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Message-ID: <7f45d93905080311152d530172@mail.gmail.com>
Date: Wed, 03 Aug 2005 18:15:00 -0000
From: Shaun Jackman <sjackman@gmail.com>
Reply-To: Shaun Jackman <sjackman@gmail.com>
To: Richard Henderson <rth@redhat.com>
Subject: Re: memcpy to an unaligned address
Cc: Ian Lance Taylor <ian@airs.com>, Andrew Pinski <pinskia@physics.uc.edu>, 
	Mike Stump <mrs@apple.com>, Dave Korn <dave.korn@artimi.com>, 
	Paul Koning <pkoning@equallogic.com>, gcc@sources.redhat.com
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	 <m3mzo0jcfm.fsf@gossamer.airs.com> <20050803180011.GA16808@redhat.com>
X-SW-Source: 2005-08/txt/msg00123.txt.bz2

On 8/3/05, Richard Henderson <rth@redhat.com> wrote:
> It is nevertheless correct.  Examine all of the parts of the expression.
>=20
> In particular, "&s->b".  What type does it have?  In an ideal world, it
> would be "pointer to unaligned integer".  But we have no such type in
> our type system, so it is "pointer to integer".  This expression is ONLY
> THEN passed to memcpy.  At which point we query the argument for its
> alignment, and get the non-intuitive result.
>=20
> If you instead pass "s" to memcpy, you should get the correct unaligned
> copy.  If that isn't happening, that's a bug.

I'm not sure I understood the last line. s is a structure, and its
address is aligned. How would you pass it to memcpy, and why would it
generate an unaligned copy?

Cheers,
Shaun