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;-----------------------------------------------------------------------------
; llrem.asm - signed long remainder routine
; Adapted from Visual Studio C runtime library
; Portions Copyright (c) Microsoft Corporation. All rights reserved.
;-----------------------------------------------------------------------------
.386
_TEXT segment use32 para public 'CODE'
public __allrem
LOWORD equ [0]
HIWORD equ [4]
;
; llrem - signed long remainder
;
; Purpose:
; Does a signed long 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 remainder (dividend%divisor)
; NOTE: this routine removes the parameters from the stack.
;
; Uses:
; ECX
;
__allrem proc near
assume cs:_TEXT
push ebx
push edi
; 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 lrem(a, b)):
;
; -----------------
; | |
; |---------------|
; | |
; |--divisor (b)--|
; | |
; |---------------|
; | |
; |--dividend (a)-|
; | |
; |---------------|
; | return addr** |
; |---------------|
; | EBX |
; |---------------|
; ESP---->| EDI |
; -----------------
;
DVND equ [esp + 12] ; stack address of dividend (a)
DVSR equ [esp + 20] ; 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 bit
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
mov edx,LOWORD(DVSR) ; lo word of b
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 ; edx <- remainder
mov eax,LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
div ecx ; edx <- final remainder
mov eax,edx ; edx:eax <- remainder
xor edx,edx
dec edi ; check result sign flag
jns short L4 ; negate result, restore stack and return
jmp short L8 ; result sign ok, 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
;
; 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.
;
mov ecx,eax ; save a copy of quotient in ECX
mul dword ptr HIWORD(DVSR)
xchg ecx,eax ; save product, get quotient in EAX
mul dword ptr 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 the original divisor from the result.
;
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:
sub eax,LOWORD(DVSR) ; subtract divisor from result
sbb edx,HIWORD(DVSR)
L7:
;
; 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 if necessary.
;
sub eax,LOWORD(DVND) ; subtract dividend from result
sbb edx,HIWORD(DVND)
;
; Now check the result sign flag to see if the result is supposed to be positive
; or negative. It is currently negated (because we subtracted in the 'wrong'
; direction), so if the sign flag is set we are done, otherwise we must negate
; the result to make it positive again.
;
dec edi ; check result sign flag
jns short L8 ; result is ok, restore stack and return
L4:
neg edx ; otherwise, negate the result
neg eax
sbb edx,0
;
; Just the cleanup left to do. edx:eax contains the quotient.
; Restore the saved registers and return.
;
L8:
pop edi
pop ebx
ret 16
__allrem endp
_TEXT ends
end