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

You need parentheses around the argument.

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

Without the parentheses, the expression will expand to:

6+5*6+5*6+5

Which you can see why it would evaluate to 71.

A safer solution would be to use an inline function instead. But, you would need to define a different one for each type. It might also be more clear to rename the macro.

static inline int cube_int (int x) { return x*x*x; }

Answer 2

Always shield your macro arguments with parenthesis:

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

Consider the evaluation without the parenthesis:

6 + 5 * 6 + 5 * 6 + 5

And recall that * has a higher precedence than +, so this is:

6 + 30 + 30 + 5 = 71;

Get to know the precedence rules if you don't already: http://en.cppreference.com/w/cpp/language/operator_precedence

Answer 3

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