Placement Papers - HP

Hewlett Packard
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Hello Friends

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 ?
  int i =1;
     case 1 : printf ("");
     case 2 : printf("");
     default : printf("");

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.