Re: [musl] [PATCH] x86/memset: avoid performing final store twice

[Denys Vlasenko <dvlasenk@redhat.com>]

On 10/11/20 2:25 AM, Rich Felker wrote:
> On Sun, Oct 04, 2020 at 12:32:09AM +0200, Denys Vlasenko wrote:
>> From: Denys Vlasenko <dvlasenk@redhat.com>
>>
>> For not very short NBYTES case:
>>
>> To handle the tail alignment, the code performs a potentially
>> misaligned word store to fill the final 8 bytes of the buffer.
>> This is done even if the buffer's end is aligned.
>>
>> Eventually code fills the rest of the buffer, which is a multiple
>> of 8 bytes now, with NBYTES / 8 aligned word stores.
>>
>> However, this means that if NBYTES *was* divisible by 8,
>> we store last word too, again.
>>
>> This patch decrements byte count before dividing it by 8,
>> making one less store in "NBYTES is divisible by 8" case,
>> and not changing anything in all other cases.
>>
...
>> --- a/src/string/x86_64/memset.s
>> +++ b/src/string/x86_64/memset.s
>> @@ -53,7 +53,7 @@ memset:
>>   2:	test $15,%edi
>>   	mov %rdi,%r8
>>   	mov %rax,-8(%rdi,%rdx)
>> -	mov %rdx,%rcx
>> +	lea -1(%rdx),%rcx
>>   	jnz 2f
>>   
>>   1:	shr $3,%rcx
>> -- 
>> 2.25.0
> 
> Does this have measurably better performance on a system you've tested
> it on?

I did not test performance, I predict it will hardly be detectable.