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

[Satyam Sharma <satyam@infradead.org>]

On Fri, 17 Aug 2007, Nick Piggin wrote:

> Satyam Sharma wrote:
> > On Fri, 17 Aug 2007, Nick Piggin wrote:
> > > Satyam Sharma wrote:
> > > 
> > > It is very obvious. msleep calls schedule() (ie. sleeps), which is
> > > always a barrier.
> > 
> > Probably you didn't mean that, but no, schedule() is not barrier because
> > it sleeps. It's a barrier because it's invisible.
> 
> Where did I say it is a barrier because it sleeps?

Just below. What you wrote:

> It is always a barrier because, at the lowest level, schedule() (and thus
> anything that sleeps) is defined to always be a barrier.

"It is always a barrier because, at the lowest level, anything that sleeps
is defined to always be a barrier".


> Regardless of
> whatever obscure means the compiler might need to infer the barrier.
> 
> In other words, you can ignore those obscure details because schedule() is
> always going to have an explicit barrier in it.

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.

> > > The "unobvious" thing is that you wanted to know how the compiler knows
> > > a function is a barrier -- answer is that if it does not *know* it is not
> > > a barrier, it must assume it is a barrier.
> > 
> > True, that's clearly what happens here. But are you're definitely joking
> > that this is "obvious" in terms of code-clarity, right?
> 
> No. If you accept that barrier() is implemented correctly, and you know
> that sleeping is defined to be a barrier,

Curiously, that's the second time you've said "sleeping is defined to
be a (compiler) barrier". How does the compiler even know if foo() is
a function that "sleeps"? Do compilers have some notion of "sleeping"
to ensure they automatically assume a compiler barrier whenever such
a function is called? Or are you saying that the compiler can see the
barrier() inside said function ... nopes, you're saying quite the
opposite below.


> then its perfectly clear. You
> don't have to know how the compiler "knows" that some function contains
> a barrier.

I think I do, why not? Would appreciate if you could elaborate on this.


Satyam
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