some error in output in using macro in C

Question

my code is:-

#include
#include

#define sq(x) x*x*x

  void main()
  {
    printf(\"Cube is : %d.\",sq(6+5));
    getch();
  }
         


        

Answer 1

If you define the macro like this:

#define sq(x) x*x*x

And call it:

sq(6+5);

The pre-processor will generate this code:

6+5*6+5*6+5

Which is, due to operator precedence, equivalent to:

6+(5*6)+(5*6)+5

That's why, the macro arguments must be parenthesized:

#define sq(x) (x)*(x)*(x)

So that pre-processor output becomes:

(6+5)*(6+5)*(6+5)

However, if you pass some arguments with side-effects such as (i++):

sq(i++)

It will be expanded to:

(i++)*(i++)*(i++)

So, be careful, perhaps you need a function