Re: [PATCH] scripts: coccinelle: check for !(un)?likely usage

[Denis Efremov <efremov@linux.com>]

On 8/28/19 4:05 PM, Rasmus Villemoes wrote:
> On 28/08/2019 14.33, Denis Efremov wrote:
>> On 8/28/19 2:33 PM, Rasmus Villemoes wrote:
>>> On 25/08/2019 21.19, Julia Lawall wrote:
>>>>
>>>>
>>>>> On 26 Aug 2019, at 02:59, Denis Efremov <efremov@linux.com> wrote:
>>>>>
>>>>>
>>>>>
>>>>>> On 25.08.2019 19:37, Joe Perches wrote:
>>>>>>> On Sun, 2019-08-25 at 16:05 +0300, Denis Efremov wrote:
>>>>>>> This patch adds coccinelle script for detecting !likely and !unlikely
>>>>>>> usage. It's better to use unlikely instead of !likely and vice versa.
>>>>>>
>>>>>> Please explain _why_ is it better in the changelog.
>>>>>>
>>>>>
>>>>> In my naive understanding the negation (!) before the likely/unlikely
>>>>> could confuse the compiler
>>>>
>>>> As a human I am confused. Is !likely(x) equivalent to x or !x?
>>>
>>> #undef likely
>>> #undef unlikely
>>> #define likely(x) (x)
>>> #define unlikely(x) (x)
>>>
>>> should be a semantic no-op. So changing !likely(x) to unlikely(x) is
>>> completely wrong. If anything, !likely(x) can be transformed to
>>> unlikely(!x).
>>
>> As far as I could understand it:
>>
>> # define likely(x)	__builtin_expect(!!(x), 1)
>> # define unlikely(x)	__builtin_expect(!!(x), 0)
>>
>> From GCC doc:
>> __builtin_expect compares the values. The semantics of the built-in are that it is expected that exp == c.
> 
> When I said "semantic" I meant from the C language point of view. Yes,
> of course, the whole reason for having these is that we can give hints
> to gcc as to which branch is more likely. Replace the dummy defines by
> #define likely(x) (!!(x)) if you like - it amounts to the same thing
> when it's only ever used in a boolean context.
> 
>> if (!likely(cond))
>> if (!__builtin_expect(!!(cond), 1))
>> if (!((!!(cond)) == 1))
> 
> You're inventing this comparison to 1. It should be "if (!(!!(cond)))",
> but it ends up being equivalent in C.
> 
>> if ((!!(cond)) != 1) and since !! could result in 0 or 1
>> if ((!!(cond)) == 0)
> 
> which in turn is equivalent to !(cond).
> 
>>
>> if (unlikely(cond))
>> if (__builtin_expect(!!(cond), 0))
>> if ((!!(cond)) == 0))
> 
> No, that last transformation is wrong. Yes, the _expectation_ is that
> !!(cond) has the value 0, but that does not mean that the whole
> condition turns into "does !!(cond) compare equal to 0?" - we _expect_
> that it does, meaning that we expect not to enter the if block. Read the
> docs, the value of __builtin_expect(whatever, foobar) is whatever, so a
> correct third line above would be
> 
>   "if (!!(cond))"
> 
> which is of course not at all the same as
> 
>   "if (!!(cond) == 0)" aka "if (!(cond))"

I get it, you are right. Thank you for the explanation.

Denis