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|
;***
;lldiv.asm - signed long divide routine
;
; Copyright (c) Microsoft Corporation. All rights reserved.
; SPDX-License-Identifier: BSD-2-Clause-Patent
;
;Purpose:
; defines the signed long divide routine
; __alldiv
;
;Original Implemenation: MSVC 14.29.30133
;
;*******************************************************************************
.686
.model flat,C
.code
;***
;lldiv - signed long divide
;
;Purpose:
; Does a signed 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:
;
;*******************************************************************************
_alldiv PROC NEAR
HIWORD EQU [4] ;
LOWORD EQU [0]
push edi
push esi
push ebx
; 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 lldiv(a, b)):
;
; -----------------
; | |
; |---------------|
; | |
; |--divisor (b)--|
; | |
; |---------------|
; | |
; |--dividend (a)-|
; | |
; |---------------|
; | return addr** |
; |---------------|
; | EDI |
; |---------------|
; | ESI |
; |---------------|
; ESP---->| EBX |
; -----------------
;
DVND equ [esp + 16] ; stack address of dividend (a)
DVSR equ [esp + 24] ; stack address of divisor (b)
; Determine sign of the result (edi = 0 if result is positive, non-zero
; otherwise) and make operands positive.
xor edi,edi ; result sign assumed positive
mov eax,HIWORD(DVND) ; hi word of a
or eax,eax ; test to see if signed
jge short L1 ; skip rest if a is already positive
inc edi ; complement result sign flag
mov edx,LOWORD(DVND) ; lo word of a
neg eax ; make a positive
neg edx
sbb eax,0
mov HIWORD(DVND),eax ; save positive value
mov LOWORD(DVND),edx
L1:
mov eax,HIWORD(DVSR) ; hi word of b
or eax,eax ; test to see if signed
jge short L2 ; skip rest if b is already positive
inc edi ; complement the result sign flag
mov edx,LOWORD(DVSR) ; lo word of a
neg eax ; make b positive
neg edx
sbb eax,0
mov HIWORD(DVSR),eax ; save positive value
mov LOWORD(DVSR),edx
L2:
;
; 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.
;
; NOTE - eax currently contains the high order word of DVSR
;
or eax,eax ; check to see if divisor < 4194304K
jnz short L3 ; 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 ; eax <- 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 ; eax <- low order bits of quotient
mov edx,ebx ; edx:eax <- quotient
jmp short L4 ; set sign, restore stack and return
;
; Here we do it the hard way. Remember, eax contains the high word of DVSR
;
L3:
mov ebx,eax ; ebx:ecx <- divisor
mov ecx,LOWORD(DVSR)
mov edx,HIWORD(DVND) ; edx:eax <- dividend
mov eax,LOWORD(DVND)
L5:
shr ebx,1 ; shift divisor right one bit
rcr ecx,1
shr edx,1 ; shift dividend right one bit
rcr eax,1
or ebx,ebx
jnz short L5 ; loop until divisor < 4194304K
div ecx ; 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 L6 ; 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 L6 ; if result > original, do subtract
jb short L7 ; if result < original, we are ok
cmp eax,LOWORD(DVND) ; hi words are equal, compare lo words
jbe short L7 ; if less or equal we are ok, else subtract
L6:
dec esi ; subtract 1 from quotient
L7:
xor edx,edx ; edx:eax <- quotient
mov eax,esi
;
; Just the cleanup left to do. edx:eax contains the quotient. Set the sign
; according to the save value, cleanup the stack, and return.
;
L4:
dec edi ; check to see if result is negative
jnz short L8 ; if EDI == 0, result should be negative
neg edx ; otherwise, negate the result
neg eax
sbb edx,0
;
; Restore the saved registers and return.
;
L8:
pop ebx
pop esi
pop edi
ret 16
_alldiv ENDP
end
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