char type in va_arg

Question

I have the following function which writes passed arguments to a binary file.

void writeFile(FILE *fp, const int numOfChars, ...)
{
   va_list ap;
   va_start(ap, numOfChars);
   for(int i = 0; i < numOfChars; i++)
   {
      const char c = va_arg(ap, char);
      putc(c, fp);
   }
   va_end(ap);
}

Upon compiling, I get the following warning from the compiler

 warning: second argument to 'va_arg' is of promotable type 'char'; this va_arg
      has undefined behavior because arguments will be promoted to 'int' [-    Wvarargs]

Now as I understand it, C wants to promote char type to int. Why does C want to do this? Second, is the best solution to cast the int back to a char?

Answer 1

Now as I understand it, C wants to promote char type to int. Why does C want to do this?

Because that's what the standard says. If you pass an integral value with conversion rank smaller than that of int (e. g. char, bool or short) to a function taking a variable number of arguments, it will be converted to int. Presumably the reason for this has its roots in performance, where it was (and in fact, often it still is nowadays) better to pass values aligned to a machine word boundary.

Second, is the best solution to cast the int back to a char?

Yes, but you don't really need a cast even, an implicit conversion will do:

char ch = va_arg(ap, int);