Output after macro expand seems not as per understanding (may be wrong understanding)

Question

Hi need to under stand execution steps for below program.

#include<stdio.h>
#define SQUARE(x) (x)*(x)
void main()
{
 int i = 5;
 printf("%d\n", SQUARE(++i));
}

As i have understanding like macro expanded like (++i)*(++i) and as result (7)*(6) = 42 but the output is 49 which i don't understand why?

'need to under stand execution steps for below program' why? If it's too difficult to easily understand, (which it is), it's not fit for purpose. Get rid of the pre-increment and add an increment as a separate line above the macro 'call'. — ThingyWotsit, Jul 02 '17 at 07:52

score 1 · Answer 1 · answered Jul 02 '17 at 05:10

That's precisely the problem with macros, if you don't pay attention, you end up with undefined behaviour. See Why are these constructs (using ++) undefined behavior?

(++i)*(++i) is undefined behaviour, the result can be anything including exploding your computer.

1 Answers1