windows-nt/Source/XPSP1/NT/base/crts/crtw32/helper/i386/llrem.asm

	title	llrem - signed long remainder routine
;***
;llrem.asm - signed long remainder routine
;
;	Copyright (c) 1985-2001, Microsoft Corporation. All rights reserved.
;
;Purpose:
;	defines the signed long remainder routine
;	    __allrem
;
;Revision History:
;	11-29-83  DFW	initial version
;	06-01-84  RN	modified to use cmacros
;	10-23-87  SKS	fixed off-by-1 error for dividend close to 2**32.
;	05-18-89  SKS	Remove redundant "MOV SP,BP" from epilog
;	11-28-89  GJF	Fixed copyright
;	11-19-93  SMK	Modified to work on 64 bit integers
;	01-17-94  GJF	Minor changes to build with NT's masm386.
;	07-22-94  GJF	Use esp-relative addressing for args. Shortened
;			conditional jumps.
;
;*******************************************************************************


.xlist
include cruntime.inc
include mm.inc
.list

;***
;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
;
;Exceptions:
;
;*******************************************************************************

	CODESEG

_allrem	PROC NEAR

	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

	end