Date: Mon, 16 Sep 2013 11:37:07 +0300
From: "Michael S. Tsirkin" <mst@redhat.com>
To: Rusty Russell <rusty@rustcorp.com.au>
Cc: torvalds@linux-foundation.org, LKML <linux-kernel@vger.kernel.org>
Subject: Re: Why does test_bit() take a volatile addr?
Message-ID: <20130916083707.GC815@redhat.com>
References: <87ioy11k8s.fsf@rustcorp.com.au>
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On Mon, Sep 16, 2013 at 01:38:35PM +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()?
> 
> Thanks,
> Rusty.

So looking at this some more, e.g. on x86 I see:

static inline int variable_test_bit(long nr, volatile const unsigned
long *addr)
{
        int oldbit;

        asm volatile("bt %2,%1\n\t"
                     "sbb %0,%0"
                     : "=r" (oldbit)
                     : "m" (*(unsigned long *)addr), "Ir" (nr));

        return oldbit;
}

and I have a vague memory that (at least for some old versions) gcc
would assume (*(unsigned long *)addr) only refers to addr[0].

OTOH constant_test_bit is
static __always_inline int constant_test_bit(long nr, const volatile unsigned long *addr)
{
        return ((1UL << (nr & (BITS_PER_LONG-1))) &
                (addr[nr >> _BITOPS_LONG_SHIFT])) != 0;
}

So there's a chance that we can drop volatile here.

I'll look at it some more.

-- 
MST
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