#define square(x) x*x void main() { int i; i = 64/square(4); printf("%d",i); } - Q4Interview" /> #define square(x) x*x void main() { int i; i = 64/square(4); printf("%d",i); } - Q4Interview" />
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C Programming :: C Preprocessor

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5. What will be output of the following "c" code?

#include
#define square(x) x*x
void main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer: Option A

Explanation :

Pre-processor will substitute square(4) 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 => 16*4 = 64

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