Messages in this thread | | | Date | Sat, 27 Apr 2024 08:51:04 +0200 | Subject | Re: [PATCH v5] mm/rmap: do not add fully unmapped large folio to deferred split list | From | David Hildenbrand <> |
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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
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