Re: call through pointer of function with variable arguments

[Stephen Biggs <xyzzy at hotpop dot com>]

On Wed, 2003-06-04 at 19:42, Geoff Keating wrote:
> Stephen Biggs <xyzzy@hotpop.com> writes:
> 
> > Given this code:
> > #include <stdio.h>
> > char buf[2];
> > 
> > int (*gfp)(char *, const char *, ...);
> > int (*gfp1)(const char *, ...);
> >      
> > void f (fp,fp1)
> >      int (*fp)(char *, const char *, ...);
> >      int (*fp1)(const char *, ...);
> > {
> >   (*fp1)("%.0f\n",5.0);
> >   (*fp)(buf, "%.0f", 5.0);
> > }
> > 
> > void h(void)
> > {
> >   gfp = sprintf;
> >   gfp1 = printf;
> > 
> >   (*gfp1)("%.0f\n",5.0);
> >   (*gfp)(buf, "%.0f", 5.0);
> > }
> > 
> > int main()
> > {
> >   f (&sprintf,&printf);
> >   h ();
> >   return 0;
> > }
> > 
> > ... in both the cases, by the time I get to the point where I receive an
> > expression for the call (in either f or g), the function pointer value
> > has already been moved to a register and no prototype information is
> > available, since all I have is the function address (in my port, the
> > arguments for the current call are already saved aside).  The
> > proprietary machine I am porting to is sensitive to whether a call has
> > variable arguments or not and has a different calling sequence based on
> > that, but how can I know?
> 
> You have to remember from the FUNCTION_ARG macro.
> 
> > Are there any other machines that already have been ported that have
> > this sensitivity?
> 
> rs6000 has similar things.  Look at CALL_V4_SET_FP_ARGS in that port.
> 
> -- 
> - Geoffrey Keating <geoffk@geoffk.org>

Thank you!