Preprocessor Directives in C

Question

Why the output of the below program is 125 and not 5?

#include<stdio.h>

#define square(x) x*x

int main()
{
  int var;
  var = 125/square(5);
  printf("%d",var);
  return 0;
}

Answer 1

This line:

var = 125/square(5);

is expanded into:

var = 125/5*5;

which evaluates from left to right to 25*5 and then to 125 ;

To fix it, parenthesize the argument in the definition of square , like this:

#define square(x) ((x)*(x))

Note also the extra parenthesis around x in order to achieve expected behavior when eg 1+2 is passed into square .

Answer 2

Note that var = 125/square(5); becomes var = 125/5*5 when you compile the code.
So the compiler calculates 125/5 before 5*5 . The result becomes (125/5)*5 = 125 .

Instead of #define square(x) x*x , put #define square(x) (x*x) .

Here is your code:

#include<stdio.h>

#define square(x) (x*x)

int main()
{
  int var;
  var = 125/square(5);
  printf("%d",var);
  return 0;
}