Variable arguments without first argument - what belongs in va_start?

Question

I want to make varargs function for freeing multiple pointers at once, mostly to clean up the code. So I have:

void free_all( ... ) {
    va_list arguments;
    /* Initializing arguments to store all values after last arg */
    // but there are no args!
    va_start ( arguments, ????? );
    /* we expect the caller to send last argument as NULL **/
    void* pointer = va_arg ( arguments, void* );
    while( (pointer = va_arg ( arguments, void* ))!=NULL ) {
        free(pointer);
    }
    va_end ( arguments );                  // Cleans up the list
}

So what to put in va_start ( arguments, ????? )?

Iharob Al Asimi · Accepted Answer · 2016-11-28 14:03:28Z

8

It's simply not possible. You MUST have a non vararg argument, always. In your case

void free_all(void *first, ...);

could work.

answered Nov 28, 2016 at 14:03

Iharob Al Asimi

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7 Comments

Tomáš Zato Over a year ago

Is that a technical limitation, or simply something not implemented? (Here I'm simply curious how it works)

StoryTeller - Unslander Monica Over a year ago

@TomášZato, it's an oversight by the standard IMO. That extra argument is a hook into the stack frame. Theoretically, it could be accomplished without it.

Iharob Al Asimi Over a year ago

@StoryTeller I was going to say it couldn't, but then I thought exactly of that. There must be a historical reason though.

StoryTeller - Unslander Monica Over a year ago

@iharob, they probably reasoned that the feature can't be useful without sending some information about the extra parameters. And that translated to requiring at least one fixed parameter.

newacct Over a year ago

@TomášZato: It's simply because the C standard says so. You could make your own extension of C that supports this somehow, but it wouldn't be standard C.

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