passing variable arguments

[Dan KoGai]

In article <353 at ankh.ftl.fl.us> chin at ankh.ftl.fl.us (Albert Chin) writes:
>How does printf() work. I believe there is some way in C to obtain the
>pointer to the argument list in a function. If so, printf() is pretty 
>simple. But how do you do this. Is it possible to get the number of
>arguments in a printf() statement?

	On C convention, arguments are pushed to the stack right to left
order before it jumps to the function.  Since you can push as many arguments
as your memory spec allows, it's quite possible, as you see on printf, to
make variable argument functions on C.
	However, variable argument functions are somewhat kludgy on C:  Bad
news is that you have to explicity include the number of arguments somewhere
in your argument and there's no other way of getting number of arguments:
in printf() and scanf(), et al., It's number of '%'s in format and it might
crash if you give wrong number of arguments.  Also note there's no way
you can tell the type of arguments in variable arg functions--printf() uses
format to specify it and even though you feed wrong type args in printf()
it uses argument type implied in format--it's a common mistake to pass
int, instead of pointer to int in scanf().
	On Ansi C, you can use <stdarg.h> macro package to write variable
argument function.  And variable argument is implied by "...".  Here's simple
example:

#include <stdarg.h>

/* multiple strcat */

char *mstrcat(int, nsrc, char *dst, char *src1, ...)	
/*... means more args to come */
{
	va_list srcs;	/* va_list is typically void ** */
	va_start(srcs, src1);	/* now srcs = &src1 */
	while(nsrc--){	/* repeat nsrc times */
		strcat(dst, va_arg(char *, srcs));
		/* va_arg(type, valist) returns *(type)valist 
		  then inclements valist */
	}
	va_end(srcs)	/* clears stack or do nothing: compiler dependent */
	return dst;
}

	You can get more details from K&R.  The most important thing to
remember is that in C you have to know how many number of what types of
arguments are there.  You C compiler will not take care of you.

----------------
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