#include <stdio.h> #define square(x) x*x int main()
{ int x;
x = 36/square(6);
printf ( "%d" , x);
return 0;
} |
(A) 1
(B) 36
(C) 0
(D) Compiler Error
Answer: (B)
Explanation: Preprocessor replaces square(6) by 6*6 and the expression becomes x = 36/6*6 and value of x is calculated as 36. Note that the macro will also fail for expressions “x = square(6-2)”
If we want correct behavior from macro square(x), we should declare the macro as
#define square(x) ((x)*(x))