Provided by www.YuvaJobs.com - HP  BANGALORE - 21  October  2003  


HP PAPER - 21 OCT 2003- BANGALORE

I  attended the test conducted on 21st of oct at their office 29,Cunningham Road. There were 80 q,s to be answered in 75mins unlike the previous ones.

there were 3 sections
PART- 1 --> 40 q,s (Fundamental computer Concepts, includes OS,N/w , protocols)
PART-2  --> 20 q,s (Purely C ) --  bit tricky (involves ADA concepts)
PART-3 -->  20 q,s (Analytical) --- very easy

I don,t remeber all the q,s.however some of them which i do have been written below. They r not in order or part of .

 Q : What is not a part of OS ?
 O :      swapper,compiler,device driver,file system.
 A : compiler.

 Q : what is the condition called when the CPU is busy swapping in and out pages of memory without doing any  useful work ?
 O :   Dining philosopher,s problem,thrashing,racearound,option d
 A: thrashing.

 Q : How are the pages got into main memory from secondary memory? DMA, Interrupts,option3, option 4
  A : as far as i know its  Interrupts --by raising a page fault exception.

Q : What is the use of Indexing ?
O : fast linear access, fast random access, sorting of records  , option 4
A : find out.

Q : in terms of both space and time which sorting is effecient. (The question is rephrased .)
O : merge sort, bubble sort, quick sort, option 4
A :  find out

which case statement will be executed in the following  code ?
main()
{
  int i =1;
 switch(i)
   {
         i++;
         case 1 : printf ("");
                     break;
         case 2 : printf("");
                      break;
         default : printf("");
                     break;
  }
}

Answer : Case1 will only be executed.

Q : In the given structure  how do you initialize the day feild?
      struct time {
          char  * day ;
         int     * mon ;
         int     * year ;
          } * times;

Options : *(times).day, *(times->day), *times->*day.

Answer : *(times->day) -- after the execution of this statement  compiler generates   
              error.i didn,t understand why.can anybody explain.

Q: The char has 1 byte boundary , short has 2 byte boundary, int has 4 byte boundary.
     what is the total no: of bytes consumed by the following structure:
   struct st {
   char a ;
   char b;
   short c ;
   int z[2] ;
   char d ;
   short f;
   int q ;
}

Options are given.
Answer : its very easy 20 and not 19 .

Apart from these there were other q,s concerning minimal spanning tree, shortest path and some complexity questions.

(Paper Submitted By : Arun)  
Provided by www.YuvaJobs.com - HP  BANGALORE - 21  October  2003