va_arg with a list of void*

Question

I would like to create a function that takes a variable number of void pointers,

val=va_arg(vl,void*);

but above doesn't work, is there portable way to achieve this using some other type instead of void*?

Works OK for me. Can you post the rest of the code that uses the va_list? Or error messages? Which compiler did you use? — Matt K, Feb 07 '11 at 19:40
possible duplicate of [Variable number of arguments in C++?](http://stackoverflow.com/questions/1657883/variable-number-of-arguments-in-c) — wilhelmtell, Feb 07 '11 at 19:50
C++? https://dbj.org/c-how-to-simply-handle-variable-number-of-function-arguments/ — Chef Gladiator, Feb 26 '22 at 12:05

score 6 · Accepted Answer · edited Jul 18 '12 at 21:37

6

#include <stdio.h>
#include <stdlib.h>
#include <stdarg.h>

void
myfunc(void *ptr, ...)
{
    va_list va;
    void *p;

    va_start(va, ptr);
    for (p = ptr; p != NULL; p = va_arg(va, void *)) {
        printf("%p\n", p);
    }
    va_end(va);
}

int
main() {
    myfunc(main,
           myfunc,
           printf,
           NULL);
    return 0;
}

I'm using Fedora 14..

edited Jul 18 '12 at 21:37

richo

8,717
3
29
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answered Feb 07 '11 at 19:59

Michael Closson

902
8
13

right, it actually works my bad, its 5 am here should have gotten some sleep before asking. – Hamza Yerlikaya Feb 07 '11 at 21:00
1

That's a bad example call, since function pointers are not convertable to `void *` (that's a GCC extension). – caf Feb 07 '11 at 23:44
Sentinel value must be used, in this case, that is `NULL` last argument. – Chef Gladiator Feb 26 '22 at 12:32

score 2 · Answer 2 · answered Feb 07 '11 at 20:29

Since you have a C++ tag, I'm going to say "don't do it this way". Instead, either use insertion operators like streams do OR just pass a (const) std::vector<void*>& as the only parameter to your function.

Then you don't have to worry about the issues with varargs.

va_arg with a list of void*

2 Answers2