Date: Mon, 10 Feb 2020 19:38:38 +0300
On 2020-02-10 19:24, Thiago Macieira via Std-Discussion wrote:
> On Monday, 10 February 2020 00:44:42 PST Andrey Semashev via Std-Discussion
> wrote:
>> Yes, x86 is one of such architectures, as is any CAS-based architectures
>> I'm aware of. The example is mentioned in the standard:
>> atomic_ref<complex<double>>. Assuming double is 64-bit and
>> complex<double> is a 128-bit structure, it must be aligned to 16 bytes
>> for atomic CAS(*), but alignof(complex<double>) is 8.
> [snip]
>> But given the definition of required_alignment in the standard, this
>> code does not guarantee that ref.is_lock_free() will be true. An
>> implementation where required_alignment is 8 and ref is implemented with
>> a lock pool is allowed by the standard. Even if such atomic_ref
>> implementation supports lock-free operations starting at alignment of 16.
>
> No, it doesn't guarantee lock-free, it just guarantees atomicity. The need to
> use a lock is usually related to size, not alignment. x86-64 will do a 16-byte
> CAS lock-free, i386 will not, regardless of alignment.
No, CMPXCHG16B requires 16-byte alignment. See CMPXCHG16B description in
Intel 64 and IA-32 Architectures Software Development Manual.
> On Monday, 10 February 2020 00:44:42 PST Andrey Semashev via Std-Discussion
> wrote:
>> Yes, x86 is one of such architectures, as is any CAS-based architectures
>> I'm aware of. The example is mentioned in the standard:
>> atomic_ref<complex<double>>. Assuming double is 64-bit and
>> complex<double> is a 128-bit structure, it must be aligned to 16 bytes
>> for atomic CAS(*), but alignof(complex<double>) is 8.
> [snip]
>> But given the definition of required_alignment in the standard, this
>> code does not guarantee that ref.is_lock_free() will be true. An
>> implementation where required_alignment is 8 and ref is implemented with
>> a lock pool is allowed by the standard. Even if such atomic_ref
>> implementation supports lock-free operations starting at alignment of 16.
>
> No, it doesn't guarantee lock-free, it just guarantees atomicity. The need to
> use a lock is usually related to size, not alignment. x86-64 will do a 16-byte
> CAS lock-free, i386 will not, regardless of alignment.
No, CMPXCHG16B requires 16-byte alignment. See CMPXCHG16B description in
Intel 64 and IA-32 Architectures Software Development Manual.
Received on 2020-02-10 10:41:20