Re: Waiver question

[Corey Minyard <minyard@acm.org>]

kilgallen@eisner.decus.org (Larry Kilgallen) writes:
> 
> In article <m2k9luibub.fsf@acm.org>, Corey Minyard <minyard@acm.org> writes:
> 
> > Actually, there is now a GCC port for the C30/C40 DSP.  Theoretically,
> > a C++ port and a GNAT port should be doable.  The C++ port should be a
> > matter of libraries.  I'm not sure what the effects of the C40's
> > minimum addressable unit being 32-bits would be on GNAT, but if that
> 
> The minimum addressable unit on DEC Alpha is 32 bits, except for the very
> latest chip where it was altered for purposes of Intel emulation. GNAT
> reportedly "does" DEC Alpha, and I presume that is without relying on
> OS traps for every unaligned instruction :-).
> 

Umm...  No.

The minimum addressable unit on the Alpha is a byte (8 bits).  The
alignment of 32-bit values (and instructions) on the the Alpha is on
32-bit boundaries, but there is no alignment restriction on bytes.
The issue is if you increment a pointer value physically by 1 how many
bits do you skip over?  For instance, the following code (and I hope
the kind readers of the newsgroup excuse me for using C, but it is
more compact for this type of stuff):

int
main()
{
   int a[4];
   char *c;

   a[0] = 0; a[1] = 0; a[2] = 0; a[3] = 0;
   c = (char *) a;
   c[0] = 0x01;
   c[1] = 0x02;
   c[2] = 0x03;
   c[3] = 0x04;
   printf("%x %x %x %x\n", a[0], a[1], a[2], a[3]);
   return 0;
}

on the Alpha will print:
	1020304 0 0 0
and on the C40 will print:
	1 2 3 4
because on the C40 a char and an int are the same size because the
minimum size of any addressable unit is 32-bits.

-- 
Corey Minyard               Internet:  minyard@acm.org
  Work: minyard@nortel.ca       UUCP:  minyard@wf-rch.cirr.com