Re: Using POPCNT and other advanced bit manipulation instructions

From: Andres Freund <andres(at)anarazel(dot)de>
To: Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us>
Cc: Alvaro Herrera <alvherre(at)2ndquadrant(dot)com>, Thomas Munro <thomas(dot)munro(at)enterprisedb(dot)com>, Andrew Gierth <andrew(at)tao11(dot)riddles(dot)org(dot)uk>, David Rowley <david(dot)rowley(at)2ndquadrant(dot)com>, PostgreSQL Hackers <pgsql-hackers(at)lists(dot)postgresql(dot)org>
Subject: Re: Using POPCNT and other advanced bit manipulation instructions
Date: 2019-02-15 16:55:13
Message-ID: 20190215165513.64ptbtt3cn3ezfxb@alap3.anarazel.de
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Hi,

On 2019-02-14 16:45:38 -0500, Tom Lane wrote:
> Andres Freund <andres(at)anarazel(dot)de> writes:
> > On 2019-02-14 15:47:13 -0300, Alvaro Herrera wrote:
> >> Hah, I just realized you have to add -mlzcnt in order for these builtins
> >> to use the lzcnt instructions. It goes from something like
> >>
> >> bsrq %rax, %rax
> >> xorq $63, %rax
>
> > I'm confused how this is a general count leading zero operation? Did you
> > use constants or something that allowed ot infer a range in the test? If
> > so the compiler probably did some optimizations allowing it to do the
> > above.
>
> No. If you compile
>
> int myclz(unsigned long long x)
> {
> return __builtin_clzll(x);
> }
>
> at -O2, on just about any x86_64 gcc, you will get
>
> myclz:
> .LFB1:
> .cfi_startproc
> bsrq %rdi, %rax
> xorq $63, %rax
> ret
> .cfi_endproc

Yea, sorry for the noise. I misremembered the bsrq mnemonic.

bsr has a latency of three cycles, xor of one. lzcnt a latency of
three. So it's mildly faster to use lzcnt (it uses fewer ports, and has
a shorter latency). But I doubt we have code where that's noticable.

Greetings,

Andres Freund

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