FRP SAMPLE QUESTIONS WITH ANSWERS

FRP Sample Questions

FRP Sample Questions

1. main()

{

float me = 1.1;

double you = 1.1;

if(me==you)

printf(”Hello”);

else

printf(”Welcome”);

}

Answer:

Welcome

Explanation:

For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

Note:

Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

2.

#define square(x) x*x

main()

{

int i;

i = 64/square(4);

printf(”%d”,i);

}

Answer:

64

Explanation:

The macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64.

3.

main()

{

int i;

printf(”%d”,scanf(”%d”,&i)); // value 10 is given as input here

}

Answer:

1

Explanation:

Scanf returns number of items successfully read and not 10. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

4.

main(int argc, char **argv)

{

printf(”enter the character”);

getchar();

sum(argv[1],argv[2]);

}

sum(num1,num2)

int num1,num2;

{

return num1+num2;

}

Answer:

Compiler error.

Explanation:

argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.atoi() is used to convert a string to integer.

5.

main()

{

char s[ ]=”man”;

int i;

for(i=0;s[ i ];i++)

printf(”\n%c%c%c%c”,s[ i ],*(s+i),*(i+s),i[s]);

}

Answer:

mmmm

aaaa

nnnn

Explanation:

s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].