main()

{

int i, j, *p;

i = 25;

j = 100;

p = &i; // Address of i is assigned to pointer p

printf("%f", i/(*p) ); // i is divided by pointer p

}

a. Runtime error.

b. 1.00000

c. Compile error

d. 0.00000

Answers were Sorted based on User's Feedback



main() { int i, j, *p; i = 25; j = 100; p = &i; // Address of i is ass..

Answer / rk

d) it will print 0.0000.

If we typecast the result to float as shown below then
expected output will be printed(i.e. 1.0000)
printf("%f",(float) i/(*p) ); // i is divided by pointer p

Is This Answer Correct ?    8 Yes 0 No

main() { int i, j, *p; i = 25; j = 100; p = &i; // Address of i is ass..

Answer / guest

c) Error becoz i/(*p) is 25/25 i.e 1 which is int & printed
as a float,

So abnormal program termination,

runs if (float) i/(*p) -----> Type Casting

Is This Answer Correct ?    6 Yes 2 No

main() { int i, j, *p; i = 25; j = 100; p = &i; // Address of i is ass..

Answer / km

the answer to this question is implementation dependent:
if tried in Turbo C++
a) Runtime error
abnormal termination
if tried in Unix using GNU C
non of the above
you get a junk result

Is This Answer Correct ?    2 Yes 0 No

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