Paper Type : Whole Testpaper
Test Date : 21 October 2003
Test Location : BANGALORE
Posted By : Arun
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.
0 comments:
Post a Comment