Re: Why does test_bit() take a volatile addr?

[Michael S. Tsirkin]

On Mon, Sep 16, 2013 at 06:02:31PM +1000, Stephen Rothwell wrote:
> Hi Michael,
> 
> On Mon, 16 Sep 2013 10:26:03 +0300 "Michael S. Tsirkin" <mst@xxxxxxxxxx> wrote:
> >
> > On Mon, Sep 16, 2013 at 04:53:44PM +1000, Stephen Rothwell wrote:
> > > 
> > > On Mon, 16 Sep 2013 13:38:35 +0930 Rusty Russell <rusty@xxxxxxxxxxxxxxx> wrote:
> > > >
> > > > Predates git, does anyone remember the rationale?
> > > > 
> > > > ie:
> > > >         int test_bit(int nr, const volatile unsigned long *addr)
> > > 
> > > Because we sometimes pass volatile pointers to it and gcc will complain
> > > if you pass a volatile to a non volatile  (I think).
> > 
> > Where are these? I did git grep -W test_bit and looked for volatile,
> > couldn't find anything.
> 
> OK, so it was a bit of a guess.  Have you really checked the type of
> every address passed to every call of test_bit()?

Yea, I have this magic tool called gcc :)

Change
-static __always_inline int constant_test_bit(long nr, const volatile unsigned long *addr)
+static __always_inline int constant_test_bit(long nr, const unsigned long *addr)

and watch for new warnings.

I didn't see any.

> Second guess:  we wanted to make the test_bit access volatile (as opposed
> to the datatypes of the objects being tested) so that things like
> 
> 	while (testbit(bit, addr)) {
> 		do_very_little();
> 	}
> 
> don't get over optimised (since we are operating in a very threaded
> environment that the compiler not might expect).
> 
> -- 
> Cheers,
> Stephen Rothwell                    sfr@xxxxxxxxxxxxxxxx


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