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Subject: Re: Possible small improvement to hacky method

Author: Frank Phillips

Date: 03:33:00 12/07/02

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On December 06, 2002 at 00:49:11, Walter Faxon wrote:

>On December 02, 2002 at 20:27:39, Gerd Isenberg wrote:
>
>>Ahh.. i see,
>>
>>a special kind of popcount for consecutive trailing "ones".
>>I love this interactive programming lessions.
>>
>>Gerd
>>
>>Btw.: Impressive assembler listing:
>>
>>00401003 8B 4C 24 0C          mov         ecx,dword ptr [esp+0Ch]
>>00401007 8B 31                mov         esi,dword ptr [ecx]
>>00401009 8B 79 04             mov         edi,dword ptr [ecx+4]
>>0040100C 8B D6                mov         edx,esi
>>0040100E 8B C7                mov         eax,edi
>>00401010 83 C2 FF             add         edx,0FFFFFFFFh
>>00401013 83 D0 FF             adc         eax,0FFFFFFFFh
>>00401016 23 F2                and         esi,edx
>>00401018 23 F8                and         edi,eax
>>0040101A 89 31                mov         dword ptr [ecx],esi
>>0040101C 89 79 04             mov         dword ptr [ecx+4],edi
>>0040101F 8B CF                mov         ecx,edi
>>00401021 33 D6                xor         edx,esi
>>00401023 33 C1                xor         eax,ecx
>>00401025 33 C2                xor         eax,edx
>>00401027 5F                   pop         edi
>>00401028 35 81 FC C5 01       xor         eax,1C5FC81h
>>0040102D 5E                   pop         esi
>>0040102E 8B D0                mov         edx,eax
>>00401030 C1 EA 10             shr         edx,10h
>>00401033 03 C2                add         eax,edx
>>00401035 33 D2                xor         edx,edx
>>00401037 8B C8                mov         ecx,eax
>>00401039 C1 E9 08             shr         ecx,8
>>0040103C 2A C1                sub         al,cl
>>0040103E 25 FF 00 00 00       and         eax,0FFh
>>00401043 8A 90 D5 62 40 00    mov         dl,byte ptr [eax+4062D5h]
>>00401049 8B C2                mov         eax,edx
>>0040104B C3                   ret
>
>
>Hi, Gerd.
>
>I have noticed that most compilers aren't too smart where it comes to bitwise
>dataflow analysis.  After the "add  eax, edx" instruction above, better (well,
>shorter!) code would be:
>
>    sub    al, ah
>    and    eax, 0FFh
>    mov    al, byte ptr [eax+4062D5h]
>    ret
>
>Perhaps if we added a little "hint" in the last 2 lines of source:
>
>inline                  // inline declaration may differ by compiler
>u8 LSB_64( u64* bb )
>    {
>    u64 t64;
>    u32 t32;
>    u8 t8;
>    t64 = *bb - 1;
>    *bb &= t64;         // omit this line to retain current LSB
>    t64 ^= *bb;
>    t32 = (u32)t64 ^ (u32)(t64 >> 32);
>    t32 ^= LSB_64_magic;
>    t32 += t32 >> 16;
>    t8 = (u8)t32 - (u8)(t32 >> 8);
>    return LSB_64_table [LSB_64_adj + t8];
>    }
>
>Of course, this really only helps because we already know that the x86 class can
>independently address both the first and 2nd bytes of a register.  And it's
>wishful thinking that much better asm code will actually be produced!  I'd be
>interested to see.
>
>If/when you run your test suite again, please try this version.  Thanks.
>
>-- Walter

Has anybody posted Walter's method with the table for 0 == highest order bit?

Frank



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