How to pass a function pointer to a function with variable arguments?

Question

I don't know how to accomplish this!
how to get the function pointer in va_list arguments?
thanks so much.

ephemient · Accepted Answer · 2009-10-20 20:37:30Z

5

Typedefs often make working with function pointers easier, but are not necessary.

#include <stdarg.h>
void foo(int count, ...) {
    va_list ap;
    int i;
    va_start(ap, count);
    for (i = 0; i < count; i++) {
        void (*bar)() = va_arg(ap, void (*)());
        (*bar)();
    }
    va_end(ap);
}

answered Oct 20, 2009 at 20:37

ephemient

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5 Comments

caf Over a year ago

In general, to create a type name (for use with va_arg, sizeof, a typecast or similar), write a declaration of a variable with that type, then remove the variable name.

Robert Gamble Over a year ago

Actually, this is (ironically) they only time that typedefs are required. Speaking of the type parameter for va_arg in section 7.15.1.1 of the C Standard: "The parameter type shall be a type name specified such that the type of a pointer to an object that has the specified type can be obtained simply by postfixing a * to type." This means that many non-trivial types aren't guaranteed to work, including your example, without a typedef.

KeatsPeeks Over a year ago

Doesn't compile on my toolset.

ephemient Over a year ago

Yeah, a couple hours after I wrote this I thought about whether it was legal, and then "well, it works with GCC..." Oh well, caf's comment is useful regardless of this particular answer.

caf Over a year ago

...except strike va_arg from the examples I gave.

KeatsPeeks · Accepted Answer · 2009-10-20 18:33:24Z

4

Use a typedef for the function pointer type.

answered Oct 20, 2009 at 18:33

KeatsPeeks

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