how to call particular function using #define macro?

Question

I want to call a particular function according to the value i passed to the macro. But it is giving me compilation error

#include <stdio.h>

#define calling(m, j) execcall ## m(j);

void execcall0 (int x) {
    printf("called 0 with arg %d\n", x);
}

void execcall1 (int x) {
    printf("called 1 with arg %d\n", x);
}

void execcall2 (int x) {
    printf("called 2 with arg %d\n", x);
}

int main () {
    int i = 0;
    for (i = 0; i < 3; i++) {
        calling(i, 1);
    }
}

Compilation error:

In function `main':
new.c:(.text+0x7a): undefined reference to `execcalli'
collect2: ld returned 1 exit status

Is it even possible whatever i am trying?

This won't work - macros are a compile-time thing - they are evaluated by the preprocessor. Use function pointers to get the behaviour you are looking for here. — Paul R
– Paul R, Commented May 13, 2014 at 11:34
@Theolodis actually it is a small picture.. in a real test case i am having thousands of function.. Is switch case/if else would be a faster approach? — Vishwadeep Singh
– Vishwadeep Singh, Commented May 13, 2014 at 11:38
Don't use a switch statement - use an array of function pointers. — Paul R
– Paul R, Commented May 13, 2014 at 11:38

John Källén · Accepted Answer · 2014-05-13 11:45:15Z

3

If you want to call a function based on the value of an integer you're better off writing an array of pointers to your functions, and indexing into the array using your integer.

void execcall0(int x); 
void execcall1(int x); 
void execcall2(int x); 

/* Array of pointers to void functions taking an int parameter. */
void (*apfn[])(int) =
{
    execcall0,
    execcall1,
    execcall2, 
};

int main()
{
   int i;
   for (i = 0; i < 3; ++i) {
      (apfn[i])(1);
   }
}

Remember to check the boundary conditions before you index!

edited May 13, 2014 at 11:45

answered May 13, 2014 at 11:38

John Källén

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Comments

perreal · Accepted Answer · 2014-05-13 11:39:34Z

2

You can use an array to store functions, and modify the macro:

#include <stdio.h>

#define calling(m, j) exec_funcs[m](j)

void execcall0 (int x) {
    printf("called 0 with arg %d\n", x);
}

void execcall1 (int x) {
    printf("called 1 with arg %d\n", x);
}

void execcall2 (int x) {
    printf("called 2 with arg %d\n", x);
}

void (*exec_funcs[3])(int) = { execcall0, execcall1, execcall2 };

int main () {
    int i = 0;
    for (i = 0; i < 3; i++) {
        calling(i, 1);
    }
}

But then you don't really need a macro.

answered May 13, 2014 at 11:39

perreal

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1 Comment

Vishwadeep Singh Over a year ago

thanks. nice workaround.. works perfectly.. i will benchmark all the answers and suggestions and will share the results...

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