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C macro argument 'storage'

Given:

#define f(x, y) (x+y)
#define g(x, y) (x*y)
#define A 1, 2
#define B 2, 3

int main() {
  int a = f(A);
  int b = g(A);
  int c = f(B);
  int d = g(B);
}

which doesn't work

how can i make it work? The main idea is that I have one list of arguments that I want to pass to two different macros, without repeating each time the list of long arguments.

Is there any way to do this? [You can change f and g; you can even change A and the way I call macros. The only requirements are: 1) the argument list can appear only once 2) it cannot be hardcoded ... so that I can call macros with different arguments

If you solution doesn’t quite work, but “almost works” (almost such a definition for you), I would also like to hear it, maybe I can fake it for work.

Thank!

: f g . .

+5
4

EDIT: , A B

#define f(x, y) (x+y) 
#define g(x, y) (x*y) 
#define A 1, 2 
#define B 2, 3

#define APPLY2(a, b) a b
#define APPLY(a, b) APPLY2(a, (b))

int main(int argc, char* argv[])
{
    int x= APPLY(f, A);
    int y= APPLY(f, B);
    int z= APPLY(g, A);
    int d= APPLY(g, B);

    return 0;
}
+3

:

static int f(int x, int y) { return (x+y); }
static int g(int x, int y) { return (x*y); }
#define A 1, 2
#define B 2, 3

C, inline, . ++,

template<T> T f(T x, T y) { return (x+y); }
template<T> t g(T x, T y) { return (x*y); }
#define A 1, 2
#define B 2, 3

, C.

f g , , C , . C.

+4

If you used C99, you can use the compound initiator syntax to do this by passing multiple arguments as a single array:

#define f(a) (a[0]+a[1])
#define g(a) (a[0]*a[1])
#define A ((int[]) {1, 2})
#define B ((int[]) {2, 3})

GCC supports this composite syntax literal in both C89 and C ++ mode.

+4
source

Perhaps this is what you want:

#include<iostream>
using namespace std;

#define A 1, 2

template <typename T>
inline T f(T a, T b) { return a + b; }

int main()
{
    cout << f(A) << endl;
    return 0;
}
0
source

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