Re: [PATCH 3/3] bitmap: Use memcmp optimisation in more situations

[Andy Shevchenko <andy.shevchenko@gmail.com>]

On Thu, Jun 8, 2017 at 4:43 PM, Rasmus Villemoes
<linux@rasmusvillemoes.dk> wrote:
> On 8 June 2017 at 14:31, Andy Shevchenko <andy.shevchenko@gmail.com> wrote:
>> On Thu, Jun 8, 2017 at 5:55 AM, Matthew Wilcox <willy@infradead.org> wrote:
>>> We only need to know if the bottom 3 bits are 0 to apply this optimisation.
>>> For example, if we have a user which does this:
>>>
>>>         nbits = 8;
>>>         if (argle)
>>>                 nbits += 8;
>>>         if (bitmap_equal(ptr1, ptr2, nbits))
>>>                 blah();
>>>
>>> then we can use memcmp() because gcc can deduce that the bottom 3 bits
>>> are never set (try it!  it works!).  We don't need nbits as a whole to
>>> be const.
>>
>> What I'm talking about is that by my opinion the both below are equivalent.
>> __builtin_constant_p(nbits)
>> __builtin_constant_p(nbits & 7)
>
> They are not. Read Matthew's example again. Assuming that argle is
> something non-constant (maybe an argument to the function), the value
> of nbits at the time of the bitmap_equal call is _not_ a
> compile-time-constant. However, if the compiler is smart (which at
> least some versions of gcc are), the compiler may deduce that nbits is
> either 8 or 16; there really are no other options. Hence it _is_
> statically known that nbits is divisible by 8, so the expression
> nbits&7 _is_ compile-time constant (0), so gcc can change the
> bitmap_equal call to a memcmp call.

Yeah, thanks for detailed explanation.
So, basically what we do, we consider
1. 3 LSBs _is_ constant, *and*
2. They are equal to 0.

> (It may then either pass a run-time value of nbits>>3 and emit a
> single memcmp call, or it may decide to unroll the two options,
> creating two memcmp calls with 1 and 2 as compile-time arguments;
> these may or may not then in turn be "inlined" to code doing roughly
> *(u8*)p1 == *(u8*)p2 and similarly for u16 casts).

-- 
With Best Regards,
Andy Shevchenko