Sorry guys, my bad, i thought it was #define sqr(x) ((x)*(x)) the out

#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 / jugal

Answer Posted / jugal

Sorry guys,
my bad,
i thought it was
#define sqr(x) ((x)*(x))

the output wud be 5
but still the value of will be 2 only

Is This Answer Correct ?

2 Yes

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