[SOLVED] wrapper for a function with variable arguments in C

wrapper for a function with variable arguments in C

Recently I wanted to implement a printf wrapper. After some search I found the vprintf is good for this need:

void my_printf(const char *fmt, ...)
{
    va_list args;
    va_start(args, fmt);
    vprintf(fmt, args);
    va_end(args);
}

But is it possible to implement such a wrapper for printf or any other similar functions with variable arguments instead of va_list?

(I mean, what if they don't provide a v version?)

Since some commenter didn't fully capture my idea, I'd better elaborate it.

Suppose you have a plain printf function as in the C library.

Some one gives you a fmt string "%d %u %f", and corresponding inputs.

Now you want to write a function similar to printf, but with all %f replaced by %.2f.

Of course you can use two statements to finish the task:

replace(fmt, "%f", "%.2f");
printf(fmt, inputs);

But if you use this function many times, probably you want to have a wrapper to save some time.

A macro can finish this task, of course. But is it possible without a macro, like:

void myprintf(fmt, ...)
{
    replace(fmt, "%f", "%.2f");
    printf(fmt, inputs);
}

The problem here is that you don't know how to feed the inner printf with the arguments ... of myprintf.

Hope this clarifies.

Solution

If you just want to use this to prepend an string to the output or so, you could use a variadic macro.

#define MYPRINT(...) printf("toto has: " __VA_ARGS__)

in that simple example this suppposes that format that you pass in is a string literal, so this is a bit restricted. But I hope you see the pattern how to use such simple macro wrappers.