How to use varargs in conjunction with function pointers in C on Win64?

Question

Consider the following C program:

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

typedef void callptr();

static void fixed(void *something, double val)
{
    printf("%f\n", val);
}

static void dynamic(void *something, ...)
{
    va_list args;
    va_start(args, something);
    double arg = va_arg(args, double);
    printf("%f\n", arg);
}

int main()
{
    double x = 1337.1337;
    callptr *dynamic_func = (callptr *) &dynamic;
    dynamic_func(NULL, x);
    callptr *fixed_func = (callptr *) &fixed;
    fixed_func(NULL, x);

    printf("%f\n", x);
}

Basically, the idea is to store a function with variable arguments in a "generic" function pointer. As a comparison, I've also included another function with fixed argument list. Now see what happens when running this on x86 Linux, amd64 Linux, Win32 and Win64:

$ gcc -m32 -o test test.c
$ file test
test: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.9, not stripped
$ ./test
1337.133700
1337.133700
1337.133700

$ gcc -o test test.c
$ file test
test: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.9, not stripped
$ ./test
1337.133700
1337.133700
1337.133700

C:\>gcc -o test.exe test.c
C:\>file test.exe
test.exe: PE32 executable for MS Windows (console) Intel 80386 32-bit
C:\>test.exe
1337.133700
1337.133700
1337.133700

C:\>x86_64-w64-mingw32-gcc -o test.exe test.c
C:\>file test.exe
test.exe: PE32+ executable for MS Windows (console) Mono/.Net assembly
C:\>test.exe
0.000000
1337.133700
1337.133700

Why does the dynamic function get a zero value from the variable argument list on Win64, but not on any of the other configurations? Is something like this even legal? I assumed it is because the compiler didn't complain.

I'm pretty sure this isn't legal; function pointers aren't convertible like this. — Oliver Charlesworth, Sep 05 '11 at 13:03

score 10 · Accepted Answer · answered Sep 05 '11 at 13:26

10

Your code is not valid. Calling a variadic function requires a prototype indicating that it's variadic, and the function pointer type you're using does not provide this. In order for the call not to invoke undefined behavior, you would have to cast the dynamic_func pointer like this to make the call:

((void (*)(void *, ...))dynamic_func)(NULL, x);

answered Sep 05 '11 at 13:26

R.. GitHub STOP HELPING ICE

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Thanks especially for the explanation. Since my project does a lot of these calls, I now wrote a small preprocessor to cast the function pointers into the correct type and it works :) – smf68 Sep 05 '11 at 19:13

score 4 · Answer 2 · answered Sep 05 '11 at 13:09

You should work with consistent function definitions, even if that means to use varargs even if not needed. The best is to be as verbose as needed.

...

typedef void myfunc_t(void *, ...);

...

myfunc_t dynamic;
void dynamic(void * something, ...)
{

...

int main()
{
    double x = 1337.1337;
    myfunc_t *callnow;
    callnow = &dynamic;
    callnow(NULL, x);

    printf("%f\n", x);
}

How to use varargs in conjunction with function pointers in C on Win64?

2 Answers2

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