linux-kernel - Re: [PATCH v5] mm/rmap: do not add fully unmapped large folio to deferred split list

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Date: Sat, 27 Apr 2024 08:51:04 +0200
From: David Hildenbrand <david@...hat.com>
To: Lance Yang <ioworker0@...il.com>
Cc: Zi Yan <ziy@...dia.com>, Andrew Morton <akpm@...ux-foundation.org>,
 linux-mm@...ck.org, "Matthew Wilcox (Oracle)" <willy@...radead.org>,
 Yang Shi <shy828301@...il.com>, Ryan Roberts <ryan.roberts@....com>,
 Barry Song <21cnbao@...il.com>, linux-kernel@...r.kernel.org
Subject: Re: [PATCH v5] mm/rmap: do not add fully unmapped large folio to
 deferred split list

On 27.04.24 06:06, Lance Yang wrote:
> On Sat, Apr 27, 2024 at 4:16 AM David Hildenbrand <david@...hat.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