stdarg.h/code generation botch?

[Spencer W. Thomas]

Ok, so I've got this little piece of code that looks like this

#include <stdarg.h>
int scanargs( argc, argv, format )
int argc; char **argv; char *format;
{
	int retval;
	va_list argl;
	va_start( argl, format );
	retval = do_scan( argc, argv, format, argl );
	va_end( argl );
	return retval;
}

It doesn't work.  The first word pointed to by argl is 0.  Why?  Let's
look at the assembly output:

scanargs:
	.option	O2
	subu	$sp, 32
	sw	$31, 20($sp)	% Store frame pointer on the stack
	sd	$4, 32($sp)	% Store args 0 & 1 on the stack
	sw	$6, 40($sp)	% Store arg 2 on the stack

Well, lo and behold, only the first 3 register arguments get put on
the stack.  The 4th is therefore garbage (on the stack) (but, since
scanargs is the first function called in the program, it's 0).

But, what's really weird, is that if I use varargs, it does work:

#include <varargs.h>
int scanargs( va_alist )
va_dcl
{
	int retval;
	va_list argl;
	int argc; char **argv; char *format;
	va_start( argl );
	argc = va_arg( argl, int );
	argv = va_arg( argl, char ** );
	format = va_arg( argl, char * );
	retval = _do_scanargs( argc, argv, format, argl );
	va_end( argl );
	return retval;
}

scanargs:
	.option	O2
	subu	$sp, 48
	sw	$31, 20($sp)	% Store frame pointer on stack
	sd	$4, 48($sp)	% Store args 0 & 1 on stack
	sd	$6, 56($sp)	% Store args 2 & 3 on stack

But va_alist is just one argument, running the above through the
preprocessor yields:

int scanargs( va_alist )
int va_alist;
...

I would expect the compiler to only save one register argument on the
stack in this case.  But it saves all 4!

So, what's up?  Why does the varargs case work, and the stdarg case
not?  Is this a compiler botch?  Is it an include file botch?  Will it
be fixed?  Should I call the Geometry Hotline?

Yours, in confusion,
--
=Spencer W. Thomas 		EECS Dept, U of Michigan, Ann Arbor, MI 48109
spencer at eecs.umich.edu		313-936-2616 (8-6 E[SD]T M-F)