variable argument type in va_arg function in c

Question

I'm getting a warning by the gcc compiler and the program aborts if the following code is executed I couldn't get why? Would be great help if someone clarified it.

#include<stdio.h>
#include<stdarg.h>
int f(char c,...);
int main()
{
   char c=97,d=98;
   f(c,d);
   return 0;
}

int f(char c,...)
{
   va_list li;
   va_start(li,c);
   char d=va_arg(li,char); 
   printf("%c\n",d);
   va_end(li);
}

GCC tells me this:

warning: 'char’ is promoted to ‘int’ when passed through ‘...’ [enabled by default]
note: (so you should pass ‘int’ not ‘char’ to ‘va_arg’)
note: if this code is reached, the program will abort

Answer 1

Arguments to variadic functions undergo default argument promotions; anything smaller than an int (such as char) is first converted to an int (and float is converted to double).

So va_arg(li,char) is never correct; use va_arg(li,int) instead.

Answer 2

Yes, this appears to be a quirk in C standard. However, it looks like this is only to do with va_arg().

You can take a look at various implementations of printf() to see how to overcome this. For example, the one in klibc is pretty easy to read.