Source code of Windows XP (NT5)
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subttl emfdiv.asm - Division
;***
;emfdiv.asm - Division
;
; Copyright (c) 1986-89, Microsoft Corporation
;
;Purpose:
; Division
;
; This Module contains Proprietary Information of Microsoft
; Corporation and should be treated as Confidential.
;
;Revision History:
; See emulator.hst
;
;*******************************************************************************
;-----------------------------------------;
; ;
; Division ;
; ;
;-----------------------------------------;
ProfBegin FDIV
RDBRQQ: ; Routine Div Both must see if we have two singles.
if fastSP
MOV BX,DX
XOR BX,Single + 256*Single
TEST BX,Single + 256*Single
JNZ RDDRQQ
MOV bx,offset TDSRQQ
JMP [bx]
endif ;fastSP
pub RDDRQQ ; Routine Division Double
; Now we have
; SI --> numerator , AX - Expon , DL - Sign
; DI --> denominator , CX - Expon , DH - Sign
if fastSP
CALL CoerceToDouble ; insure that both args are double
endif ;fastSP
STC ; exponent will be difference - 1
SBB AX,CX ; compute result exponent
; AH has the (tentative) true exponent of the result. It is correct if the
; result does not need normalizing. If normalizing is required, then this
; must be incremented to give the correct result exponent
XOR DH,DL ; Compute sign
PUSH ebp
PUSH edx ; Save sign
PUSH esi
PUSH edi
ADD esi,6
ADD edi,6
MOV ecx,4
STD
REP CMPS word ptr [esi],word ptr [edi] ; compare numerator mantissa
CLD ; with denominator mantissa
POP edi
POP esi
PUSHF ; save the flags from the compare
MOV BP,AX ; save the exponent
LODS word ptr [esi] ; Load up numerator
MOV CX,AX
LODS word ptr [esi]
MOV BX,AX
LODS word ptr [esi]
MOV DX,AX
LODS word ptr [esi]
XCHG AX,DX
; Move divisor to DAC so we can get at it easily.
MOV esi,edi ; Move divisor to DAC
MOV edi,offset DAC
ifdef i386
MOVSD
MOVSD
else
MOVSW
MOVSW
MOVSW
MOVSW
endif
; Now we're all set:
; DX:AX:BX:CX has dividend
; DAC has divisor (in normal format)
; Both are 64 bits with zeros and have implied bit set.
; Top of stack has sign and tentative exponent.
XOR DI,DI
POPF ; numerator mantissa < denominator?
; 80286 errata for POPF shouldn't
; apply because interrupts should be
; turned on in this context
JB short DivNoShift ; if so bypass numerator shift
SHR DX,1 ; Make sure dividend is smaller than divisor
RCR AX,1 ; by dividing it by two
RCR BX,1
RCR CX,1
RCR DI,1
INC BP ; increment result exponent
pub DivNoShift
PUSH ebp ; save result exponent
MOV [REMLSW],DI ; Save lsb of remainder
CALL DIV16 ; Get a quotient digit
PUSH edi
MOV [REMLSW],0 ; Turn off the shifted bit
CALL DIV16
PUSH edi
CALL DIV16
PUSH edi
CALL DIV16
MOV BP,8001H ; turn round and sticky on
SHL CX,1
RCL BX,1
RCL AX,1
RCL DX,1 ; multiply remainder by 2
JC short BPset ; if overflow, then round,sticky valid
MOV esi,offset DAC
CMP DX,[esi+6]
JNE short RemainderNotHalf
CMP AX,[esi+4]
JNE short RemainderNotHalf
CMP BX,[esi+2]
JNE short RemainderNotHalf
CMP CX,[esi] ; compare 2*remainder with denominator
;Observe, oh wondering one, how you can assume the result of this last
;compare is not equality. Use the following notation: n=numerator,
;d=denominator,q=quotient,r=remainder,b=base(2^64 here). If
;initially we had n < d then there was no shift and we will find q and r
;so that q*d+r=n*b, if initially we had n >= d then there was a shift and
;we will find q and r so that q*d+r=n*b/2. If we have equality here
;then r=d/2 ==> n={possibly 2*}(2*q+1)*d/(2*b), since this can only
;be integral if d is a multiple of b, but by definition b/2 <= d < b, we
;have a contradiction. Equality is thus impossible at this point.
pub RemainderNotHalf ; if 2*remainder > denominator
JAE short BPset ; then round and sticky are valid
OR AX,DX
OR AX,CX
OR AX,BX
OR AL,AH ; otherwise or sticky bits into AL
XOR AH,AH ; clear round bit
MOV BP,AX ; move round and sticky into BP
pub BPset
MOV DX,DI ; get low 16 bits into proper location
POP ecx
POP ebx
POP edi
POP esi ; Now restore exponent
JMP ROUND ; Result is normalized, round it
; Remainder in DX:AX:BX:CX:REMLSW
pub DIV16
MOV SI,[DAC+6] ; Get high word of divisor
XOR DI,DI ; Initialize quotient digit to zero
CMP DX,SI ; Will we overflow?
JAE MAXQUO ; If so, go handle special
OR DX,DX ; Is dividend small?
JNZ short DDIV
CMP SI,AX ; Will divisor fit at all?
JA short ZERQUO ; No - quotient is zero
pub DDIV
DIV SI ; AX is our digit "guess"
PUSH edx ; Save remainder -
PUSH ebx ; top 32 bits
XCHG AX,DI ; Quotient digit in DI
XOR BP,BP ; Initialize quotient * divisor
MOV SI,BP
MOV AX,[DAC]
OR AX,AX ; If zero, save multiply time
JZ short REM2
MUL DI ; Begin computing quotient * divisor
MOV SI,DX
pub REM2
PUSH eax ; Save lowest word of quotient * divisor
MOV AX,[DAC+2]
OR AX,AX
JZ short REM3
MUL DI
ADD SI,AX
ADC BP,DX
pub REM3
MOV AX,[DAC+4]
OR AX,AX
JZ short REM4
MUL DI
ADD BP,AX
ADC DX,0
XCHG AX,DX
; Remainder - Quotient * divisor
; [SP+4]:[SP+2]:CX:REMLSW - AX:BP:SI:[SP]
pub REM4
MOV DX,[REMLSW] ; Low word of remainder
POP ebx ; Recover lowest word of quotient * divisor
SUB DX,BX
SBB CX,SI
POP ebx
SBB BX,BP
POP ebp ; Remainder from DIV
SBB BP,AX
XCHG AX,BP
pub ZERQUO ; Remainder in AX:BX:CX:DX
XCHG AX,DX
XCHG AX,CX
XCHG AX,BX
JNC short DRET ; Remainder in DX:AX:BX:CX
pub RESTORE
DEC DI ; Drop quotient since it didn't fit
ADD CX,[DAC] ; Add divisor back in until remainder goes +
ADC BX,[DAC+2]
ADC AX,[DAC+4]
ADC DX,[DAC+6]
JNC RESTORE ; Loop is performed at most twice
pub DRET
RET
pub MAXQUO
DEC DI ; DI=FFFF=2**16-1, DX:AX:BX:CX is remainder,
SUB CX,[DAC] ; DX = [DAC+6], d = divisor = [DAC]
SBB BX,[DAC+2]
SBB AX,[DAC+4] ; subtract 2^16*d from DX:AX:BX:CX:0000H
ADD CX,[DAC+2] ; (DX-[DAC+6] = 0 is implied)
ADC BX,[DAC+4]
ADC AX,DX ; add high 48 bits of d to AX:BX:CX:0000H
MOV DX,[DAC] ; add low 16 bits of d to zero giving DX
CMC ; DI should be FFFEH if no carry from add
JMP ZERQUO
ProfEnd FDIV