Re: Skybuck presents Faster Div 7

"rio" wrote:

> you can try this

> ; uns32  div7(uns32   )
> div7:
> mov eax, 0x92492493
;> mov edx, [esp+4]
;> mul edx
mul [esp+4]        ;why not ?
> mov eax, edx
> shr eax, 2
> ret

Ok, you use 24542670027 which is (rounded up) 2**34/7
which then asks for the additional shr 2.

I now see the difference in the results when I use 2**32/7

24924924h * 7  = 0_FFFF_FFFCh  ;wrong
So yes, we forgot to round it up yet.

24924925h * 7  = 1_0000_0003h  ;ok
24924925h * ffff_ffffh = 29429425_00000000
                        -00000000_29429425
((2**32)-1)/7          = 29429424_d6bd6bdb  ;fits here too

now Robert's code should work as well:

;input in eax
MOV edx, 0x24924925  ;(2**32 /7)+1
MUL edx
;result in edx.eax

and it becomes a single instruction with a reciprocal_LUT:
;divisor in edx  (limited to table size)
;dividend in eax (unsigned dword)
MUL [Lut+edx*4]
;quotient in edx.eax

__
wolfgang



> "Skybuck Flying" <BloodyShame@[EMAIL PROTECTED]
> ha scritto nel messaggio
> news:6bdd5$48134d79$541983fa$31840@[EMAIL PROTECTED]
> > Nope still verification failure:
> >
> > // doesn't work. verification failure.
> > function AsmDiv7( Value : longword ) : longword;
> > asm
> > MOV  edx, 613566756     // or faster in a LUT
> > MUL  edx                // dividend in eax
> > XCHG edx, eax           // result in eax
> > end;
> >
> > You really should write a verification program.
> >
> > Try verifieing yourself for:
> >
> > AsmDiv( 7 );
> >
> > Correct output is ofcourse:
> >
> > 7 div 7 = 1
> >
> > Your method outputs zero which is incorrect ;)
> >
> > Please try again and then come again :)
> >
> > Other example:
> >
> > 70 div 7 = 10.
> >
> > Bye,
> >  Skybuck :)