* ( long ) disk.df_bsec
* ( long ) disk.df_sclus ;
printf ( "\nThe current drive %c: has %ld bytes
available as free space\n", ,A, + dr, freesp ) ;
}


Use of array indices...
If we wish to store a character in a char variable ch and the character to be stored depends on the value of another variable say color (of type int), then the code would be as shown below:
switch ( color )
{
case 0 :
ch = ,R, ;
break ;
case 1 :
ch = ,G, ;
break ;
case 2 :
ch = ,B, ;
break ;
}
In place of switch-case we can make use of the value in color as an index for a character array. How to do this is shown in following code snippet.
char *str = "RGB, ;
char ch ;
int color ;

Texas Placement Paper : Reasoning

Texas Instruments
If the date is written as MMDDYYYY, and then 10022001, ie)oct 2 2001 is a palindrome. which is the immediate palindrome before that date.


There r 2 trains travelling at 100 MPH each. there is a 200Mile tunnel. the trains travel in opposite directions. when the trains enter, a supersonic bee starts at the tunnel, touches the other train, comes back touches the first train, turns again and touches the other train and so on until the trains collide. Find the distance travelled by the bee.


It is 12.00 o clock. Find how often the minute hand meets the hour hand.


Find the maximum product using positive integers the sum of which is 100. Dont know if i am sure with this one. sorry. i didnt do this.


In a train Art is found dead. hours later 4 people are queried,

Blonde says i am innocent, i didnt speak with Art
White says i am innocent, Blonde spoke with Art
Old says i am innocent, the brunette killed Art
Brunette says i am innocent, one of the men killed Art
The inspector says simple, 4 true statements, 4 false statements.
I know the killer, very simple!!!
Find out the killer. (only one is involved.)


2 maths geniuses meet after 20 yrs.one says i am married. and have 3 daughters. the product of their ages is 72. the sum is the same as the house no. other says, but I cant find it out. The other says. . simple. ok, my eldest daughter has just started the piano classes.Find their ages.This one is there in previous question papers also.


5 sea pirates have 100 gold coins and want to share it. They propose a plan. The senior most one has to propose an idea, if at least 50 % agree, the coins r shared accordingly. else the senior is killed and the next senior most is asked to present a plan and so on .note, all the guys r very clever and very greedy and dont want to lose the coins, and dont want to die. Form a way to share the money.
I didnt do this....no time

ANSWERS

I worked it out like this.. before 2001, 2000 cannot be taken as MMDD becomes 0002.also anything after 14 century cant be considered as 1401 would mean 1041, there r only a max of 31 days a month. so the century nos must be only 13(reverse for 31)so the yr is 13xx. now 1399to 1391 is not possible as reverse means 99 to 19 but max no of months is 12. so 1390 becomes 09311390... but 9th month has only 30 days(lucky i remembered it).. so 1380 is the most recent palindrome ... which beocomes 08311380.SOln is 08 31 1380.


This one i really screwed up royally. it took me 25 minutes to solve.. i took the average speeds, the distance travelled by the trains and the bee, took ratio of speeds and distance, subtracting it from the tunnels length for every time the bee touched the trains. phew lot of calculation.. and got the ans as 1000.But when i rechecked it a thought struck me... so silly of me. The trains collide at the middle. ie)after 100 miles. that means after 1 hour. so the bee travels for 1 hour touching each train till they collide. its speed being 1000 mph, it travelled a 1000 miles..


At 12.00 the minute and hour hand are together. but at 1.00 they are 5 mins apart. so in 60 minutes the gain is 55 mins for the min hand. to be together it has to gain another 5 more mins. so the time taken to gain 5 mins is 60*5/55==5.454 So the time taken to meet again is 65.454 So the hands meet every 65.454 minutes


Sorry no time to see this and dont remember it correctly.

Texas Placement Paper : Technical - C & C++

Texas Instruments

Interview Procedure

The test is followed by a Technical and a HR interview. The technical interview is highly specialized and covers almost all subjects you have done in your curriculum. Some puzzles may also be asked in the interview. Special emphasis is laid on C and Data Structures.

Written Test paper there was 20 questions as follows in 60 minutes second part consists of 36 queue. in 30 minutes all questions are diagramatical.(figurs)..

Texas

if a 5-stage pipe-line is flushed and then we have to execute 5 and 12 instructions respectively then no. of cycles will be
a. 5 and 12
b. 6 and 13
c. 9 and 16
d.none


k-map
ab
----------
c 1 x 0 0
1 x 0 x
solve it

a. A.B
B. ~A
C. ~B
D. A+B


CHAR A[10][15] AND INT B[10][15] IS DEFINED WHAT,S THE ADDRESS OF A[3][4] AND B[3][4] IF ADDRESS OD A IS OX1000 AND B IS 0X2000

A. 0X1030 AND 0X20C3
B. OX1031 AND OX20C4
AND SOME OTHERS..


int f(int *a)
{
int b=5;
a=&b;
}

main()
{
int i;
printf("\n %d",i);
f(&i);
printf("\n %d",i);
}

what,s the output .

1.10,5
2,10,10
c.5,5
d. none


main()
{
int i;
fork();
fork();
fork();
printf("----");
}

how many times the printf will be executed .
a.3
b. 6
c.5
d. 8


void f(int i)
{
int j;
for (j=0;j<16;j++)
{
if (i & (0x8000>>j))
printf("1");
else
printf("0");
}
}
what,s the purpose of the program

a. its output is hex representation of i
b. bcd
c. binary
d. decimal


#define f(a,b) a+b
#define g(a,b) a*b

main()
{

int m;
m=2*f(3,g(4,5));
printf("\n m is %d",m);
}

what,s the value of m
a.70
b.50
c.26
d. 69


main()
{
char a[10];
strcpy(a,"\0");
if (a==NULL)
printf("\a is null");
else
printf("\n a is not null");}

what happens with it .
a. compile time error.
b. run-time error.
c. a is null
d. a is not null.


char a[5]="hello"

a. in array we can,t do the operation .
b. size of a is too large
c. size of a is too small
d. nothing wrong with it .


local variables can be store by compiler
a. in register or heap
b. in register or stack
c .in stack or heap .
d. global memory.


average and worst time complexity in a sorted binary tree is


a tree is given and ask to find its meaning (parse-tree)
(expression tree)
ans. ((a+b)-(c*d)) ( not confirmed)


convert 40.xxxx into binary .


global variable conflicts due to multiple file occurance
is resolved during
a. compile-time
b. run-time
c. link-time
d. load-time


two program is given of factorial. one with recursion and one without recursion .question was which program won,t run for very big no. input because
of stack overfow .
a. i only (ans.)
b. ii only
c. i& ii both .
c. none


struct a
{
int a;
char b;
int c;
}
union b
{
char a;
int b;
int c;
};
which is correct .
a. size of a is always diff. form size of b.(ans.)
b. size of a is always same form size of b.
c. we can,t say anything because of not-homogeneous (not in ordered)
d. size of a can be same if ...

Texas Placement Paper : Technical - C & C++ - 8 September 2005

Texas Instruments : 8/9/05


Pattern
Technical Test
Contains three sections (Data Structures and Algorithms, Analog Circuits, Digital Circuits) we have to choose any two sections among the three. Each section has 10 questions each and the total time for the two sections is 45 minutes. There is negative marking.
Aptitude Test
75 questions – 1 hour – Negative marking is there.
Technical Interview
Behavioral Interview
(1) Which of the following program is likely to create a Stack overflow for higher values of n
(i) (ii)
int fact(int n) int fact_ii(int n)
{ {
if (n >0) int fact_val = 1, i = 0;
return n*fact (n-1); for (i = 1; i <= n ; i++)
else fact_val = fact_val*i;
return 1; return fact_val;
} }
(a) (i) only
(b) (ii) only
(c) both
(d) None of these


(2) Which of the following is a semantic error?
(a) Division by zero
(b) Missing of a semicolon at the end of a statement
(c) Assigning a single precision real value to a long integer
(d) All the above

(3) If the following code segment is to count the no of zero’s in the given integer ‘x’ in its binary representation, what is to replaced by CONDITION?
int i, count, x;
for(i=0, count=0;i<16;i++)
if(CONDITION)
count++;
(x & (1 << i))
! (x & (1 << i))
! (x && (1 << i))
None of these

(4) A graph was given and asked to find out the minimum no of colors required to color the graph?
1
2
3
The graph can’t be colored

(5) Inorder traversal of a binary tree is d c e b a and preorder traversal of a binary tree is b c d e a. Find out the post order traversal?
(6) Which is not an advantage of using subroutines?
(a) Easier maintenance
(b) Runtime reduces
(c) Storage space reduces
(d) Modularity

(7) A k-diagonal matrix is a n *n square matrix in which the elements on the principal diagonal and k diagonals above the principal diagonal and k diagonals below the principal diagonal only have none zero elements. Other elements are zero’s. In order to save the space, the non zero elements are stored in a one dimensional array. The no of locations in this array are:
(a) n*(n-k-1)/2
(b) n*(n-1) – (n-k)(n-k-2)
(c) n*(n-1) – (n-k)(n-k-2)
(d) n*n – (n-k-1)

(8)A 1D - array A whose size is N is given and is divided into P partitions and an element x is to be searched in the array. Each partition is given to one processor. The elements are searched with in the partition using a binary search. What is the time complexity of the algorithm?
(a) O (N/P)
(b) O (P)
(c) O ( log (N/P))
(d) O ( log (N/P)) + O (P log P)

(9) Write the code segment to insert an element p into the linked list after an element q? Make necessary pointer adjustments?
Ans:
p.next = q.next;
q.next = p;
(10) Given a matrix A. What is the minimum no of multiplications do u need to compute A10 ?
(a) 9
(b) 5
(c) 4
(d) 6


Texas Placement Paper : Technical - Electronics

Sample Questions


(steeper transition) by:

Increasing W/L of PMOS transistor

Increasing W/L of NMOS transistor

Increasing W/L of both transistors by the same factor

Decreasing W/L of both transistor by the same factor

Minimum number of 2-input NAND gates that will be required to implement the function: Y = AB + CD + EF is

4

5

6

7

Consider a two-level memory hierarchy system M1 & M2. M1 is accessed first and on miss M2 is accessed. The access of M1 is 2 nanoseconds and the miss penalty (the time to get the data from M2 in case of a miss) is 100 nanoseconds. The probability that a valid data is found in M1 is 0.97. The average memory access time is:


4.94 nanoseconds

3.06 nanoseconds

5.00 nanoseconds

5.06 nanoseconds

Interrupt latency is the time elapsed between:

Occurrence of an interrupt and its detection by the CPU

Assertion of an interrupt and the start of the associated ISR

Assertion of an interrupt and the completion of the associated ISR

Start and completion of associated ISR

Which of the following is true for the function (A.B + A’.C + B.C)

This function can glitch and can be further reduced

This function can neither glitch nor can be further reduced

This function can glitch and cannot be further reduced

This function cannot glitch but can be further reduced

For the two flip-flop configuration below, what is the relationship of the output at B to the clock frequency?

Output frequency is 1/4th the clock frequency, with 50% duty cycle

Output frequency is 1/3rd the clock frequency, with 50% duty cycle

Output frequency is 1/4th the clock frequency, with 25% duty cycle

Output frequency is equal to the clock frequency






The voltage on Node B is:

0
10
–10


8.A CPU supports 250 instructions. Each instruction op-code has these fields:

The instruction type (one among 250)
A conditional register specification
3 register operands
Addressing mode specification for both source operands
9. The CPU has 16 registers and supports 5 addressing modes. What is the instruction op-code length in bits?

32

24
30
36
10. In the iterative network shown, the output Yn of any stage N is 1 if the total number of 1s at the inputs starting from the first stage to the Nth stage is odd. (Each identical box in the iterative network has two inputs and two outputs). The optimal logic structure for the box consists of:

One AND gate and one NOR gate

One NOR gate and one NAND gate
Two XNOR gates
One XOR gate


11.Consider a circuit with N logic nets. If each net can be stuck-at either values 0 and 1, in how many ways can the circuit be faulty such that only one net in it can be faulty, and such that up-to all nets in it can be faulty?

2 and 2N

N and 2^N
2N and 3^N-1
2N and 3N
12. In the circuit shown, all the flip-flops are identical. If the set-up time is 2 ns, clock->Q delay is 3 ns and hold time is 1 ns, what is the maximum frequency of operation for the circuit?


a. 200 MHz
b. 333 MHz
c. 250 MHz
d. None of the above

13. Which of the following statements is/are true?
I. Combinational circuits may have feedback, sequential circuits do not.

II. Combinational circuits have a ‘memory-less’ property, sequential circuits do not.

III. Both combinational and sequential circuits must be controlled by an external clock.

I only
II and III only
I and II only
II only
14.Consider an alternate binary number representation scheme, wherein the number of ones M, in a word of N bits, is always the same. This scheme is called the M-out-of-N coding scheme. If M=N/2, and N=8, what is the efficiency of this coding scheme as against the regular binary number representation scheme? (As a hint, consider that the number of unique words represent able in the latter representation with N bits is 2^N. Hence the efficiency is 100%)

Close to 30%

Close to 50%
Close to 70%
Close to 100%
15.A CPU supports 4 interrupts- I1, I2, I3 and I4. It supports priority of interrupts. Nested interrupts are allowed if later interrupt is higher priority than previous one. During a certain period of time, we observe the following sequence of entry into and exit from the interrupt service routine:

I1-start---I2-start---I2-end---I4-start---I3-start---I3-end---I4-end---I1-end

From this sequence, what can we infer about the interrupt routines?

I3 > I4 > I2 > I1
I4 > I3 > I2 > I1
I2 > I1; I3 > I4 > I1
I2 > I1, I3 > I4 > I2 > I1
16. I decide to build myself a small electric kettle to boil my cup of tea. I need 200 ml of water for my cup of tea. Assuming that typical tap water temperature is 25 C and I want the water boiling in exactly one minute, then what is the wattage required for the heating element?

[Assume: Boiling point of water is 100 C, 1 Calorie (heat required to change 1 gm of water by 1 C)= 4 joules, 1 ml of water weighs 1 gm.]
Data given is insufficient
800 W
300 W
1000 W
250 W
17.The athletics team from REC Trichy is traveling by train. The train slows down, (but does not halt) at a small wayside station that has a 100 mts long platform. The sprinter (who can run 100 mts in 10 sec) decides to jump down and get a newspaper and some idlis. He jumps out just as his compartment enters the platform and spends 5 secs buying his newspaper that is at the point where he jumped out. He then sprints along the platform to buy idlis that is another 50 mts. He spends another 5 secs buying the idlis. He is now just 50 mts from the other end of the platform where the train is moving out. He begins running in the direction of the train and the only other open door in his train is located 50 mts behind the door from where he jumped. At what(uniform) speed should the train be traveled if he just misses jumping into the open door at the very edge of the platform?

Make the following assumptions

He always runs at his peak speed uniformly

The train travels at uniform speed
He does not wait (other than for the idlis & newspaper) or run baclwards
Data given is insufficient
4 m/s

5 m/s
7.5 m/s
10 m/s
18.State which of the following gate combinations does not form a universal logic set:

2-input AND + 2-input OR
2-to-1 multiplexer
2-input XOR + inverter
3-input NAND
19.For the circuit shown below, what should the function F be, so that it produces an output of the same frequency (function F1), and an output of double the frequency (function F2).




F1= NOR gate and F2= OR gate
F1=NAND gate and F2= AND gate

F1=AND gate and F2=XOR gate
None of the above
20.The FSM (finite state machine) below starts in state Sa, which is the reset state, and detects a particular sequence of inputs leading it to state Sc. FSMs have a few characteristics. An autonomous FSM has no inputs. For a Moore FSM, the output depends on the present state alone. For a Mealy FSM, the output depends on the present state as well as the inputs. Which of the statements best describes the FSM below?

It has two states and is autonomous
The information available is insufficient

It is a Mealy machine with three states
It is a Moor machine with three states


21In the circuit given below, the switch is opened at time t=0. Voltage across the capacitor at t=infinity is:

2V
3V
5V
7V






22.What is the functionality represented by the following circuit?

y= ! (b+ac)
y= ! (a+bc)
y= ! (a(b+c))
y= ! (a+b+c)







23.The value (0xdeadbeef) needs to stored at address 0x400. Which of the below ways will the memory look like in a big endian machine: 0x403 0x402 0x401 0x400

a. be ef de ad

b. ef be ad de

c. fe eb da ed

d. ed da eb fe

24. In a given CPU-memory sub-system, all accesses to the memory take two cycles. Accesses to memories in two consecutive cycles can therefore result in incorrect data transfer. Which of the following access mechanisms guarantees correct data transfer?

A read operation followed by a write operation in the next cycle.
A write operation followed by a read operation in the next cycle.

A NOP between every successive reads & writes
None of the above
25.An architecture saves 4 control registers automatically on function entry (and restores them on function return). Save of each registers costs 1 cycle (so does restore). How many cycles are spent in these tasks (save and restore) while running the following un-optimized code with n=5:

Void fib(int n)

{

if((n==0) || (n==1)) return 1;
return(fib(n-1) + fib(n-2));

}

120
80
125
128
26.The maximum number of unique Boolean functions F(A,B), realizable for a two input (A,B) and single output (Z) circuit is:

2
6
8

Texas Placement Paper : General - other Chennai - 8 September 2005

Texas Instruments Test at Chennai : 8 Sept 2005

Test Paper 01

1. Can we declare a static function as virtual?
Ans: No. The virtual function mechanism is used on the specific object that determines which virtual function to call. Since the static functions are not any way related to objects, they cannot be declared as virtual.

2. Can user-defined object be declared as static data member of another class?
Ans: Yes. The following code shows how to initialize a user-defined object.
#include
class test
{
int i ;
public :
test ( int ii = 0 )
{
i = ii ;
}
} ;
class sample
{
static test s ;
} ;
test sample::s ( 26 ) ;
Here we have initialized the object s by calling the one-argument constructor. We can use the same convention to initialize the object by calling multiple-argument constructor.

3. What is forward referencing and when should it be used?
Ans: Consider the following program:
class test
{
public :
friend void fun ( sample, test ) ;
} ;
class sample
{
public :
friend void fun ( sample, test ) ;
} ;
void fun ( sample s, test t )
{
// code
}
void main( )
{
sample s ;
test t ;
fun ( s, t ) ;
}
This program would not compile. It gives an error that sample is undeclared identifier in the statement friend void fun ( sample, test ) ; of the class test. This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample. Forward referencing is generally required when we make a class or a function as a friend.

4. The istream_withassign class has been derived from the istream class and overloaded assignment operator has been added to it. The _withassign classes are much like their base classes except that they include overloaded assignment operators. Using these operators the objects of the _withassign classes can be copied. The istream, ostream, and iostream classes are made uncopyable by making their overloaded copy constructor and assignment operators private.

5. How do I write my own zero-argument manipulator that should work same as hex?
Ans: This is shown in following program.
#include
ostream& myhex ( ostream &o )
{
o.setf ( ios::hex) ;
return o ;
}
void main( )
{
cout << endl << myhex << 2000 ;
}

6.We all know that a const variable needs to be initialized at the time of declaration. Then how come the program given below runs properly even when we have not initialized p?
#include
void main( )
{
const char *p ;
p = "A const pointer" ;
cout << p ;
}
Ans: The output of the above program is ,A const pointer,. This is because in this program p is declared as ,const char*, which means that value stored at p will be constant and not p and so the program works properly

7. How do I refer to a name of class or function that is defined within a namespace?
Ans: There are two ways in which we can refer to a name of class or function that is defined within a namespace: Using scope resolution operator through the using keyword. This is shown in following example:

namespace name1
{
class sample1
{
// code
} ;
}
namespace name2
{
class sample2
{
// code
} ;
}
using namespace name2 ;
void main( )
{
name1::sample1 s1 ;
sample2 s2 ;
}
Here, class sample1 is referred using the scope resolution operator. On the other hand we can directly refer to class sample2 because of the statement using namespace name2 ; the using keyword declares all the names in the namespace to be in the current scope. So we can use the names without any qualifiers.

8. While overloading a binary operator can we provide default values?
Ans: No!. This is because even if we provide the default arguments to the parameters of the overloaded operator function we would end up using the binary operator incorrectly. This is explained in the following example:

sample operator + ( sample a, sample b = sample (2, 3.5f ) )
{
}
void main( )
{
sample s1, s2, s3 ;
s3 = s1 + ; // error
}

9. How do I carry out conversion of one object of user-defined type to another?
Ans: To perform conversion from one user-defined type to another we need to provide conversion function. Following program demonstrates how to provide such conversion function.
class circle
{
private :
int radius ;
public:
circle ( int r = 0 )
{
radius = r ;
}
} ;
class rectangle
{
private :
int length, breadth ;
public :
rectangle( int l, int b )
{
length = l ;
breadth = b ;
}
operator circle( )
{
return circle ( length ) ;
}
} ;
void main( )
{
rectangle r ( 20, 10 ) ;
circle c;
c = r ;
}
Here, when the statement c = r ; is executed the compiler searches for an overloaded assignment operator in the class circle which accepts the object of type rectangle. Since there is no such overloaded assignment operator, the conversion operator function that converts the rectangle object to the circle object is searched in the rectangle class. We have provided such a conversion function in the rectangle class. This conversion operator function returns a circle object. By default conversion operators have the name and return type same as the object type to which it converts to. Here the type of the object is circle and hence the name of the operator function as well as the return type is circle.

10. How do I write code that allows to create only one instance of a class?
Ans: This is shown in following code snippet.

#include
class sample
{
static sample *ptr ;
private:
sample( )
{
}
public:
static sample* create( )
{
if ( ptr == NULL )
ptr = new sample ;
return ptr ;
}
} ;
sample *sample::ptr = NULL ;
void main( )
{
sample *a = sample::create( ) ;
sample *b = sample::create( ) ;
}
Here, the class sample contains a static data member ptr, which is a pointer
to the object of same class. The constructor is private which avoids us from creating objects outside the class. A static member function called create( ) is used to create an object of the class. In this function the condition is checked whether or not ptr is NULL, if it is then an object is created dynamically and its address collected in ptr is returned. If ptr is not NULL, then the same address is returned. Thus, in main( ) on execution of the first statement one object of sample gets created whereas on execution of second statement, b holds the address of the first object. Thus, whatever number of times you call create( ) function, only one object of sample class will be available.

11. How do I write code to add functions, which would work as get and put properties of a class?
Ans: This is shown in following code.
#include
class sample
{
int data ;
public:
__declspec ( property ( put = fun1, get = fun2 ) )
int x ;
void fun1 ( int i )
{
if ( i < 0 )
data = 0 ;
else
data = i ;
}
int fun2( )
{
return data ;
}
} ;
void main( )
{
sample a ;
a.x = -99 ;
cout << a.x ;
}
Here, the function fun1( ) of class sample is used to set the given integer value into data, whereas fun2( ) returns the current value of data. To set these functions as properties of a class we have given the statement as shown below:
__declspec ( property ( put = fun1, get = fun2 )) int x ;

As a result, the statement a.x = -99 ; would cause fun1( ) to get called to set the value in data. On the other hand, the last statement would cause fun2( ) to get called to return the value of data.

12. How do I write code to make an object work like a 2-D array?
Ans: Take a look at the following program.
#include
class emp
{
public :
int a[3][3] ;
emp( )
{
int c = 1 ;
for ( int i = 0 ; i <= 2 ; i++ )
{
for ( int j = 0 ; j <= 2 ; j++ )
{
a[i][j] = c ;
c++ ;
}
}
}
int* operator[] ( int i )
{
return a[i] ;
}
} ;
void main( )
{
emp e ;
cout << e[0][1] ;
}
The class emp has an overloaded operator [ ] function. It takes one argument an integer representing an array index and returns an int pointer. The statement cout << e[0][1] ; would get converted into a call to the overloaded [ ] function as e.operator[ ] ( 0 ). 0 would get collected in i. The function would return a[i] that represents the base address of the zeroeth row. Next the statement would get expanded as base address of zeroeth row[1] that can be further expanded as *( base address + 1 ). This gives us a value in zeroth row and first column.

13. What are formatting flags in ios class?
Ans: The ios class contains formatting flags that help users to format the stream data. Formatting flags are a set of enum definitions. There are two types of formatting flags:
On/Off flags
Flags that work in-group
The On/Off flags are turned on using the setf( ) function and are turned off using the unsetf( ) function. To set the On/Off flags, the one argument setf( ) function is used. The flags working in groups are set through the two-argument setf( ) function. For example, to left justify a string we can set the flag as,
cout.setf ( ios::left ) ;
cout << "KICIT Nagpur" ;
To remove the left justification for subsequent output we can say,
cout.unsetf ( ios::left ) ;
The flags that can be set/unset include skipws, showbase, showpoint,
uppercase, showpos, unitbuf and stdio. The flags that work in a group can have only one of these flags set at a time.

14. What is the purpose of ios::basefield in the following statement?
cout.setf ( ios::hex, ios::basefield ) ;
Ans: This is an example of formatting flags that work in a group. There is a flag for each numbering system (base) like decimal, octal and hexadecimal. Collectively, these flags are referred to as basefield and are specified by ios::basefield flag. We can have only one of these flags on at a time. If we set the hex flag as setf ( ios::hex ) then we will set the hex bit but we won,t clear the dec bit resulting in undefined behavior. The solution is to call setf( ) as setf ( ios::hex, ios::basefield ). This call first clears all the bits and then sets the hex bit.

15. Can we get the value of ios format flags?
Ans: Yes! The ios::flags( ) member function gives the value format flags. This function takes no arguments and returns a long ( typedefed to fmtflags) that contains the current format flags.

16. Is there any function that can skip certain number of characters present in the input stream?
Ans: Yes! This can be done using cin::ignore( ) function. The prototype of this function is as shown below:
istream& ignore ( int n = 1, int d =EOF ) ;
Sometimes it happens that some extra characters are left in the input stream while taking the input such as, the ?\n? (Enter) character. This extra character is then passed to the next input and may pose problem.

To get rid of such extra characters the cin::ignore( ) function is used. This is equivalent to fflush ( stdin ) used in C language. This function ignores the first n characters (if present) in the input stream, stops if delimiter d is encountered.

17. Write a program that implements a date class containing day, month and year as data members. Implement assignment operator and copy constructor in this class.
Ans: This is shown in following program:
#include
class date
{
private :
int day ;
int month ;
int year ;
public :
date ( int d = 0, int m = 0, int y = 0 )
{
day = d ;
month = m ;
year = y ;
}
// copy constructor
date ( date &d )
{
day = d.day ;
month = d.month ;
year = d.year ;
}
// an overloaded assignment operator
date operator = ( date d )
{
day = d.day ;
month = d.month ;
year = d.year ;
return d ;
}
void display( )
{
cout << day << "/" << month << "/" << year ;
}
} ;
void main( )
{
date d1 ( 25, 9, 1979 ) ;
date d2 = d1 ;
date d3 ;
d3 = d2 ;
d3.display( ) ;
}

18. When should I use unitbuf flag?
Ans: The unit buffering (unitbuf) flag should be turned on when we want to ensure that each character is output as soon as it is inserted into an output stream. The same can be done using unbuffered output but unit buffering provides a better performance than the unbuffered output.

19.What are manipulators?
Ans: Manipulators are the instructions to the output stream to modify the output in various ways. The manipulators provide a clean and easy way for formatted output in comparison to the formatting flags of the ios class. When manipulators are used, the formatting instructions are inserted directly into the stream. Manipulators are of two types, those that take an argument and those that don?t.

20. What is the difference between the manipulator and setf( ) function?
Ans: The difference between the manipulator and setf( ) function are as follows:

The setf( ) function is used to set the flags of the ios but manipulators directly insert the formatting instructions into the stream. We can create user-defined manipulators but setf( ) function uses data members of ios class only. The flags put on through the setf( ) function can be put off through unsetf( ) function. Such flexibility is not available with manipulators.

21. How do I get the current position of the file pointer?
Ans: We can get the current position of the file pointer by using the tellp( ) member function of ostream class or tellg( ) member function of istream class. These functions return (in bytes) positions of put pointer and get pointer respectively.

22. What are put and get pointers?
Ans: These are the long integers associated with the streams. The value present in the put pointer specifies the byte number in the file from where next write would take place in the file. The get pointer specifies the byte number in the file from where the next reading should take place.

23. What do the nocreate and noreplace flag ensure when they are used for opening a file?
Ans: nocreate and noreplace are file-opening modes. A bit in the ios class defines these modes. The flag nocreate ensures that the file must exist before opening it. On the other hand the flag noreplace ensures that while opening a file for output it does not get overwritten with new one unless ate or app is set. When the app flag is set then whatever we write gets appended to the existing file. When ate flag is set we can start reading or writing at the end of existing file.

24. What is the limitation of cin while taking input for character array?
Ans: To understand this consider following statements,
char str[5] ;
cin >> str ;
While entering the value for str if we enter more than 5 characters then there is no provision in cin to check the array bounds. If the array overflows, it may be dangerous. This can be avoided by using get( ) function. For example, consider following statement,
cin.get ( str, 5 ) ;
On executing this statement if we enter more than 5 characters, then get( ) takes only first five characters and ignores rest of the characters. Some more variations of get( ) are available, such as shown below:
get ( ch ) ? Extracts one character only
get ( str, n ) ? Extracts up to n characters into str
get ( str, DELIM ) ? Extracts characters into array str until specified delimiter (such as ,\n,). Leaves delimiting character in stream.
get ( str, n, DELIM ) ? Extracts characters into array str until n characters or DELIM character, leaving delimiting character in stream.

25. What is the purpose of istream class?
Ans: The istream class performs activities specific to input. It is derived from the ios class. The most commonly used member function of this class is the overloaded >> operator which can extract values of all basic types. We can extract even a string using this operator.

26. Would the following code work?
#include
void main( )
{
ostream o ;
o << "Dream. Then make it happen!" ;
}
Ans: No! This is because we cannot create an object of the ostream class since its constructor and copy constructor are declared private.

27. Can we use this pointer inside static member function?
Ans: No! The this pointer cannot be used inside a static member function. This is because a static member function is never called through an object.

28. What is strstream?
Ans: strstream is a type of input/output stream that works with the memory. It allows using section of the memory as a stream object. These streams provide the classes that can be used for storing the stream of bytes into memory. For example, we can store integers, floats and strings as a stream of bytes. There are several classes that implement this in-memory formatting. The class ostrstream derived from ostream is used when output is to be sent to memory, the class istrstream derived from istream is used when input is taken from memory and strstream class derived from iostream is used for memory objects that do both input and output. Ans: When we want to retrieve the streams of bytes from memory we can use istrestream. The following example shows the use of istrstream class.
#include
void main( )
{
int age ;
float salary ;
char name[50] ;
char str[] = "22 12004.50 K. Vishwanatth" ;
istrstream s ( str ) ;
s >> age >> salary >> name ;
cout << age << endl << salary << endl << name ;
cout << endl << s.rdbuf( ) ;
}
Here, s is the object of the class istrstream. When we are creating the object s, the constructor of istrstream gets called that receives a pointer to the zero terminated character array str. The statement s >> age >> salary >> name ; extracts the age, salary and the name from the istrstream object s. However, while extracting the name, only the first word of name gets extracted. The balance is extracted using rdbuf( ).

29. When the constructor of a base class calls a virtual function, why doesn,t the override function of the derived class gets called?
Ans: While building an object of a derived class first the constructor of the base class and then the constructor of the derived class gets called. The object is said an immature object at the stage when the constructor of base class is called. This object will be called a matured object after the execution of the constructor of the derived class. Thus, if we call a virtual function when an object is still immature, obviously, the virtual function of the base class would get called. This is illustrated in the following example.
#include
class base
{
protected :
int i ;
public :
base ( int ii = 0 )
{
i = ii ;
show( ) ;
}
virtual void show( )
{
cout << "base,s show( )" << endl ;
}
} ;
class derived : public base
{
private :
int j ;
public :
derived ( int ii, int jj = 0 ) : base ( ii )
{
j = jj ;
show( ) ;
}
void show( )
{
cout << "derived,s show( )" << endl ;
}
} ;

void main( )
{
derived dobj ( 20, 5 ) ;
}
The output of this program would be:
base,s show( )
derived,s show( )

30. Can I have a reference as a data member of a class? If yes, then how do I initialise it?
Ans: Yes, we can have a reference as a data member of a class. A reference as a data member of a class is initialized in the initialization list of the constructor. This is shown in following program.
#include
class sample
{
private :
int& i ;
public :
sample ( int& ii ) : i ( ii )
{
}
void show( )
{
cout << i << endl ;
}
} ;
void main( )
{
int j = 10 ;
sample s ( j ) ;
s.show( ) ;
}
Here, i refers to a variable j allocated on the stack. A point to note here is that we cannot bind a reference to an object passed to the constructor as a value. If we do so, then the reference i would refer to the function parameter (i.e. parameter ii in the constructor), which would disappear as soon as the function returns, thereby creating a situation of dangling reference.

31. Why does the following code fail?
#include
class sample
{
private :
char *str ;
public :
sample ( char *s )
{
strcpy ( str, s ) ;
}
~sample( )
{
delete str ;
}
} ;
void main( )
{
sample s1 ( "abc" ) ;
}
Ans: Here, through the destructor we are trying to deal locate memory, which has been allocated statically. To remove an exception, add following statement to the constructor.
sample ( char *s )
{
str = new char[strlen(s) + 1] ;
strcpy ( str, s ) ;
}
Here, first we have allocated memory of required size, which then would get deal located through the destructor.

32. assert( ) macro...
We can use a macro called assert( ) to test for conditions that should not occur in a code. This macro expands to an if statement. If test evaluates to 0, assert prints an error message and calls abort to abort the program.
#include
#include
void main( )
{
int i ;
cout << "\nEnter an integer: " ;
cin >> i ;
assert ( i >= 0 ) ;
cout << i << endl ;
}

33. Why it is unsafe to deal locate the memory using free( ) if it has been
allocated using new?
Ans: This can be explained with the following example:
#include
class sample
{
int *p ;
public :
sample( )
{
p = new int ;
}
~sample( )
{
delete p ;
}
} ;
void main( )
{
sample *s1 = new sample ;
free ( s1 ) ;
sample *s2 = ( sample * ) malloc ( sizeof ( sample ) ) ;
delete s2 ;
}
The new operator allocates memory and calls the constructor. In the constructor we have allocated memory on heap, which is pointed to by p. If we release the object using the free( ) function the object would die but the memory allocated in the constructor would leak. This is because free( ) being a C library function does not call the destructor where we have deal located the memory.

As against this, if we allocate memory by calling malloc( ) the constructor would not get called. Hence p holds a garbage address. Now if the memory is deal located using delete, the destructor would get called where we have tried to release the memory pointed to by p. Since p contains garbage this may result in a runtime error.

34. Can we distribute function templates and class templates in object libraries?
Ans: No! We can compile a function template or a class template into object code (.obj file). The code that contains a call to the function template or the code that creates an object from a class template can get compiled. This is because the compiler merely checks whether the call matches the declaration (in case of function template) and whether the object definition matches class declaration (in case of class template). Since the function template and the class template definitions are not found, the compiler leaves it to the linker to restore this. However, during linking, linker doesn,t find the matching definitions for the function call or a matching definition for object creation. In short the expanded versions of templates are not found in the object library. Hence the linker reports error.

35. What is the difference between an inspector and a mutator ?
Ans: An inspector is a member function that returns information about an object,s state (information stored in object,s data members) without changing the object,s state. A mutator is a member function that changes the state of an object. In the class Stack given below we have defined a mutator and an inspector.
class Stack
{
public :
int pop( ) ;
int getcount( ) ;
}
In the above example, the function pop( ) removes top element of stack thereby changing the state of an object. So, the function pop( ) is a mutator. The function getcount( ) is an inspector because it simply counts the number of elements in the stack without changing the stack.

36. Namespaces:
The C++ language provides a single global namespace. This can cause problems with global name clashes. For instance, consider these two C++ header files:
// file1.h
float f ( float, int ) ;
class sample { ... } ;
// file2.h
class sample { ... } ;
With these definitions, it is impossible to use both header files in a single program; the sample classes will clash. A namespace is a declarative region that attaches an additional identifier to any names declared inside it. The additional identifier thus avoids the possibility that a name will conflict with names declared elsewhere in the program. It is possible to use the same name in separate namespaces without conflict even if the names appear in the same translation unit. As long as they appear in separate namespaces, each name will be unique because of the addition of the namespace identifier. For example:
// file1.h
namespace file1
{
float f ( float, int ) ;
class sample { ... } ;
}
// file2.h
namespace file2
{
class sample { ... } ;
}

Now the class names will not clash because they become file1::sample and file2::sample, respectively.

37. What would be the output of the following program?
#include
class user
{
int i ;
float f ;
char c ;
public :
void displaydata( )
{
cout << endl << i << endl << f << endl << c ;
}
} ;
void main( )
{
cout << sizeof ( user ) ;
user u1 ;
cout << endl << sizeof ( u1 ) ;
u1.displaydata( ) ;
}
Ans: The output of this program would be,
9 or 7
9 or 7
Garbage
Garbage
Garbage
Since the user class contains three elements, int, float and char its size would be 9 bytes (int-4, float-4, char-1) under Windows and 7 bytes (int-2, float-4, char-1) under DOS. Second output is again the same because u1 is an object of the class user. Finally three garbage values are printed out because i, f and c are not initialized anywhere in the program.

Note that if you run this program you may not get the answer shown here. This is because packing is done for an object in memory to increase the access efficiency. For example, under DOS, the object would be aligned on a 2-byte boundary. As a result, the size of the object would be reported as 6 bytes. Unlike this, Windows being a 32-bit OS the object would be aligned on a 4-byte boundary. Hence the size of the object would be reported as 12 bytes. To force the alignment on a 1-byte boundary, write the following statement before the class declaration.
#pragma pack ( 1 )

38. Write a program that will convert an integer pointer to an integer and vice-versa.
Ans: The following program demonstrates this.
#include
void main( )
{
int i = 65000 ;
int *iptr = reinterpret_cast ( i ) ;
cout << endl << iptr ;
iptr++ ;
cout << endl << iptr ;
i = reinterpret_cast ( iptr ) ;
cout << endl << i ;
i++ ;
cout << endl << i ;
}

39. What is a const_cast?
Ans. The const_cast is used to convert a const to a non-const. This is shown in the following
program:
#include
void main( )
{
const int a = 0 ;
int *ptr = ( int * ) &a ; //one way
ptr = const_cast_ ( &a ) ; //better way
}
Here, the address of the const variable a is assigned to the pointer to a non-const variable. The const_cast is also used when we want to change the data members of a class inside the const member functions. The following code snippet shows this:
class sample
{
private:
int data;
public:
void func( ) const
{
(const_cast (this))->data = 70 ;
}
} ;

40. What is forward referencing and when should it be used?
Ans: Forward referencing is generally required when we make a class or a function as a friend.
Consider following program:
class test
{
public:
friend void fun ( sample, test ) ;
} ;
class sample
{
public:
friend void fun ( sample, test ) ;
} ;
void fun ( sample s, test t )
{
// code
}
void main( )
{
sample s ;
test t ;
fun ( s, t ) ;
}
On compiling this program it gives error on the following statement of test class. It gives an error that sample is undeclared identifier. friend void fun ( sample, test );
This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample.
class sample ;

41. How would you give an alternate name to a namespace?
Ans: An alternate name given to namespace is called a namespace-alias. namespace-alias is generally used to save the typing effort when the names of namespaces are very long or complex. The following syntax is used to give an alias to a namespace.
namespace myname = my_old_very_long_name ;

42. Using a smart pointer can we iterate through a container?
Ans: Yes. A container is a collection of elements or objects. It helps to properly organize and store the data. Stacks, linked lists, arrays are examples of containers. Following program shows how to iterate through a container using a smart pointer.
#include
class smartpointer
{
private :
int *p ; // ordinary pointer
public :
smartpointer ( int n )
{
p = new int [ n ] ;
int *t = p ;
for ( int i = 0 ; i <= 9 ; i++ )
*t++ = i * i ;
}
int* operator ++ ( int )
{
return p++ ;
}
int operator * ( )
{
return *p ;
}
} ;
void main( )
{
smartpointer sp ( 10 ) ;
for ( int i = 0 ; i <= 9 ; i++ )
cout << *sp++ << endl ;
}
Here, sp is a smart pointer. When we say *sp, the operator * ( ) function gets called. It returns the integer being pointed to by p. When we say sp++ the operator ++ ( ) function gets called. It increments p to point to The next element in the array and then returns the address of this new location.

43. Can objects read and write themselves?
Ans: Yes! This can be explained with the help of following example:
#include
#include
class employee
{
private :
char name [ 20 ] ;
int age ;
float salary ;
public :
void getdata( )
{
cout << "Enter name, age and salary of employee : " ;
cin >> name >> age >> salary ;
}
void store( )
{
ofstream file ;
file.open ( "EMPLOYEE.DAT", ios::app | ios::binary ) ;
file.write ( ( char * ) this, sizeof ( *this ) ) ;
file.close( ) ;
}
void retrieve ( int n )
{
ifstream file ;
file.open ( "EMPLOYEE.DAT", ios::binary ) ;
file.seekg ( n * sizeof ( employee ) ) ;
file.read ( ( char * ) this, sizeof ( *this ) ) ;
file.close( ) ;
}
void show( )
{
cout << "Name : " << name
<< endl << "Age : " << age
<< endl << "Salary :" << salary << endl ;
}
} ;
void main( )
{
employee e [ 5 ] ;
for ( int i = 0 ; i <= 4 ; i++ )
{
e [ i ].getdata( ) ;
e [ i ].store( ) ;
}
for ( i = 0 ; i <= 4 ; i++ )
{
e [ i ].retrieve ( i ) ;
e [ i ].show( ) ;
}
}
Here, employee is the class whose objects can write and read themselves. The getdata( ) function has been used to get the data of employee and store it in the data members name, age and salary. The store( ) function is used to write an object to the file. In this function a file has been opened in append mode and each time data of current object has been stored after the last record (if any) in the file. Function retrieve( ) is used to get the data of a particular employee from the file. This retrieved data has been stored in the data members name, age and salary. Here this has been used to store data since it contains the address of the current object. The function show( ) has been used to display the data of employee.

44. Why is it necessary to use a reference in the argument to the copy constructor?
Ans : If we pass the copy constructor the argument by value, its copy would get constructed using the copy constructor. This means the copy constructor would call itself to make this copy. This process would go on and on until the compiler runs out of memory. This can be explained with the help of following example:
class sample
{
int i ;
public :
sample ( sample p )
{
i = p.i ;
}
} ;
void main( )
{
sample s ;
sample s1 ( s ) ;
}
While executing the statement sample s1 ( s ), the copy constructor would get called. As the copy construct here accepts a value, the value of s would be passed which would get collected in p. We can think of this statement as sample p = s. Here p is getting created and initialized. Means again the copy constructor would get called. This would result into recursive calls. Hence we must use a reference as an argument in a copy constructor.

45.



46. Virtual Multiple Inheritance:
A class b is defined having member variable i. Suppose two classes d1 and d2 are derived from class b and a class multiple is derived from both d1 and d2. If variable i is accessed from a member function of multiple then it gives error as ,member is ambiguous,. To avoid this error derive classes d1 and d2 with modifier virtual as shown in the following program.
#include
class b
{
public :
int i ;
public :
fun( )
{
i = 0 ;
}
} ;
class d1 : virtual public b
{
public :
fun( )
{
i = 1 ;
}
} ;
class d2 : virtual public b
{
public :
fun( )
{
i = 2 ;
}
} ;
class multiple : public d1, public d2
{
public :
fun( )
{
i = 10 ;
}
} ;
void main( )
{
multiple d ;
d.fun( ) ;
cout << d.i ;
}

46. Can we use this pointer in a class specific, operator-overloading function for new operator?
Ans: No! The this pointer is never passed to the overloaded operator new() member function because this function gets called before the object is created. Hence there is no question of the this pointer getting passed to operator new( ).

47. Can we allocate memory dynamically for a reference?
Ans: No! It is not possible to allocate memory dynamically for a reference. This is because, when we create a reference, it gets tied with some variable of its type. Now, if we try to allocate memory dynamically for a reference, it is not possible to mention that to which variable the reference would get tied.

48. When should I overload new operator on a global basis or a class basis?
Ans: We overload operator new in our program, when we want to initialize a data item or a class object at the same place where it has been allocated memory. The following example shows how to overload new operator on global basis.
#include
#include
void * operator new ( size_t s )
{
void *q = malloc ( s ) ;
return q ;
}
void main( )
{
int *p = new int ;
*p = 25 ;
cout << *p ;
}
When the operator new is overloaded on global basis it becomes impossible to initialize the data members of a class as different classes may have different types of data members. The following example shows how to overload new operator on class-by-class basis.
#include
#include
class sample
{
int i ;
public :
void* operator new ( size_t s, int ii )
{
sample *q = ( sample * ) malloc ( s ) ;
q -> i = ii ;
return q ;
}
} ;
class sample1
{
float f ;
public :
void* operator new ( size_t s, float ff )
{
sample1 *q = ( sample1 * ) malloc ( s ) ;
q -> f = ff ;
return q ;
}
} ;
void main( )
{
sample *s = new ( 7 ) sample ;
sample1 *s1 = new ( 5.6f ) sample1 ;
}
Overloading the operator new on class-by-class basis makes it possible to allocate memory for an object and initialize its data members at the same place.

49. How would you define a pointer to a data member of the type pointer to pointer?
Ans: The following program demonstrates this...
#include
class sample
{
public :
sample ( int **pp )
{
p = pp ;
}
int **p ;
} ;
int **sample::*ptr = &sample::p ;
void main( )
{
int i = 9 ;
int *pi = &i ;
sample s ( &pi ) ;
cout << ** ( s.*ptr ) ;
}
Here, ptr is the pointer to data member p of class sample, which in turn is a pointer pointing to an int.

50. How do I write a code to catch multiple types of exceptions in one single catch block?
Ans: The following program demonstrates the use of a single catch block to catch multiple exceptions.
#include
class test
{
} ;
class sample
{
public :
void fun1( )
{
throw 99 ;
}
void fun2( )
{
throw 3.14f ;
}
void fun3( )
{
throw "error" ;
}
void fun4( )
{
throw test( ) ;
}
} ;
void main( )
{
try
{
sample s ;
s.fun4( ) ;
s.fun1( ) ;
s.fun2( ) ;
s.fun3( ) ;
}
catch ( ... )
{
cout << "strange" ;
}
}
Here, different types of exceptions are thrown by the member functions of the class sample. While catching the exception instead of four different catch blocks we can as well define one single catch block. Note the syntax for defining the catch block, where we have used three dots (?) in the formal parameter list. This indicates that any thrown exception should get caught in the same catch block. When the exception is thrown from the fun4( ) control reaches the catch block, ignoring the rest of the calls.

51. Can we return an error value from the constructor of a class?
Ans: No. We cannot return any error value from the constructor, as the constructor doesn,t have any return type. However, by throwing an exception we can pass value to catch block. This is shown in the following example:
#include
class sample
{
public :
sample ( int i )
{
if ( i == 0 )
throw "error" ;
}
} ;
void main( )
{
try
{
sample s ( 0 ) ;
}
catch ( char * str )
{
cout << str ;
}
}
In this program, the statement throw "error" ; would throw an exception when an object s of the class sample would get created. The catch block would collect the string error.

52. How do I define the member function of a template class, which has to be defined outside the template class. The function receives an object of its own class as a parameter and returns the value of the same type.
Ans: The following example shows how we can define such a function.
sample sample::fun ( sample s )
{
// code
}
Here, the first sample indicates the return type of the function and the next sample is used for the scope of function.

53. How name mangling can be prevented?
Ans: To avoid name mangling the function should be declared with an extern "C" attribute. Functions declared as extern "C" are treated as C-style functions. Hence the compiler does not mangle them. The following code snippet shows how to declare such a function.
#include
extern "C" void display( )
{
cout << "See the effect of C in C++ " ;
}
void main( )
{
display( ) ;
}

54. Can we allocate memory dynamically for a reference?
Ans: No, it is not possible to allocate memory dynamically for a reference. A reference is initialized at the time of creation. Trying to allocate memory dynamically for a reference creates a problem in initializing it. Thus, the compiler does not allow us to dynamically allocate the memory for references.

55. What is RTTI?
Ans: RTTI stands for ,Run Time Type Information,. We use virtual function mechanism where we can call derived class,s member functions using base class,s pointer. However, many times we wish to know the exact type of the object. We can know the type of the object using RTTI. A function that returns the type of the object is known as RTTI functions. C++ supports two ways to obtain information about the object,s class at run time, they are typeid( ) operator and dynamic_cast operator.

56. What is Data Conversion?
Ans: Assignments between types whether they are basic or user-defined, are handled by the compiler. If the variables are of different basic types compiler calls a special routine to convert the value. But if we want to convert between user-defined data type and basic types we have to write conversion routine ourselves. A conversion routine to convert user-defined data type string to integer is shown below:
class string
{
private :
char str[20] ;
public :
string( )
{
}
string ( char *s )
{
strcpy ( str, s ) ;
}
operator int( )
{
return 123 ; // Write logic to convert string to integer
}
} ;
main( )
{
string s2 = "123" ;
int i1 = int ( s2 ) ;
cout << endl << i1 ;
}

57. How to obtain type information using typeid( ) operator?
Ans: typeid( ) operator takes an object, a reference or a pointer and returns its type. Following program shows how to use the typeid( ) operator.
#include
#include
class Base
{
public :
virtual void show( )
{
}
};

class Der1 : public Base
{
} ;
void main( )
{
Base *b1 ;
cout << endl << typeid ( b1 ).name( ) ;
Der1 d1 ;
b1 = &d1 ;
cout << endl << typeid ( *b1 ).name( ) ;
cout << endl << typeid ( 12 ).name( ) << endl << typeid ( 12.5 ).name( ) ;
}
The output of this program will be
Base*
Der1
int
double
RTTI operators must be used for polymorphic class (class having virtual function) only. For non-polymorphic class static type information is returned.

58. How to use RTTI with class templates?
Ans: Templates can generate different classes. We may wish to get the type of class, which we are working in. The following program shows how to use RTTI operator typeid( ) with class template.
#include
#include
template
class base
{
public :
base( )
{
cout << typeid ( *this ).name( ) << "Constructor" << endl ;
}
T add ( T a, T b )
{
return a + b ;
}
~base( )
{
cout << typeid ( *this ).name( ) << "Destructor" << endl ;
}
} ;
void main( )
{
base b1 ;
cout << b1.add ( 10, 20 ) << endl ;
base b2 ;
cout << b2.add ( 5.5, 10.5 ) << endl ;
}

59. We can use following C++ operators for typecasting. static_cast is used for castless conversions, narrowing conversions, conversion from void* and implicit type conversions. const_cast is used to convert a const to a non-const. reinterpret_cast is used to assign one kind of pointer to another.

60. What will be the output of the following program?
#include
class A
{
public :
A( )
{
cout << "Reached in Constructor\n" ;
}
} ;
void main( )
{
A a( ) ;
A b ;
}
Output : Reached in Constructor
Constructor gets called only once when the object b is created. When the
statement A a( ) ; gets executed constructor does not get called. This is because compiler takes this statement as a prototype declaration of function a( ) that returns an object of class A. However, if we pass arguments like
A a ( 10 ) ;
Compiler would search for one argument constructor and if not found would flash an error.

61. What is a container?
Ans: A container is an object that holds other objects. Various collection classes like List, Hash Table, AbstractArray, etc. are the examples of containers. We can use the classes to hold objects of any derived classes. The containers provide various methods using which we can get the number of objects stored in the container and iterate through the objects stored in it.

62. Function template overloading
One can declare several function templates with the same name and even declare a combination of function templates and ordinary functions with the same name. When an overloaded function is called, overload resolution is necessary to find the right function or template function to invoke.
For example:
template < class T > T sqrt ( T ) ;
template < class T > complex < T > sqrt ( complex < T > ) ;double sqrt ( double ) ;
void f ( complex < double > z )
{
sqrt ( 2 ) ; // sqrt < int > ( int )
sqrt ( 2.0 ) ; // sqrt ( double )
sqrt ( z ) ; // sqrt < complex < double > ( complex < double > )
}
In the same way that a template function is a generalization of the notion of a function, the rules for resolution in the presence of function templates are generalizations of the function overload resolution rules. Basically, for each template we find the specialization that is best for the set of function arguments. Then we apply the usual function overload resolution rules to these specializations and all ordinary functions.

63. Exception Handling in C++
In C++ we can handle run-time errors generated by c++ classes by using three new keywords: throw, catch, and try. We also have to create an exception class. If during the course of execution of a member function of this class a run-time error occurs, then this member function informs the application that an error has occurred. This process of informing is called ,throwing, an exception. The following code shows how to deal with exception handling.
class sample
{
public :
class errorclass
{
} ;
void fun( )
{
if ( some error occurs )
throw errorclass( ) // throws exception
}
} ;
//application
void main( )
{
try
{
sample s ;
s.fun( ) ;
}
catch ( sample::errorclass )
{
// do something about the error
}
}

64. Consider the following code:
#include
class base
{
public :
int data ;
} ;
class d1 : public base
{
} ;
class d2 : public base
{
} ;
class der : public d1, public d2
{
public :
void showdata( )
{
cout << data ;
}
} ;
void main( )
{
der d ;
d.showdata( ) ;
}
If you run this program it is bound to give you errors. This is because of the rules of inheritance:
1. Each base class not specified virtual will have its own sub-object representing it. In the above program, if we create object of d1 it will have a sub-object of class base containing a data member data. If we create an object of class der it will have sub-objects of classes d1 and d2 and both the sub-objects will refer to a separate copy of data. Hence, to access data from class der we will have to mention the class name. For example, d1::data or d2::data.
2. If we want that only one sub-object should exist we must use the concept of virtual base class. The single object of this will represent every base class of given name that is specified to be virtual class. After making d1 and d2 as virtual base class if we create an object of der only one sub-object would exist and so accessing data would no longer give us errors.

65. How to declare a pointer to a member function?
Ans: Suppose, I wish to declare a pointer to a member function that receives an int and returns an int. I will have to declare it as int (A::* ) ( int ). Following is an example.
#include
class A
{
public :
int fun ( int f )
{
cout << "in fun\n" ;
return f * f ;
}
} ;
typedef int ( A:: *pfun ) ( int ) ;
void main( )
{
pfun p = A::fun ;
A a ;
int s = ( a.*p ) ( 6 ) ;
cout << s ;
}

66. What is the disadvantage of a template function?
Ans: A template function cannot be distributed in the obj form. This is because, with which parameters the template function is going to be called is decided at the run time only. Therefore an obj form of a template function cannot be made by merely compiling it.

67. How to declare a pointer to the data members of a class?
Ans: Following program shows how to declare a pointer to non-function members of a class.
#include
class A
{
public :
int a ;
void print( )
{
cout << a ;
}
} ;
void main( )
{
int A::*pa = &A::a ;
A obj ;
obj.*pa = 20 ;
obj.print( ) ;
}
Here, we have initialised the data member a using the pointer pa.

68. How to allocate memory for a multidimensional array dynamically?
Ans: Many times we need to allocate memory for a multidimensional array dynamically. Because of complexity of pointers many find this difficult. Following program allocates memory for a 3 x 3 array dynamically, copies contents of a 3 x 3 array in it and prints the contents using the pointer.
#include
#include
int a[ ][3] = {
1, 2, 3,
4, 5, 6,
7, 8, 9
} ;
void main( )
{
int **p ;
p = new int *[3] ;
for ( int i = 0 ; i < 3 ; i++ )
p[i] = new int[3] ;
for ( i = 0 ; i < 3 ; i++ )
for ( int j = 0 ; j < 3 ; j++ )
p[i][j] = a[i][j] ;
for ( i = 0 ; i < 3 ; i++ )
{
for ( j = 0 ; j < 3 ; j++ )
cout << p[i][j] ;
cout << "\n" ;
}
}

69. When should we use the :: ( scope resolution ) operator to invoke the virtual functions?
Ans: Generally, :: operator is used to call a virtual function from constructor or destructor. This is because, if we call a virtual function from base class constructor or destructor the virtual function of the base class would get called even if the object being constructed or destroyed would be the object of the derived class. Thus, whenever we want to bypass the dynamic binding mechanism we must use the :: operator to call a virtual function.

70. How do I use operators .* and ->* in a program?
Ans: The following code snippet demonstrates the use of .* and ->* operators.
#include
class sample
{
public :
int i ;
void fun( )
{
cout << "fun" << endl ;
}
} ;
void ( sample::*pf )( ) = &sample::fun ;
int sample::*pdm = &sample::i ;
void main( )
{
sample s ;
sample *p = new sample ;
( s .* pf )( ) ;
( p ->* pf )( ) ;
s .* pdm = 1 ;
p ->* pdm = 2 ;
cout << s .* pdm << endl ;
cout << p ->* pdm << endl ;
}
In the above program pf is a pointer to a function fun( ) of class sample, and
pdm is a pointer to a data member i of the same class sample. The object s of the class sample is created statically. Next, p is a pointer to an object created dynamically. The using the operator .* and ->* the member functions are called and also the public data member is accessed.

71. What happens when we add an int value to a user defined type of object?
Ans: Whenever an int value is added to an object of user defined type, the object would search for an overloaded operator int( ). This operator must be defined in such a way that it always returns an int value. However, we need not specify the return type as on doing so the compiler flashes an error.
#include
class sample
{
int i ;
public :
sample ( )
{
i = 10 ;
}
operator int( )
{
return this -> i ;
}
} ;
void main( )
{
sample s ;
int i ;
i = s + 10 ;
cout << i ;
}
In the above program on adding 10 to an object s, the value of i would become 20.

72. Can we have a reference to an array?
Ans: Yes, we can have a reference to an array.
int a[ ] = { 8, 2, 12, 9 } ;
int ( &r ) [ 4 ] = a ; // reference to an array
Here, r is a reference to an array of four elements. We can even print the elements of array with the help of reference. This is shown in the following code segment:
for ( int i = 0 ; i < 4 ; i++ )
cout << r [i] << endl ;

73. When friend function becomes indispensable...
Ans: Consider the following program.
#include
class distance
{
private :
int feet ;
public :
distance( )
{
feet = 0 ;
}
distance ( int f )
{
feet = f ;
}
distance operator + ( distance x )
{
int f = feet + x.feet ;
return distance ( f ) ;
}
} ;
void main( )
{
distance d1 ( 20 ), d2, d3 ;
d2 = d1 + 10 ;
d3 = 10 + d2 ;
}
If you run this program it is bound to give errors. The error lies in the statement d3 = 10 + d2 ; We may think that since we have overloaded + operator this statement would add 10 to d2. But this does not happen. This is because the specified statement will get converted as d3 = 10.operator+ ( d2 ) ; This means that this statement should call the operator+( ) function that takes an object of distance class as parameter written in the float class, which is not possible. The solution is to write operator+( ) as a ,friend, function. Declare operator+ function in distance class as given below:
friend distance operator + ( distance x1, distance x2 ) ;
and define it outside the class as shown below:
distance operator + ( distance x1, distance x2 )
{
int f = x1.feet + x2.feet ;
return distance ( f ) ;
}
When compiler would see that the ,friend, operator+( ) function is available it would convert the statement d3 = 10 + d2 as operator+ (10, d2 ). Now since 10 is passed as a parameter not as a calling object there would be no error. Thus in such cases ,friend, function becomes indispensable.

74. How to use a memory as a stream?
Ans: Suppose, details of an employee such as name, designation, age, etc. are stored in different types of variables. Now, if we wish to concatenate these details in a character array we will have to use various string manipulation functions like strcpy( ) and strcat( ). Instead of using these functions we can use more easy and clean way to gather the details in the char array in the form of streams. We can declare the memory allocated for the array as stream and use the << operator to store variables having different types in this memory. Following program shows how to achieve this.
#include
void main( )
{
char buff [50] ;
char str[ ] = "Sanjay" ;
char desig[ ] = "Manager" ;
char jd[ ] = "27/12/1995" ;
int age = 35 ;
ostrstream o ( buff, sizeof ( buff ) ) ;
o << str << endl << desig << endl << jd << endl << age << ends ;
cout << buff ;
}
As shown in the program we can also use the manipulators and formatting flags. The output of this program will be:
Sanjay
Manager
27/12/1995
35

75. How would you declare and initialize reference to a data member?
Ans: Sometimes we may need to declare a data member, which is a reference to another data member of the class as shown below:
class A
{
public :
char *p ;
char *&rp ;
} ;
We can,t initialize a reference to a data member at the time of declaration. It
should be initialized using ,member wise initialization as shown below.
#include
class A
{
public :
char *p ;
char *&rp ;
A( ) : rp ( p )
{
p = "" ;
}
A ( char *s ) : rp ( p )
{
p = s ;
}
} ;
void main( )
{
A a ( "abcd" ) ;
cout << a.rp ;
}

76. iostream library has made it easy to read data from various input devices and write data to the output devices. The following program shows how to print a disk file ,data.dat, on the printer using stream classes. Every hardware device has a familiar name given by the operating system. The printer is generally connected to the first parallel port. So, the file name for the printer should be PRN or lpt1.
#include
void main( )
{
ifstream i ( "data.dat" ) ;
ofstream o ;
o.open ( "PRN" ) ;
char ch ;
while ( 1 )
{
i.get ( ch ) ;
if ( i.eof( ) )
break ;
o.put ( ch ) ;
}
o.put ( ,\x0C, ) ;
}

77. We know that a destructor is automatically called when an object of a class
goes out of scope. There is another case where destructor is called automatically. If an object is created in a try block and an exception is thrown after the object is created, then the destructor is called automatically.

78. Can a function call be at the left hand side of the assignment operator?
Ans: Yes. Following program shows how it is possible.
#include
class ref
{
private :
struct data
{
int a ; char *p ;
} d1, d2 ;
public :
data &set ( )
{
return d1 ;
}
data &get ( )
{
cin >> d2.a >> d2.p ;
return d2 ;
}
} ;
void main( )
{
ref r ;
r.set( ) = r.get( ) ;
r.print( ) ;
}
In the above program the functions get( ) and set( ) both return a reference to the object of the structure data. We have assigned the reference returned by get( ) to the reference returned by set( ) function. That is, we are assigning d2 to d1. So, the values of d2 would get assigned to d1. You can check this out by printing the values of d1.

79. If a class contains a virtual function a pointer called VPTR is created. This VPTR becomes a part of every object of that class. The first two bytes (in DOS) are occupied by VPTR. We can prove this by displaying the first two bytes of memory allocated for the objects. Following program shows how this can be achieved.
#include
class vir
{
public :
virtual void f( )
{
}
} ;
void main( )
{
vir v, v1 ;
int *p1 = ( int* ) &v ;
int *p2 = ( int* ) &v1 ;
cout << endl << *p1 << " " << *p2 ;
}

80. Exception Handling in C++
In C++ we can handle run-time errors generated by c++ classes by using three new keywords: throw, catch, and try. We also have to create an exception class. If during the course of execution of a member function of this class a run-time error occurs, then this member function informs the application that an error has occurred. This process of informing is called ,throwing, an exception. The following code shows how to deal with exception handling.
class sample
{
public :
class errorclass
{
} ;
void fun( )
{
if ( some error occurs )
throw errorclass( ) // throws exception
}
} ;
//application
void main( )
{
try
{
sample s ;
s.fun( ) ;
}
catch ( sample::errorclass )
{
// do something about the error
}
}

81. Accessing a private data member from a different Object...Different objects of the same class can access each other,s members, even if these members are private. For example:
#include < iostream.h >
class sample
{
float f ;
public :
sample ( float ff )
{
f = ff ;
}
void fun ( sample* objptr )
{
objptr -> n = 0 ;
cout << "Value of this objects f is : " << f << endl ;
cout << "Value of other objects f" << objptr -> n << endl ;
} // another object,s private member!
} ;
void main( )
{
sample s1 ( 6.5f ) , s2 ( 2.5f ) ;
s1.f ( &s2 ) ; // s1 changes s2,s n
}
Typically, this coding style should be avoided. However, you should be aware
that private members of an object can be changed by another object of the same
type. Therefore, in certain special conditions, this coding style may be useful.

82. Can you access private data members of a class from out side the class?
Ans: Yes. This program shows how.
#include
class emp
private :
int i ;
public :
emp( )
{
i = 10 ;
}
} ;
void main( )
emp *p = new emp ;
int *pi = (int*) p ;
cout << *pi ;
*pi = 20 ;
cout << *pi ;
}
The pointer to the class is typecasted in an integer pointer. With the help of this pointer private data member ,i, is accessed in main( ).

83. Why creating array of references is not possible?
Ans: The array name always refers or points to the zeroeth element. If array is of references then the array name would point to the zeroeth element which happens to be a reference. Creating pointer to a reference is not valid. So, creating array of references too is not possible.

84. How do I call a virtual function of a class using a pointer to a function ?
Ans :
#include
class Cvirtual
{
public :
virtual float vfun( )
{
cout << "from vfun" << endl ;
return 7.03f ;
}
} ;
void main( )
{
Cvirtual obj ;
int **p = ( int ** ) &obj ;
float ( *pf1 ) ( ) ;
pf1 = ( float ( * ) ( ) ) **p ;
float f = ( *pf1 ) ( ) ;
cout << "return val = " << f << endl ;
}
In the above program class Cvirtual consists of a virtual function vfun(). In variable p we have stored the address of an object of class Cvirtual. While doing so, we have type casted the address of obj to int **, because obj holds a hidden data member called vptr, which in turn holds the address of virtual function vfun( ). In pf1, a pointer to a function, we are collecting the address of the virtual function vfun( ). Thus the value returned by vfun( ) would then get collected in f.

85. Why an overloaded new operator defined in a class is static?
Ans: An overloaded new function is by default static even if it is not declared so. This is because non-static member functions can be called through an object only. But when an overloaded new operator function gets called the object doesn,t stand created. Since new operator function itself is responsible for creating the object. Hence to be able to call a function without an object, the function must be static.

86. What is a pure virtual destructor?
Ans: Like a pure virtual function we can also have a pure virtual destructor. If a base class contains a pure virtual destructor it becomes necessary for the derived classes to implement the destructor. An ordinary pure virtual function does not have a body but pure virtual destructor must have a body. This is because all the destructors in the hierarchy of inheritance are always called as a part of destruction.

87. When we are required to find offset of an element within a structure? or, how do we call the function of an outer class from a function in the inner class? (The inner class is nested in the outer class)
Ans:
#include
class outer
{
int i ;
float f ;
public :
class inner
{
public :
infunc( )
{
outer *pout ;
pout = (outer*) this - ( size_t ) &( ( ( outer* ) 0 ) -> in ) ;
pout -> outfunc( ) ;
}
} ;
inner in ;
outfunc( )
{
cout << "in outer class,s function" ;
}
} ;
void main( )
{
outer out ;
out.in.infunc( )
}
In the above example we are calling outer::outfunc( ) from inner::infunc(). To call outfunc( ) we need a pointer to the outer class. To get the pointer we have subtracted offset of the inner class,s object (base address of outer class,s object - address of inner class,s object) from address of inner class,s object.

88. void f ( float n, int i = 10 ) ;
void f ( float a ) ;
void main( )
{
f ( 12.6 ) ;
}
void f ( float n, int i )
{
}
void f ( float n )
{
}
The above program results in an error (ambiguous call) since without the default argument the two functions have arguments that are matching in number, order and type.

89. Some programs need to exercise precise control over the memory areas where data is placed. For example, suppose we wish to read the contents of the boot sector into a structure. For this the byte arrangement of the structure elements must match the arrangement of various fields in the boot sector of the disk.
The #pragma pack directives offer a way to fulfill this requirement. The #pragma pack directive specifies packing alignment for structure and union members. The #pragma takes effect at the first structure or union declaration after the #pragma is seen. Consider the following structure:
#pragma pack (1)
struct emp
{
int a ;
float s ;
char ch ;
} ;
#pragma pack( )
Here, #pragma pack ( 1 ) lets each structure element to begin on a 1-byte boundary. Hence the size of the structure will be 9. (int - 4, float - 4, char - 1). If we use #pragma pack ( 2 ) each structure element can begin on a 2-byte boundary. Hence the size of the structure will be 10. (int - 4, float - 4, char - 2).

90. How to restrict a friend class,s access to the private data members?
Ans: If we declare a class as a friend of our class the friend class can access the private data members of our class. However, if we want we can restrict this access to some selective functions of the class. Following program shows how to achieve this:
#include
class X
{
public :
void print ( class Z &z ) ;
} ;
class Z
{
private :
int i ;
public :
Z ( int ii )
{
i = ii ;
}
friend X::print ( class Z &z ) ;
} ;
void X::print ( Z &z1 )
{
cout << z1.i ;
}
main( )
{
Z z ( 10 ) ;
X x ;
x.print ( 10 ) ;
}
In the above program only the X::print( ) function can access the private data members of class Z.

91. What is name mangling?
Ans: C++ enables you to assign the same function name to more than one functions but with different parameter types. This feature is called function overloading. But when we give several functions the same name, how does the compiler decide which particular function is to be called? C++ solves this problem by applying a process called name mangling. Name mangling applies a decorated name to the function. The mangled name includes tokens that identify the functions, return type and the types of its arguments.
class test
{
public :
void fun ( int a, char b ) ;
void fun ( char *c, float y ) ;
} ;
void main( )
{
test s1 ;
s1.fun ( 65, ,A, ) ;
s1.fun ( "Anil", 5.5f ) ;
}
At the time of resolving the calls to fun( ) function the linker would not be able to find the definition of the overloaded function fun( ) and it would report an error. If you look at these errors you will see the mangled names like, (?fun@test@@QAEXJJ@Z) and (?fun@test@@QAEXMM@Z). Note that different compilers may use different name mangling schemes.

92. How would you call a C function from C++ code?
Ans: Using extern "C".
The function prototype must be preceded by extern "C". More than one C functions can be grouped inside braces as shown below:
extern "C"
{
void f( ) ;
void f1( ) ;
}
// In cfunc.c
#include
void f( )
{
printf ( "in f( )" ) ;
}
// In func.cpp
#include
extern "C" void f( ) ;
void main( )
{
f( ) ;
}
Ensure that both .c and .cpp files are in the same project.

93. How to restrict the number of floating-point digits displayed ?
Ans: When we display floating-point values, we can use the setprecision manipulator to specify the desired number of digits to the right of the decimal point.
For example,
cout << setprecision ( 3 ) << 12.34678 ;
This statement would give the output as 12.347.

94. What is a wild pointer ?
Ans: A wild pointer is the one that points to a garbage value. For example, an uninitialized pointer that contains garbage value or a pointer that refers to something that no longer exists.

95. How friend function helps to increase the versatility of overloaded operators?
Ans: Consider the following statement,
s2 = s1 * 2 ;
where, s1 and s2 are objects of sample class. This statement would work if the overloaded operator * ( sample s ) or conversion function is provided in the class. Internally this statement would get converted to,
s2 = s1.operator * ( 2 ) ;
The function materializes because it is called with an object s1. The this pointer of s1 would get passed implicitly. To collect 2 in s, first the compiler would call the one-argument constructor, then it would build a nameless object, which then would get collected in s. However, if we write the above statement as,
s2 = 2 * s1 ;
then it won,t compile. This is because the call now would get treated as,
s2 = 2.operator * ( s1 ) ;
and 2 is not an object. The friend function helps to get rid of such a situation. This is shown in the following program.
#include
class sample
{
private :
int i ;
public :
sample ( int ii = 0 )
{
i = ii ;
}
void showdata( )
{
cout << i << endl ;
}
friend sample operator * ( sample, sample ) ;
} ;
sample operator * ( sample s1, sample s2 )
{
sample temp ;
temp.i = s1.i * s2.i ;
return ( temp ) ;
}
void main( )
{
sample s1 ( 10 ), s2 ;
s2 = s1 * 2 ;
s2.showdata( ) ;
s1 = 2 * s2 ;
s1.showdata( ) ;
}
Here the operator *( ) function takes two parameters. This is because the operator function is no longer a member function of the class. It is a friend of the class sample. Thus the statement s2 = s1 * 2 ; would not take the form s2.operator * ( 2 ). This example shows that using friend permits the overloaded operators to be more versatile.

96. What is a const_cast?
Ans: The const_cast is used to convert a const to a non-const. This is shown in the following program.
#include
void main( )
{
const int a = 0 ;
int *ptr = ( int * ) &a ; // one way
ptr = const_cast ( &a ) ; // better way
}
Here, the address of the const variable a is assigned to the pointer to a non-const variable. The const_cast is also used when we want to change the data members of a class inside the const member functions. The following code snippet shows how to do this.
class sample
{
private :
int data ;
public :
void fun( ) const
{
( const_cast ( this ) ) -> data = 70 ;
}
} ;

97. Using a smart pointer we can make an object appear like a pointer.
If a class overloads the operator -> then any object of that class can appear like a pointer when the operator -> ( ) is called. The following program illustrates this.
#include
class test
{
public :
void fun( )
{
cout << "fun of smart pointer" ;
}
} ;
class smartpointer
{
test t ;
public :
test* operator ->( )
{
return &t ;
}
} ;
void main( )
{
smartpointer sp ;
sp -> fun( ) ;
}
The beauty of overloading operator -> is that even though sp is an object we can make it work like a pointer. The operator -> ( ) returns the address of the object of the type test. Using this address of the test object the function fun( ) of the class test gets called. Thus even though fun( ) is not a member of smartpointer class we can still call it using sp.

98. Can we apply delete on this pointer inside a member function?
Ans : Yes! If the member function of a class is called using a pointer to an object, which is allocated dynamically, the object would get deleted. But if the member function is called using the object, which is allocated statically, then a runtime error would occur. This is because we cannot call delete on statically allocated objects. This is illustrated in the following example.
class sample
{
private :
int i ;
public :
void fun( )
{
delete this ;
}
} ;
void main( )
{
sample *s = new sample ;
s -> fun( ) ; // no error

sample s1 ;
s1.fun( ) ; // would throw a runtime error
}

99. Why can,t data members of a class be initialized at the time of declaration as given in the following code?
class emp
{
private :
int j = 10 ;
} ;
Ans: Memory for data members of a class is allocated only when object of
that class is created. One cannot store data in a memory location, which does not exist at all. Therefore initialization at the time of declaration is not possible.

100. Why in a copy constructor an object is collected in a reference to object as shown below?
#include
class emp
{
public :
emp( )
{
}
emp ( emp& )
{
cout << "copy" ;
}
} ;
void main( )
{
emp e ;
emp e1 = e ;
}
Ans: A copy constructor is called when an object is created and initialised at the same time. It is also called when object is passed to a function. So, If we pass the object to copy constructor copy constructor would get called recursively. Thus it will stuck up in an infinite loop.

101. What is Early Binding and Dynamic Binding?
Ans: The term binding refers to the connection between a function call and the actual code executed as a result of the call. Early Binding: If which function is to be called is known at the compile-time it is known as static or early binding. Dynamic Binding: If which function is to be called is decided at run time it is called as late or dynamic binding. Dynamic binding is so called because the actual function called at run-time depends on the contents of the pointer. For example, call to virtual functions, call to functions to be linked from dlls use late binding.

102. When can we use the function ostrstream::freeze( )?
Ans: While outputting data to memory in the in-memory formatting we need to create an object of the class ostrstream. The constructor of ostrstream receives the address of the buffer but if we want that the ostrstream object should do its own memory management then we need to create an ostrstream object with no constructor arguments as:
ostrstream s ;
Now s will do its own memory management. We can stuff as many bytes into it as we want. If it falls short of memory, it will allocate more memory. If it cannot, it may even move the block of memory. When the object goes out of scope, the heap storage is automatically released. This is a more flexible approach if we do not know how much space we are going to need. If we want the physical address of the memory used by s we can obtain it by calling the str( ) member function:
char* p = s.str( ) ;
Once str( ) has been called then the block of memory allocated by ostrstream cannot be moved. This is logical. It can,t move the block since we are now expecting it to be at a particular location. In such a case we say that ostrstream has freezed itself. Once frozen we can,t add any more characters to it. Adding characters to a frozen ostrstream results in undefined behavior. In addition, the ostrstream is no longer responsible for cleaning up the storage. You took over that responsibility when you asked for the char * with str( ). We can clean the storage in two ways: Using the delete operator as shown below:
ostrstream s ;
char *p ;
p = s.str( ) ;
delete p ;
By unfreezing the ostrstream. You do this by calling freeze( ), with an argument 1. During freezing it is called with the default argument of 0.

Texas Placement Paper : Technical - C & C++


1. if a 5-stage pipe-line is flushed and then we have to execute 5 and 12
instructions respectively then no. of cycles will be
a. 5 and 12
b. 6 and 13
c. 9 and 16
d.none


2. k-map
ab
----------
c 1 x 0 0
1 x 0 x
solve it

a. A.B
B. ~A
C. ~B
D. A+B

3.CHAR A[10][15] AND INT B[10][15] IS DEFINED WHAT,S THE ADDRESS OF A[3][4] AND B[3][4] IF ADDRESS OD A IS OX1000 AND B IS 0X2000
A. 0X1030 AND 0X20C3
B. OX1031 AND OX20C4
AND SOME OTHERS..

4. int f(int *a)
{
int b=5;
a=&b;
}

main()
{
int i;
printf("\n %d",i);
f(&i);
printf("\n %d",i);
}
what,s the output .
1.10,5
2,10,10
c.5,5
d. none

5. main()
{
int i;
fork();
fork();
fork();
printf("----");
}
how many times the printf will be executed .
a.3
b. 6
c.5
d. 8

6.
void f(int i)
{
int j;
for (j=0;j<16;j++)
{
if (i & (0x8000>>j))
printf("1");
else
printf("0");
}
}
what,s the purpose of the program
a. its output is hex representation of i
b. bcd
c. binary
d. decimal

7.#define f(a,b) a+b
#define g(a,b) a*b
main()
{
int m;
m=2*f(3,g(4,5));
printf("\n m is %d",m);
}

what,s the value of m
a.70
b.50
c.26
d. 69

8.main()
{
char a[10];
strcpy(a,"\0");
if (a==NULL)
printf("\a is null");
else
printf("\n a is not null");}

what happens with it .
a. compile time error.
b. run-time error.
c. a is null
d. a is not null.

9. char a[5]="hello"
a. in array we can,t do the operation .
b. size of a is too large
c. size of a is too small
d. nothing wrong with it .

10. local variables can be store by compiler
a. in register or heap
b. in register or stack
c .in stack or heap .
d. global memory.

11. average and worst time complexity in a sorted binary tree is

12. a tree is given and ask to find its meaning (parse-tree)
(expression tree)
ans. ((a+b)-(c*d)) ( not confirmed)

13. convert 40.xxxx into binary .

14. global variable conflicts due to multiple file occurrence is resolved during
a. compile-time
b. run-time
c. link-time
d. load-time


15.Two program is given of factorial. one with recursion and one without recursion . question was which program won,t run for very big no. input because
of stack overfow .
a. i only (ans.)
b. ii only
c. i& ii both .
c. none


16.
struct a
{
int a;
char b;
int c;
}

union b
{
char a;
int b;
int c;
};
which is correct .
a. size of a is always diff. form size of b.(ans.)
b. size of a is always same form size of b.
c. we can,t say anything because of not-homogeneous (not in ordered)
d. size of a can be same if

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