Messages in this thread | | | From | David Laight <> | Subject | RE: how many memset(,0,) calls in kernel ? | Date | Tue, 14 Sep 2021 08:23:40 +0000 |
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From: Willy Tarreau > Sent: 13 September 2021 17:10 > > On Mon, Sep 13, 2021 at 04:03:09PM +0000, David Laight wrote: > > > 36: b9 06 00 00 00 mov $0x6,%ecx > > > 3b: 4c 89 e7 mov %r12,%rdi > > > 3e: f3 ab rep stos %eax,%es:(%rdi) > > > > > > The last line does exactly "memset(%rdi, %eax, %ecx)". Just two bytes > > > for some code that modern processors are even able to optimize. > > > > Hmmm I'd bet that 6 stores will be faster on ~everything. > > 'modern' processors do better than some older ones [1], but 6 > > writes isn't enough to get into the really fast paths. > > So you'll still take a few cycles of setup. > > The exact point is, here it's up to the compiler to decide thanks to > its builtin what it considers best for the target CPU. It already > knows the fixed size and the code is emitted accordingly. It may > very well be a call to the memset() function when the size is large > and a power of two because it knows alternate variants are available > for example. > > The compiler might even decide to shrink that area if other bytes > are written just after the memset(), leaving only holes touched by > memset().
You might think the compiler will make sane choices for the target CPU. But it often makes a complete pig's breakfast of it. I'm pretty sure 6 'rep stos' is slower than 6 write an absolutely everything - with the possible exception of an 8088.
By far the worst ones are when the compiler decides to pessimise a loop by using the simd (eg avx512) instructions to do 4 (or 8) loop iterations in one pass. It might be fine if the loop count is in the 100s - but not when it is 3.
One compiler I've used nicely converted any byte copy loop into a 'rep movsb' instruction. That was contemporary with P4 netburst - where it was terribly slow.
David
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