C | Macro & Preprocessor | Question 6

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#include <stdio.h>
#define square(x) x*x
int main()
{
  int x;
  x = 36/square(6);
  printf("%d", x);
  return 0;
}

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(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))  


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