Thread: printf like function

Hmmm - not sure if I understand you here - Are you wanting to know how to make a function that will accept an 'infinite' amount of arguments?

if so, it would go something like this:

Code:

void foo( float stuff, int more_stuff, ... )
{
   statements
   declarations      
}

Its the ellipsis notation (the '...') that you should be concerned with - using this notation, the number of arguments passed to the function's parameter list must be equal to, or exceed, the amount of parameters given in the parameter list. This is different from a regular parmater list, as you no doubt already know, where the number of arguments passed must be the same as the number of parameters.

~/

Originally posted by tyouk
Okay but how would you check to see how many arguments are actually passed? And how would you access them?

Something like this maybe (not my code, it's nicked from some old site).

Code:

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

void  test_fn(const char *msg, const char *types, ...);

int main(void)
{
  printf("VA...TEST\n");
  test_fn("PARAMETERS: 1, \"abc\", 546", "isi", 1, "abc", 546);
  test_fn("PARAMETERS: \"def\", 789", "si", "def", 789);
  return 0;
}

static void test_fn
(
  const char  *msg,   /* message to be printed     */
  const char  *types, /* parameter types (i,s)     */
  ...
)                     /* variable arguments     */
{
  va_list     argument;
  int         arg_int;
  char        *arg_string;
  const char  *types_ptr;

  types_ptr = types;
  printf("\n%s -- %s\n", msg, types);
  va_start(argument, types);
  while (*types_ptr != '\0')
  {
    if (*types_ptr == 'i')
    {
      arg_int = va_arg(argument, int);
      printf("integer: %d\n", arg_int);
    }
    else if (*types_ptr == 's')
    {
      arg_string = va_arg(argument, char *);
      printf("string:  %s\n", arg_string);
    }

    ++types_ptr;
  }

  va_end(argument);
}