SiemarglНам нужно деление с остатком 64бит на 32бит.
Оно реализуется 32-битными операциями, например так div_s64_rem()
а SIMD каким боком тогда?
вариация на тему _ui64toa получается
что касается реализации деления в msvc для x86 -
деление - _aulldvrm (msvc runtime, ветка для короткого делителя нас интересует)
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title ulldvrm - unsigned long divide and remainder routine
;***
;ulldvrm.asm - unsigned long divide and remainder routine
;
; Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
; defines the unsigned long divide and remainder routine
; __aulldvrm
;
;*******************************************************************************
.xlist
include vcruntime.inc
include mm.inc
.list
;***
;ulldvrm - unsigned long divide and remainder
;
;Purpose:
; Does a unsigned long divide and remainder of the arguments. Arguments
; are not changed.
;
;Entry:
; Arguments are passed on the stack:
; 1st pushed: divisor (QWORD)
; 2nd pushed: dividend (QWORD)
;
;Exit:
; EDX:EAX contains the quotient (dividend/divisor)
; EBX:ECX contains the remainder (divided % divisor)
; NOTE: this routine removes the parameters from the stack.
;
;Uses:
; ECX
;
;Exceptions:
;
;*******************************************************************************
CODESEG
_aulldvrm PROC NEAR
.FPO (1, 4, 0, 0, 0, 0)
push esi
; Set up the local stack and save the index registers. When this is done
; the stack frame will look as follows (assuming that the expression a/b will
; generate a call to aulldvrm(a, b)):
;
; -----------------
; | |
; |---------------|
; | |
; |--divisor (b)--|
; | |
; |---------------|
; | |
; |--dividend (a)-|
; | |
; |---------------|
; | return addr** |
; |---------------|
; ESP---->| ESI |
; -----------------
;
DVND equ [esp + 8] ; stack address of dividend (a)
DVSR equ [esp + 16] ; stack address of divisor (b)
;
; Now do the divide. First look to see if the divisor is less than 4194304K.
; If so, then we can use a simple algorithm with word divides, otherwise
; things get a little more complex.
;
mov eax,HIWORD(DVSR) ; check to see if divisor < 4194304K
or eax,eax
jnz short L1 ; nope, gotta do this the hard way
mov ecx,LOWORD(DVSR) ; load divisor
mov eax,HIWORD(DVND) ; load high word of dividend
xor edx,edx
div ecx ; get high order bits of quotient
mov ebx,eax ; save high bits of quotient
mov eax,LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
div ecx ; get low order bits of quotient
mov esi,eax ; ebx:esi <- quotient
;
; Now we need to do a multiply so that we can compute the remainder.
;
mov eax,ebx ; set up high word of quotient
mul dword ptr LOWORD(DVSR) ; HIWORD(QUOT) * DVSR
mov ecx,eax ; save the result in ecx
mov eax,esi ; set up low word of quotient
mul dword ptr LOWORD(DVSR) ; LOWORD(QUOT) * DVSR
add edx,ecx ; EDX:EAX = QUOT * DVSR
jmp short L2 ; complete remainder calculation
;
; Here we do it the hard way. Remember, eax contains DVSRHI
;
L1:
mov ecx,eax ; ecx:ebx <- divisor
mov ebx,LOWORD(DVSR)
mov edx,HIWORD(DVND) ; edx:eax <- dividend
mov eax,LOWORD(DVND)
L3:
shr ecx,1 ; shift divisor right one bit; hi bit <- 0
rcr ebx,1
shr edx,1 ; shift dividend right one bit; hi bit <- 0
rcr eax,1
or ecx,ecx
jnz short L3 ; loop until divisor < 4194304K
div ebx ; now divide, ignore remainder
mov esi,eax ; save quotient
;
; We may be off by one, so to check, we will multiply the quotient
; by the divisor and check the result against the orignal dividend
; Note that we must also check for overflow, which can occur if the
; dividend is close to 2**64 and the quotient is off by 1.
;
mul dword ptr HIWORD(DVSR) ; QUOT * HIWORD(DVSR)
mov ecx,eax
mov eax,LOWORD(DVSR)
mul esi ; QUOT * LOWORD(DVSR)
add edx,ecx ; EDX:EAX = QUOT * DVSR
jc short L4 ; carry means Quotient is off by 1
;
; do long compare here between original dividend and the result of the
; multiply in edx:eax. If original is larger or equal, we are ok, otherwise
; subtract one (1) from the quotient.
;
cmp edx,HIWORD(DVND) ; compare hi words of result and original
ja short L4 ; if result > original, do subtract
jb short L5 ; if result < original, we are ok
cmp eax,LOWORD(DVND) ; hi words are equal, compare lo words
jbe short L5 ; if less or equal we are ok, else subtract
L4:
dec esi ; subtract 1 from quotient
sub eax,LOWORD(DVSR) ; subtract divisor from result
sbb edx,HIWORD(DVSR)
L5:
xor ebx,ebx ; ebx:esi <- quotient
L2:
;
; Calculate remainder by subtracting the result from the original dividend.
; Since the result is already in a register, we will do the subtract in the
; opposite direction and negate the result.
;
sub eax,LOWORD(DVND) ; subtract dividend from result
sbb edx,HIWORD(DVND)
neg edx ; otherwise, negate the result
neg eax
sbb edx,0
;
; Now we need to get the quotient into edx:eax and the remainder into ebx:ecx.
;
mov ecx,edx
mov edx,ebx
mov ebx,ecx
mov ecx,eax
mov eax,esi
;
; Just the cleanup left to do. edx:eax contains the quotient.
; Restore the saved registers and return.
;
pop esi
ret 16
_aulldvrm ENDP
end
и _aulldiv (остаток восстанавливать в вызывающем коде)
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title ulldiv - unsigned long divide routine
;***
;ulldiv.asm - unsigned long divide routine
;
; Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
; defines the unsigned long divide routine
; __aulldiv
;
;*******************************************************************************
.xlist
include vcruntime.inc
include mm.inc
.list
;***
;ulldiv - unsigned long divide
;
;Purpose:
; Does a unsigned long divide of the arguments. Arguments are
; not changed.
;
;Entry:
; Arguments are passed on the stack:
; 1st pushed: divisor (QWORD)
; 2nd pushed: dividend (QWORD)
;
;Exit:
; EDX:EAX contains the quotient (dividend/divisor)
; NOTE: this routine removes the parameters from the stack.
;
;Uses:
; ECX
;
;Exceptions:
;
;*******************************************************************************
CODESEG
_aulldiv PROC NEAR
.FPO (2, 4, 0, 0, 0, 0)
push ebx
push esi
; Set up the local stack and save the index registers. When this is done
; the stack frame will look as follows (assuming that the expression a/b will
; generate a call to uldiv(a, b)):
;
; -----------------
; | |
; |---------------|
; | |
; |--divisor (b)--|
; | |
; |---------------|
; | |
; |--dividend (a)-|
; | |
; |---------------|
; | return addr** |
; |---------------|
; | EBX |
; |---------------|
; ESP---->| ESI |
; -----------------
;
DVND equ [esp + 12] ; stack address of dividend (a)
DVSR equ [esp + 20] ; stack address of divisor (b)
;
; Now do the divide. First look to see if the divisor is less than 4194304K.
; If so, then we can use a simple algorithm with word divides, otherwise
; things get a little more complex.
;
mov eax,HIWORD(DVSR) ; check to see if divisor < 4194304K
or eax,eax
jnz short L1 ; nope, gotta do this the hard way
mov ecx,LOWORD(DVSR) ; load divisor
mov eax,HIWORD(DVND) ; load high word of dividend
xor edx,edx
div ecx ; get high order bits of quotient
mov ebx,eax ; save high bits of quotient
mov eax,LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
div ecx ; get low order bits of quotient
mov edx,ebx ; edx:eax <- quotient hi:quotient lo
jmp short L2 ; restore stack and return
;
; Here we do it the hard way. Remember, eax contains DVSRHI
;
L1:
mov ecx,eax ; ecx:ebx <- divisor
mov ebx,LOWORD(DVSR)
mov edx,HIWORD(DVND) ; edx:eax <- dividend
mov eax,LOWORD(DVND)
L3:
shr ecx,1 ; shift divisor right one bit; hi bit <- 0
rcr ebx,1
shr edx,1 ; shift dividend right one bit; hi bit <- 0
rcr eax,1
or ecx,ecx
jnz short L3 ; loop until divisor < 4194304K
div ebx ; now divide, ignore remainder
mov esi,eax ; save quotient
;
; We may be off by one, so to check, we will multiply the quotient
; by the divisor and check the result against the orignal dividend
; Note that we must also check for overflow, which can occur if the
; dividend is close to 2**64 and the quotient is off by 1.
;
mul dword ptr HIWORD(DVSR) ; QUOT * HIWORD(DVSR)
mov ecx,eax
mov eax,LOWORD(DVSR)
mul esi ; QUOT * LOWORD(DVSR)
add edx,ecx ; EDX:EAX = QUOT * DVSR
jc short L4 ; carry means Quotient is off by 1
;
; do long compare here between original dividend and the result of the
; multiply in edx:eax. If original is larger or equal, we are ok, otherwise
; subtract one (1) from the quotient.
;
cmp edx,HIWORD(DVND) ; compare hi words of result and original
ja short L4 ; if result > original, do subtract
jb short L5 ; if result < original, we are ok
cmp eax,LOWORD(DVND) ; hi words are equal, compare lo words
jbe short L5 ; if less or equal we are ok, else subtract
L4:
dec esi ; subtract 1 from quotient
L5:
xor edx,edx ; edx:eax <- quotient
mov eax,esi
;
; Just the cleanup left to do. edx:eax contains the quotient.
; Restore the saved registers and return.
;
L2:
pop esi
pop ebx
ret 16
_aulldiv ENDP
end
беда в том,что самая затратная операция в алгориме - деление |
