Question

This is a very simplified version of some code I just ran into at work:

#include <stdio.h>

#define F(G) G(1)
#define G(x) x+1

int main() {
  printf("%d\n", F(G));
}

prints 2.

Now, I can see that F(G) expands to G(1) and then G(1) expands to 2, but its not clear to me why. I would have expected to get an error that G is not a function from the printf line.

How does the pre-processor parse code like this?

Answer 1

A function-like macro is only invoked if its name is followed by a (.

In F(G), G is not followed by a (, so the G there is not a macro invocation.

In F(G) G(1), G is a macro parameter and thus is not macro-replaced directly (this is a very confusing macro you’ve got :-O). In G(1), G is replaced by the argument corresponding to the parameter G, which also happens to be G. That replacement is then rescanned and G(1) is evaluated to 1 + 1.

If we rewrite your macros so that you aren’t using G in multiple different ways, it’s far easier to understand:

#define F(x) x(1)
#define G(x) x + 1

Here, F(G) is replaced by G(1). This is then rescanned, and the invocation of G is evaluated, yielding 1 + 1.