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

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Message-ID: <75f5210f-43a3-ab0d-912a-6dff6163fd9a@rasmusvillemoes.dk>
Date:   Wed, 28 Aug 2019 15:05:08 +0200
From:   Rasmus Villemoes <linux@...musvillemoes.dk>
To:     efremov@...ux.com, Julia Lawall <julia.lawall@...6.fr>
Cc:     Joe Perches <joe@...ches.com>, linux-kernel@...r.kernel.org,
        Gilles Muller <Gilles.Muller@...6.fr>,
        Nicolas Palix <nicolas.palix@...g.fr>,
        Michal Marek <michal.lkml@...kovi.net>
Subject: Re: [PATCH] scripts: coccinelle: check for !(un)?likely usage

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@...ux.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))"

Rasmus