#include <stdio.h> #define sqr(x) (x*x) int main() { int x=2; pri

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#include <stdio.h>
#define sqr(x) (x*x)
int main()
{
int x=2;
printf("value of x=%d",sqr(x+1));
}

What is the value of x?

Question Posted / mittal

16 Answers
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#include <stdio.h> #define sqr(x) (x*x) int main() { int x=2; printf("value of x..

Answer / jugal

The OUTPUT of the program wud be
"value of x=9"

NOTE:
#define sqr(x) (x*x) and
#define sqr(x) x*x
are two very different things
what Divakar and all are saying is referring to the 2nd
one, where as in this case the 1st one is given

Coming to the actual question
The value of x will remain 2, since its value is not being
changed anywhere in the program, its just being passed to a
macro, but NOT modified there either.
Try adding a line at the end of the program
printf("x=%d",x);

Is This Answer Correct ?

1 Yes

1 No

#include <stdio.h> #define sqr(x) (x*x) int main() { int x=2; printf("value of x..

Answer / poorna

Warning: Function Should have return value........

Is This Answer Correct ?

4 Yes

5 No

#include <stdio.h> #define sqr(x) (x*x) int main() { int x=2; printf("value of x..

Answer / guest

the x itself is 2 and the print out is 9

Is This Answer Correct ?

14 Yes

16 No

#include <stdio.h> #define sqr(x) (x*x) int main() { int x=2; printf("value of x..

Answer / c harlee jain

Value of x will be 2

Is This Answer Correct ?

2 Yes

4 No

#include <stdio.h> #define sqr(x) (x*x) int main() { int x=2; printf("value of x..

Answer / grbr

Is This Answer Correct ?

3 Yes

8 No

#include <stdio.h> #define sqr(x) (x*x) int main() { int x=2; printf("value of x..

Answer / anil

Is This Answer Correct ?

3 Yes

16 No

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