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Date: Wed, 11 Apr 2001 11:27:28 -0700 (PDT)
From: Linus Torvalds <torvalds@transmeta.com>
To: Bernd Schmidt <bernds@redhat.com>
cc: Andreas Franck <afranck@gmx.de>,
        David Howells <dhowells@cambridge.redhat.com>, <andrewm@uow.edu.au>,
        <bcrl@redhat.com>, <alan@lxorguk.ukuu.org.uk>,
        <linux-kernel@vger.kernel.org>
Subject: Re: [PATCH] 2nd try: i386 rw_semaphores fix
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On Wed, 11 Apr 2001, Bernd Schmidt wrote:
> >
> > The example in there compiles out-of-the box and is much easier to
> > experiment on than the whole kernel :-)
>
> That example seems to fail because a "memory" clobber only tells the compiler
> that memory is written, not that it is read.

The above makes no sense. The notion of "memory is written" is a true
superset of the "memory is read", and must be a complete memory barrier
(ie telling the compiler that "we read memory" is non-sensical: it can't
give the compiler any more information).

Because the memory clobber doesn't tell _what_ memory is clobbered, you
cannot consider memory dead after the instruction. As such, the compiler
HAS to treat a clobber as a "read-modify-write" - because on a very
fundamental level it _is_. Clobbering memory is logically 100% equivalent
to reading all of memory, modifying some of it, and writing the modified
information back.

(This is different from a "clobber specific register" thing, btw, where
the compiler can honestly assuem that the register is dead after the
instruction and contains no useful data. That is a write-only dependency,
and means that gcc can validly use optimization techniques like removing
previous dead writes to the register.)

See? Do you see why a "memory" clobber is _not_ comparable to a "ax"
clobber? And why that non-comparability makes a memory clobber equivalent
to a read-modify-write cycle?

In short: I disagree 100%. A "memory" clobber -does- effectively tell the
compiler that memory is read. If the compiler doesn't realize that, then
it's a compiler bug waiting to happen. No ifs, buts of maybes.

			Linus

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