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Re: FAIL: gcc.c-torture/execute/950628-1.c execution, -O1
- To: Richard Kenner <kenner at vlsi1 dot ultra dot nyu dot edu>
- Subject: Re: FAIL: gcc.c-torture/execute/950628-1.c execution, -O1
- From: Richard Henderson <rth at redhat dot com>
- Date: Wed, 24 Jan 2001 09:51:36 -0800
- Cc: gcc-patches at gcc dot gnu dot org
- References: <10101241239.AA29983@vlsi1.ultra.nyu.edu>
On Wed, Jan 24, 2001 at 07:38:28AM -0500, Richard Kenner wrote:
> If there are more, the second store can be moved ahead of a
> preceeding read because anti_dependence evaluates false.
>
> That's not the way the code looks to me. It looks like they conflict if
> both are unchanging. I think a lot of things depend on that.
Look again.
/* If MEM is an unchanging read, then it can't possibly conflict with
the store to X, because there is at most one store to MEM, and it must
have occurred somewhere before MEM. */
if (! writep && RTX_UNCHANGING_P (mem))
return 0;
r~