Macro and function with same name

Question

I have the following code

#define myfunc(a,b) myfunc(do_a(a), do_b(b))

void myfunc(int a, int b)
{
  do_blah(a,b);
}
int main()
{
    int x = 6, y = 7;
    myfunc(x,y);

    return 0;
}

I want the pre-processor to expand function myfunc only at calling. Required code after pre-processing looks like this:

void myfunc(int a, int b)
{
  do_blah(a,b);
}
int main()
{
    int x = 6, y = 7;
    myfunc(do_a(x),do_b(y));

    return 0;
}

The problem is that function definition is expanded also like this

void myfunc(do_a(int a), do_b(int b))
{
  do_blah(a,b);
}

Is there any way to make macro expands only if we are expanding a function call? I tried many solutions, and it seems impossible but I hope that some one saw situation like this..

NOTE: please don't tell me to rename the macro or function names :D

Update1: Thanks for you help. But I can only change the definition of the macro, I can't change its position and I can't change function implementation.

Answer 1

Use () to stop the preprocessor from expanding the function definition:

#include <stdio.h>

#define myfunc(a, b) myfunc(do_a(a), do_b(b))
/* if you have a preprocessor that may be non-standard
 * and enter a loop for the previous definition, define
 * myfunc with an extra set of parenthesis:
#define myfunc(a, b) (myfunc)(do_a(a), do_b(b))
 ******** */

int (myfunc)(int a, int b) /* myfunc does not get expanded here */
{
    printf("a=%d; b=%d\n", a, b);
    return 0;
}

int do_a(int a)
{
    return a * 2;
}

int do_b(int b)
{
    return b - 5;
}

int main(void)
{
    myfunc(4, 0);
    return 0;
}