From: David Hildenbrand <david@redhat.com>
To: Lance Yang <ioworker0@gmail.com>
Cc: Zi Yan <ziy@nvidia.com>,
Andrew Morton <akpm@linux-foundation.org>,
linux-mm@kvack.org,
"Matthew Wilcox (Oracle)" <willy@infradead.org>,
Yang Shi <shy828301@gmail.com>,
Ryan Roberts <ryan.roberts@arm.com>,
Barry Song <21cnbao@gmail.com>,
linux-kernel@vger.kernel.org
Subject: Re: [PATCH v5] mm/rmap: do not add fully unmapped large folio to deferred split list
Date: Sat, 27 Apr 2024 08:51:04 +0200 [thread overview]
Message-ID: <436316d3-7fee-4843-93f3-1c07860566c8@redhat.com> (raw)
In-Reply-To: <CAK1f24=PnZ-Q=TrG9+SKWhwNHF_=EbVBWVcexqsEU-dctrAOFA@mail.gmail.com>
On 27.04.24 06:06, Lance Yang wrote:
> On Sat, Apr 27, 2024 at 4:16 AM David Hildenbrand <david@redhat.com> wrote:
>>
>> On 26.04.24 21:20, Zi Yan wrote:
>>> On 26 Apr 2024, at 15:08, David Hildenbrand wrote:
> [...]
>>>>> + bool partially_mapped = false;
> [...]
>>>>> +
>>>>> + partially_mapped = !!nr && !!atomic_read(mapped);
>>>>
>>>> Nit: The && should remove the need for both !!.
>>>
>>> My impression was that !! is needed to convert from int to bool and I do
>>> find "!!int && !!int" use in the kernel.
>>
>> I might be wrong about this, but if you wouldn't write
>
> I think you're correct.
>
>>
>> if (!!nr && !!atomic_read(mapped))
>>
>> then
>>
>> bool partially_mapped = nr && atomic_read(mapped);
>>
>> is sufficient.
>
> +1
>
>>
>> && would make sure that the result is either 0 or 1, which
>> you can store safely in a bool, no matter which underlying type
>> is used to store that value.
>>
>> But I *think* nowdays, the compiler will always handle that
>> correctly, even without the "&&" (ever since C99 added _Bool).
>>
>> Likely, also
>>
>> bool partially_mapped = nr & atomic_read(mapped);
>>
>> Would nowadays work, but looks stupid.
>>
>>
>> Related: https://lkml.org/lkml/2013/8/31/138
>>
>> ---
>> #include <stdio.h>
>> #include <stdbool.h>
>> #include <stdint.h>
>> #include <inttypes.h>
>>
>> volatile uint64_t a = 0x8000000000000000ull;
>>
>> void main (void) {
>> printf("uint64_t a = a: 0x%" PRIx64 "\n", a);
>>
>> int i1 = a;
>> printf("int i1 = a: %d\n", i1);
>>
>> int i2 = !!a;
>> printf("int i2 = !!a: %d\n", i2);
>>
>> bool b1 = a;
>> printf("bool b1 = a: %d\n", b1);
>>
>> bool b2 = !!a;
>> printf("bool b2 = !!a: %d\n", b2);
>> }
>> ---
>> $ ./test
>> uint64_t a = a: 0x8000000000000000
>> int i1 = a: 0
>> int i2 = !!a: 1
>> bool b1 = a: 1
>> bool b2 = !!a: 1
>> ---
>>
>> Note that if bool would be defined as "int", you would need the !!, otherwise you
>> would lose information.
>
> Agreed. We need to be careful in this case.
>
>>
>> But even for b1, the gcc generates now:
>>
>> 40118c: 48 8b 05 7d 2e 00 00 mov 0x2e7d(%rip),%rax # 404010 <a>
>> 401193: 48 85 c0 test %rax,%rax
>> 401196: 0f 95 c0 setne %al
>>
>>
>> My stdbool.h contains
>>
>> #define bool _Bool
>>
>> And I think C99 added _Bool that makes that work.
>>
>> But I didn't read the standard, and it's time for the weekend :)
>
> I just read the C99 and found some interesting information as follows:
>
> 6.3.1.2 Boolean type
> When any *scalar value* is converted to _Bool, the result is 0 if the
> value compares equal to 0; otherwise, the result is 1.
>
> 6.2.5 Types
> 21. Arithmetic types and pointer types are collectively called *scalar
> types*. Array and structure types are collectively called aggregate types.
>
> 6.5.13 Logical AND operator
> Semantics
> The && operator shall yield 1 if both of its operands compare unequal to
> 0; otherwise, it yields 0. The result has type int.
>
> 6.5.10 Bitwise AND operator
> Constraints
> Each of the operands shall have integer type.
> Semantics
> The result of the binary & operator is the bitwise AND of the operands
> (that is, each bit in the result is set if and only if each of the
> corresponding
> bits in the converted operands is set).
>
> && would ensure that the result is either 0 or 1, as David said, so no worries.
>
My example was flawed: I wanted to express that "if any bit is set, the
bool value will be 1. That works for "| vs ||" but not for "& vs &&",
obviously :)
> We defined partially_mapped as a bool(_Bool). IIUC, "partially_mapped
> = int & int;"
> would work correctly as well. However, "partially_mapped = long &
> int;" might not.
Implicit type conversion would convert "long & int" to "long & long"
first, which should work just fine. I think really most concerns
regarding the bool type are due to < C99 not supporting _Bool.
Great weekend everybody!
--
Cheers,
David / dhildenb
next prev parent reply other threads:[~2024-04-27 6:51 UTC|newest]
Thread overview: 15+ messages / expand[flat|nested] mbox.gz Atom feed top
2024-04-26 19:02 [PATCH v5] mm/rmap: do not add fully unmapped large folio to deferred split list Zi Yan
2024-04-26 19:08 ` David Hildenbrand
2024-04-26 19:20 ` Zi Yan
2024-04-26 20:15 ` David Hildenbrand
2024-04-26 20:22 ` Zi Yan
2024-04-27 4:06 ` Lance Yang
2024-04-27 6:51 ` David Hildenbrand [this message]
2024-04-27 9:32 ` Barry Song
2024-04-26 20:42 ` Yang Shi
2024-04-27 4:09 ` Lance Yang
2024-05-01 13:24 ` Alexander Gordeev
2024-05-01 13:38 ` Zi Yan
2024-05-01 15:54 ` David Hildenbrand
2024-05-02 13:18 ` Alexander Gordeev
2024-05-02 13:20 ` Zi Yan
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