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From: Allan Sandfeld Jensen <linux@carewolf.com>
To: gcc@gcc.gnu.org
Cc: Stefan Kanthak <stefan.kanthak@nexgo.de>
Subject: Re: Peephole optimisation: isWhitespace()
Date: Mon, 17 Aug 2020 16:31:35 +0200
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On Freitag, 14. August 2020 18:43:12 CEST Stefan Kanthak wrote:
> Hi @ll,
> 
> in his ACM queue article <https://queue.acm.org/detail.cfm?id=3372264>,
> Matt Godbolt used the function
> 
> | bool isWhitespace(char c)
> | {
> | 
> |     return c == ' '
> |     
> |       || c == '\r'
> |       || c == '\n'
> |       || c == '\t';
> | 
> | }
> 
> as an example, for which GCC 9.1 emits the following assembly for AMD64
> 
> processors (see <https://godbolt.org/z/acm19_conds>):
> |    xor    eax, eax              ; result = false
> |    cmp    dil, 32               ; is c > 32
> |    ja     .L4                   ; if so, exit with false
> |    movabs rax, 4294977024       ; rax = 0x100002600
> |    shrx   rax, rax, rdi         ; rax >>= c
> |    and    eax, 1                ; result = rax & 1
> |
> |.L4:
> |    ret
> 
No it doesn't. As your example shows if you took the time to read it, it is 
what gcc emit when generating code to run on a _haswell_ architecture. If you 
remove -march=haswell from the command line you get:

        xor     eax, eax
        cmp     dil, 32
        ja      .L1
        movabs  rax, 4294977024
        mov     ecx, edi
        shr     rax, cl
        and     eax, 1

It uses one mov more, but no shrx. 

'Allan