forward args of variadic function in C

Question

I want to forward args of variadic function, I have already find the some topic.

Forward an invocation of a variadic function in C

When I start to practice, I found a problem.

#include <stdio.h>
#include <stdarg.h>
void fun1(const char *msg, ...) // try to forward printf
{
    va_list arg_list;
    va_start(arg_list, msg);
    vprintf(msg, arg_list);
    va_end(arg_list);
}

void fun2(const char *msg, ...) // try to forward fun1
{
    va_list arg_list;
    va_start(arg_list, msg);
    fun1(msg, arg_list);
    va_end(arg_list);
}

int main()
{
    fun1("this is int %d, float %f\n", 1, 2.3);
    fun2("this is int %d, float %f\n", 1, 2.3);
    return 0;
}

I compile code with gcc main.c and the output shown that

this is int 1, float 2.300000
this is int 6684168, float 2.300000

I can not understand why the fun2 not forward the args of fun1 correctly. Why the int 1 goes to another number but 2.3 still good. How can I modify my code to implement the forward?

Thanks for your time.

`vprintf` is not itself a var arg function. Therein lies the difference — StoryTeller - Unslander Monica, Jan 31 '23 at 08:35
Thanks sir. Can I make the fun1 be a var arg function, and fun2 still a var arg function? In my project, fun1 is debug output and fun2 is another level debug output. Thus I want to both be var arg function. — Xu Hui, Jan 31 '23 at 08:38
The typical way is to create two versions. `printf` is usually implemented in terms of `vprintf`. You can do the same for the added flexibility, `fun1` and `vfun1`. — StoryTeller - Unslander Monica, Jan 31 '23 at 08:39

Chris Dodd · Accepted Answer · 2023-02-01T00:06:21.113

fun1 needs a list of arguments to match its format, but when call it from fun2 you give it a va_list. To call it that way you need to rewrite it to take a va_list rather than a ...:

void fun1(const char *fmt, va_list args) {
    vfprintf(fmt, args);
}

If you're using gcc, you can avoid many problems of this nature by using -Wall and adding a declaration for your functions that take ...:

extern void fun2(const char *fmt, ...) __attribute__((format(printf, 1, 2)));

This tells gcc that fun2 takes printf-style arguments starting from the second argument with the first argument as the format. With this, it will warn you if the arguments passed to fun1 don't match the format string.

score 0 · Answer 2 · answered Jan 31 '23 at 08:53

As suggtested by StoryTeller - Unslander Monica and Chris Dodd, My finial code looks like follows.

// before modify
//   vprintf  <-- fun1  <-- fun2


//    after modify

//    vprintf  <-- vfun1 <--- fun1
//                         |- fun2

#include <stdio.h>
#include <stdarg.h>
void vfun1(const char *msg, va_list arg_list)
{
    printf("this is vfun1\n");
    vprintf(msg, arg_list);
}

void fun1(const char *msg, ...) // try to forward vprintf, but implemented by call vfun1
{
    printf("this is fun1\n");
    va_list arg_list;
    va_start(arg_list, msg);
    vfun1(msg, arg_list);
    va_end(arg_list);
}

void fun2(const char *msg, ...) // try to forward fun1, but implemented by call vfun1
{
    printf("this is fun2\n");
    va_list arg_list;
    va_start(arg_list, msg);
    vfun1(msg, arg_list);
    va_end(arg_list);
}

int main()
{
    printf("------------------------------------\n");
    fun1("this is int %d, float %f\n", 5, 2.3);
    
    printf("------------------------------------\n");
    fun2("this is int %d, float %f\n", 5, 2.3);
    return 0;
}

forward args of variadic function in C

2 Answers2