netdev - Re: [PATCH 0/24] make atomic_read() behave consistently across all architectures

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Date:	Sat, 18 Aug 2007 07:45:53 +0530 (IST)
From:	Satyam Sharma <satyam@...radead.org>
To:	Nick Piggin <piggin@...erone.com.au>
cc:	Stefan Richter <stefanr@...6.in-berlin.de>,
	paulmck@...ux.vnet.ibm.com,
	Herbert Xu <herbert@...dor.apana.org.au>,
	Paul Mackerras <paulus@...ba.org>,
	Christoph Lameter <clameter@....com>,
	Chris Snook <csnook@...hat.com>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	linux-arch@...r.kernel.org,
	Linus Torvalds <torvalds@...ux-foundation.org>,
	netdev@...r.kernel.org, Andrew Morton <akpm@...ux-foundation.org>,
	ak@...e.de, heiko.carstens@...ibm.com, davem@...emloft.net,
	schwidefsky@...ibm.com, wensong@...ux-vs.org, horms@...ge.net.au,
	wjiang@...ilience.com, cfriesen@...tel.com, zlynx@....org,
	rpjday@...dspring.com, jesper.juhl@...il.com,
	segher@...nel.crashing.org
Subject: Re: [PATCH 0/24] make atomic_read() behave consistently across all
 architectures

On Fri, 17 Aug 2007, Nick Piggin wrote:

> Satyam Sharma wrote:
> 
> > I didn't quite understand what you said here, so I'll tell what I think:
> > 
> > * foo() is a compiler barrier if the definition of foo() is invisible to
> >  the compiler at a callsite.
> > 
> > * foo() is also a compiler barrier if the definition of foo() includes
> >  a barrier, and it is inlined at the callsite.
> > 
> > If the above is wrong, or if there's something else at play as well,
> > do let me know.
> 
> [...]
> If a function is not completely visible to the compiler (so it can't
> determine whether a barrier could be in it or not), then it must always
> assume it will contain a barrier so it always does the right thing.

Yup, that's what I'd said just a few sentences above, as you can see. I
was actually asking for "elaboration" on "how a compiler determines that
function foo() (say foo == schedule), even when it cannot see that it has
a barrier(), as you'd mentioned, is a 'sleeping' function" actually, and
whether compilers have a "notion of sleep to automatically assume a
compiler barrier whenever such a sleeping function foo() is called". But
I think you've already qualified the discussion to this kernel, so okay,
I shouldn't nit-pick anymore.
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