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

[Oliver Neukum <oliver@neukum.org>]

On Mon, 2013-09-16 at 11:44 +0300, Michael S. Tsirkin wrote:
> On Mon, Sep 16, 2013 at 10:40:00AM +0200, Oliver Neukum wrote:
> > On Mon, 2013-09-16 at 13:38 +0930, Rusty Russell wrote:
> > > Predates git, does anyone remember the rationale?
> > > 
> > > ie:
> > >         int test_bit(int nr, const volatile unsigned long *addr)
> > > 
> > > I noticed because gcc failed to elimiate some code in a patch I was
> > > playing with.
> > > 
> > > I'm nervous about subtle bugs involved in ripping it out, even if noone
> > > knows why.  Should I add __test_bit()?
> > 
> > It seems to me that if you do
> > 
> > b = *ptr & 0xf;
> > c = b << 2;
> > if (test_bit(1, ptr))
> > 
> > the compiler could optimize away the memory access on ptr without
> > the volatile. We'd have to add a lot of mb().
> > 
> > 	Regards
> > 		Oliver
> 
> What is this code supposed to do?
> Any specific examples?
> 

Often you see

while (test_bit(...) && condition) ... ;

If the compiler can show that you don't change the memory you
do the test_bit on, it can change this to:

if (test_bit(...))
	while (condition) ...;

That must not happen.

	Regards
		Oliver