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.

pmg · Accepted Answer · 2009-12-23T18:16:10.313

24

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;
}

edited Dec 23 '09 at 18:16

answered Dec 23 '09 at 13:30

pmg

106,608
13
126
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I think some preprocessors may balk at the the recursive use of the form myfunc(xxx,xxx). I'd do #define myfunc(a,b) (myfunc)(do_a(a), do_b(b)) int (myfunc)(int a, int b) &c – Tim Schaeffer Dec 23 '09 at 17:39
3

Good point, Tim, thanks! But I think it's safe to assume the preprocessor will never enter a recursive loop (cf Standard 6.10.3.4/2 "... if any nested replacements encounter the name of the macro being replaced, it is not replaced ..."). – pmg Dec 23 '09 at 18:11
1

If you can change the function to put () around it's name, why can you not change the function's name? – Pete Kirkham Dec 23 '09 at 23:20
I've been wondering about that too, but didn't have the guts to ask the question directly :) – pmg Dec 23 '09 at 23:36
3

To change the function's name you have to - get approval from all the managers - update all documents - re-run test cases – Yousf Dec 24 '09 at 06:16

score 9 · Answer 2 · answered Dec 23 '09 at 10:23

I see three possible solutions:

define your macro after function definition.
define, before the function definition, do_a() and do_b() such that they return their argument, and redefine them at your will after function definition

perform do_a() and do_b() inside the function:

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

I have a strong preference for the latter.

score 4 · Answer 3 · answered Dec 23 '09 at 10:16

4

Define the macro after the defintion of the function.

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

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

int main()
{
    int x = 6, y = 7;
    myfunc(x,y);

    return 0;
}

answered Dec 23 '09 at 10:16

Gunther Piez

29,760
6
71
103

score 1 · Answer 4 · answered Dec 23 '09 at 10:16

Define the macro after you define the function.

Alternatively, use a pattern like this:

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

.
.

#undef myfunc
void myfunc(int a, int b)
{
  do_blah(a,b);
}
#define myfunc(a,b) myfunc_macro(a,b)

.
.

int main()
{
    int x = 6, y = 7;
    myfunc(x,y);

    return 0;
}

Macro and function with same name

4 Answers4

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