474 lines
17 KiB
NASM
474 lines
17 KiB
NASM
subttl emfdiv.asm - Division
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page
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; Copyright (c) Microsoft Corporation 1991
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; All Rights Reserved
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;emfdiv.asm - long double divide
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; by Tim Paterson
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;Purpose:
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; Long double division.
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;Inputs:
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; ebx:esi = op1 mantissa
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; ecx = op1 sign in bit 15, exponent in high half
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; edi = pointer to op2 and result location
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; [Result] = edi
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; Exponents are unbiased. Denormals have been normalized using
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; this expanded exponent range. Neither operand is allowed to be zero.
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;Outputs:
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; Jumps to [RoundMode] to round and store result.
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;Revision History:
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; [] 09/05/91 TP Initial 32-bit version.
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;Dispatch tables for division
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;One operand has been loaded into ecx:ebx:esi ("source"), the other is
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;pointed to by edi ("dest"). edi points to dividend for fdiv,
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;to divisor for fdivr.
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;Tag of source is shifted. Tag values are as follows:
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.erre TAG_SNGL eq 0 ;SINGLE: low 32 bits are zero
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.erre TAG_VALID eq 1
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.erre TAG_ZERO eq 2
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.erre TAG_SPCL eq 3 ;NAN, Infinity, Denormal, Empty
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;dest = dest / source
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tFdivDisp label dword ;Source (reg) Dest (*[di])
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dd DivSingle ;single single
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dd DivSingle ;single double
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dd XorDestSign ;single zero
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dd DivSpclDest ;single special
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dd DivDouble ;double single
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dd DivDouble ;double double
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dd XorDestSign ;double zero
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dd DivSpclDest ;double special
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dd DivideByZero ;zero single
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dd DivideByZero ;zero double
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dd ReturnIndefinite ;zero zero
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dd DivSpclDest ;zero special
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dd DivSpclSource ;special single
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dd DivSpclSource ;special double
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dd DivSpclSource ;special zero
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dd TwoOpBothSpcl ;special special
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dd ReturnIndefinite ;Two infinities
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;dest = source / dest
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tFdivrDisp label dword ;Source (reg) Dest (*[di])
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dd DivrSingle ;single single
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dd DivrDouble ;single double
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dd DivideByZero ;single zero
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dd DivrSpclDest ;single special
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dd DivrSingle ;double single
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dd DivrDouble ;double double
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dd DivideByZero ;double zero
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dd DivrSpclDest ;double special
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dd XorSourceSign ;zero single
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dd XorSourceSign ;zero double
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dd ReturnIndefinite ;zero zero
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dd DivrSpclDest ;zero special
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dd DivrSpclSource ;special single
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dd DivrSpclSource ;special double
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dd DivrSpclSource ;special zero
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dd TwoOpBothSpcl ;special special
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dd ReturnIndefinite ;Two infinities
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EM_ENTRY eFIDIV16
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eFIDIV16:
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push offset DivSetResult
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jmp Load16Int ;Returns to DivSetResult
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EM_ENTRY eFIDIVR16
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eFIDIVR16:
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push offset DivrSetResult
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jmp Load16Int
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EM_ENTRY eFIDIV32
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eFIDIV32:
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push offset DivSetResult
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jmp Load32Int
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EM_ENTRY eFIDIVR32
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eFIDIVR32:
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push offset DivrSetResult
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jmp Load32Int
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EM_ENTRY eFDIV32
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eFDIV32:
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push offset DivSetResult
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jmp Load32Real ;Returns to DivSetResult
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EM_ENTRY eFDIVR32
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eFDIVR32:
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push offset DivrSetResult ;Returns to DivrSetResult
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jmp Load32Real
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EM_ENTRY eFDIV64
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eFDIV64:
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push offset DivSetResult
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jmp Load64Real ;Returns to DivSetResult
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EM_ENTRY eFDIVR64
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eFDIVR64:
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push offset DivrSetResult
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jmp Load64Real ;Returns to DivrSetResult
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EM_ENTRY eFDIVRPreg
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eFDIVRPreg:
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push offset PopWhenDone
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EM_ENTRY eFDIVRreg
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eFDIVRreg:
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xchg esi,edi
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EM_ENTRY eFDIVRtop
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eFDIVRtop:
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mov ecx,EMSEG:[esi].ExpSgn
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mov ebx,EMSEG:[esi].lManHi
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mov esi,EMSEG:[esi].lManLo
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DivrSetResult:
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;cl has tag of dividend
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mov ebp,offset tFdivrDisp
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mov EMSEG:[Result],edi ;Save result pointer
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mov ah,cl
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mov al,EMSEG:[edi].bTag
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and ah,not 1 ;Ignore single vs. double on dividend
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cmp ax,1
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.erre bTAG_VALID eq 1
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.erre bTAG_SNGL eq 0
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jz DivrDouble ;Divisor was double
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ja TwoOpResultSet
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;.erre DivrSingle eq $ ;Fall into DivrSingle
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DivrSingle:
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;Computes op1/op2
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;Op1 is double, op2 is single (low 32 bits are zero)
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mov edx,ebx
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mov eax,esi ;Mantissa in edx:eax
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mov ebx,EMSEG:[edi].ExpSgn
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mov edi,EMSEG:[edi].lManHi
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jmp DivSingleReg
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SDivBigUnderflow:
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;Overflow flag set could only occur with denormals (true exp < -32768)
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or EMSEG:[CURerr],Underflow
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test EMSEG:[CWmask],Underflow ;Is exception masked?
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jnz UnderflowZero ;Yes, return zero (in emfmul.asm)
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add ecx,Underbias shl 16 ;Fix up exponent
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jmp ContSdiv ;Continue with multiply
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EM_ENTRY eFDIVPreg
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eFDIVPreg:
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push offset PopWhenDone
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EM_ENTRY eFDIVreg
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eFDIVreg:
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xchg esi,edi
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EM_ENTRY eFDIVtop
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eFDIVtop:
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mov ecx,EMSEG:[esi].ExpSgn
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mov ebx,EMSEG:[esi].lManHi
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mov esi,EMSEG:[esi].lManLo
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DivSetResult:
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;cl has tag of divisor
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mov ebp,offset tFdivDisp
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mov EMSEG:[Result],edi ;Save result pointer
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mov al,cl
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mov ah,EMSEG:[edi].bTag
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and ah,not 1 ;Ignore single vs. double on dividend
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cmp ax,1
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.erre bTAG_VALID eq 1
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.erre bTAG_SNGL eq 0
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jz DivDouble ;Divisor was double
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ja TwoOpResultSet
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;.erre DivSingle eq $ ;Fall into DivSingle
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DivSingle:
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;Computes op2/op1
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;Op2 is double, op1 is single (low 32 bits are zero)
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xchg edi,ebx ;Mantissa in edi, op2 ptr to ebx
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xchg ebx,ecx ;ExpSgn to ebx, op2 ptr to ecx
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mov edx,EMSEG:[ecx].lManHi
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mov eax,EMSEG:[ecx].lManLo
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mov ecx,EMSEG:[ecx].ExpSgn ;Op2 loaded
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DivSingleReg:
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;dividend mantissa in edx:eax, exponent in high ecx, sign in ch bit 7
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;divisor mantissa in edi, exponent in high ebx, sign in bh bit 7
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xor ch,bh ;Compute result sign
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xor bx,bx ;Clear out sign and tag
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sub ecx,1 shl 16 ;Exponent adjustment needed
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sub ecx,ebx ;Compute result exponent
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.erre TexpBias eq 0 ;Exponents not biased
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jo SDivBigUnderflow ;Dividing denormal by large number
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ContSdiv:
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;If dividend >= divisor, the DIV instruction will overflow. Check for
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;this condition and shift the dividend right one bit if necessary.
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;In previous versions of this algorithm for 24-bit and 53-bit mantissas,
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;this shift was always performed without a test. This meant that a 1-bit
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;normalization might be required at the end. This worked fine because
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;32 or 64 bits were calculated, so extra precision was available for
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;normalization. However, this version needs all 64 bits that are calculated,
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;so we can't afford a normalization shift at the end. This test tells us
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;up front how to align so we'll be normalized.
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xor ebx,ebx ;Extend dividend
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cmp edi,edx ;Will DIV overflow?
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ja DoSdiv ;No, we're safe
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shrd ebx,eax,1
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shrd eax,edx,1
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shr edx,1
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add ecx,1 shl 16 ;Bump exponent to account for shift
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DoSdiv:
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div edi
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xchg ebx,eax ;Save quotient in ebx, extend remainder
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div edi
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mov esi,eax
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;We have a 64-bit quotient in ebx:esi. Now compare remainder*2 with divisor
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;to compute round and sticky bits.
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mov eax,-1 ;Set round and sticky bits
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shl edx,1 ;Double remainder
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jc RoundJmp ;If too big, round & sticky set
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cmp edx,edi ;Is remainder*2 > divisor?
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ja RoundJmp
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;Observe, oh wondering one, how you can assume the result of this last
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;compare is not equality. Use the following notation: n=numerator,
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;d=denominator,q=quotient,r=remainder,b=base(2^64 here). If
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;initially we had n < d then there was no shift and we will find q and r
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;so that q*d+r=n*b, if initially we had n >= d then there was a shift and
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;we will find q and r so that q*d+r=n*b/2. If we have equality here
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;then r=d/2 ==> n={possibly 2*}(2*q+1)*d/(2*b), since this can only
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;be integral if d is a multiple of b, but by definition b/2 <= d < b, we
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;have a contradiction. Equality is thus impossible at this point.
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cmp edx,1 ;Check for zero remainder
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sbb eax,-2 ;eax==0 if CY, ==1 if NC (was -1)
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RoundJmp:
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jmp EMSEG:[RoundMode]
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DDivBigUnderflow:
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;Overflow flag set could only occur with denormals (true exp < -32768)
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or EMSEG:[CURerr],Underflow
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test EMSEG:[CWmask],Underflow ;Is exception masked?
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jnz UnderflowZero ;Yes, return zero (in emfmul.asm)
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add ecx,Underbias shl 16 ;Fix up exponent
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jmp ContDdiv ;Continue with multiply
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DivrDoubleSetFlag:
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;Special entry point used by FPATAN to set bit 6 of flag dword pushed
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;on stack before call.
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or byte ptr [esp+4],40H
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DivrDouble:
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;Computes op1/op2
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mov edx,ebx
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mov eax,esi ;Mantissa in edx:eax
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mov ebx,EMSEG:[edi].ExpSgn
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mov esi,EMSEG:[edi].lManHi
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mov edi,EMSEG:[edi].lManLo
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jmp short DivDoubleReg
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HighHalfEqual:
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;edx:eax:ebp = dividend
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;esi:edi = divisor
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;ecx = exponent and sign of result
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;High half of dividend is equal to high half of divisor. This will cause
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;the DIV instruction to overflow. If whole dividend >= whole divisor, then
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;we just shift the dividend right 1 bit.
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cmp eax,edi ;Is dividend >= divisor?
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jae ShiftDividend ;Yes, divide it by two
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;DIV instruction would overflow, so skip it and calculate the effective
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;result. Assume a quotient of 2^32-1 and calculate the remainder. See
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;detailed comments under MaxQuo below--this is a copy of that code.
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push ecx ;Save exp. and sign
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mov ebx,-1 ;Max quotient digit
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sub eax,edi ;Calculate correct remainder
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;Currently edx == esi, but the next instruction ensures that is no longer
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;true, since eax != 0. This will allow us to skip the MaxQuo check at
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;DivFirstDigit.
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add edx,eax ;Should set CY if quotient fit
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mov eax,edi ;ecx:eax has new remainder
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jc ComputeSecond ;Remainder was positive
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;Quotient doesn't fit. Note that we can no longer ensure that edx != esi
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;after making a correction.
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mov ecx,edx ;Need remainder in ecx:eax
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jmp DivCorrect1
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DivDouble:
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;Computes op2/op1
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mov eax,edi ;Move op2 pointer
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mov edi,esi
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mov esi,ebx ;Mantissa in esi:edi
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mov ebx,ecx ;ExpSgn to ebx
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mov ecx,EMSEG:[eax].ExpSgn ;Op2 loaded
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mov edx,EMSEG:[eax].lManHi
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mov eax,EMSEG:[eax].lManLo
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DivDoubleReg:
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;dividend mantissa in edx:eax, exponent in high ecx, sign in ch bit 7
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;divisor mantissa in esi:edi, exponent in high ebx, sign in bh bit 7
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xor ch,bh ;Compute result sign
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xor bx,bx ;Clear out sign and tag
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sub ecx,1 shl 16 ;Exponent adjustment needed
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sub ecx,ebx ;Compute result exponent
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.erre TexpBias eq 0 ;Exponents not biased
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jo DDivBigUnderflow ;Dividing denormal by large number
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ContDdiv:
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;If dividend >= divisor, we must shift the dividend right one bit.
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;This will ensure the result is normalized.
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;In previous versions of this algorithm for 24-bit and 53-bit mantissas,
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;this shift was always performed without a test. This meant that a 1-bit
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;normalization might be required at the end. This worked fine because
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;32 or 64 bits were calculated, so extra precision was available for
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;normalization. However, this version needs all 64 bits that are calculated,
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;so we can't afford a normalization shift at the end. This test tells us
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;up front how to align so we'll be normalized.
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xor ebp,ebp ;Extend dividend
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cmp esi,edx ;Dividend > divisor
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ja DoDdiv
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jz HighHalfEqual ;Go compare low halves
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ShiftDividend:
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shrd ebp,eax,1
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shrd eax,edx,1
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shr edx,1
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add ecx,1 shl 16 ;Bump exponent to account for shift
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DoDdiv:
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push ecx ;Save exp. and sign
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;edx:eax:ebp = dividend
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;esi:edi = divisor
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;Division algorithm from Knuth vol. 2, p. 237, using 32-bit "digits":
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;Guess a quotient digit by dividing two MSDs of dividend by the MSD of
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;divisor. If divisor is >= 1/2 the radix (radix = 2^32 in this case), then
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;this guess will be no more than 2 larger than the correct value of that
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;quotient digit (and never smaller). Divisor meets magnitude condition
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;because it's normalized.
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div esi ;Guess first quotient "digit"
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;Check out our guess.
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;Currently, remainder in edx = dividend - (quotient * high half divisor).
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;The definition of remainder is dividend - (quotient * all divisor). So
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;if we subtract (quotient * low half divisor) from edx, we'll get
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;the true remainder. If it's negative, our guess was too big.
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mov ebx,eax ;Save quotient
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mov ecx,edx ;Save remainder
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mul edi ;Quotient * low half divisor
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sub ebp,eax ;Subtract from dividend extension
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sbb ecx,edx ;Subtract from remainder
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mov eax,ebp ;Low remainder to eax
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jnc DivFirstDigit ;Was quotient OK?
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DivCorrect1:
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dec ebx ;Quotient was too big
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add eax,edi ;Add divisor back into remainder
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adc ecx,esi
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jnc DivCorrect1 ;Repeat if quotient is still too big
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DivFirstDigit:
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cmp ecx,esi ;Would DIV instruction overflow?
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jae short MaxQuo ;Yes, figure alternate quotient
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mov edx,ecx ;Remainder back to edx:eax
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;Compute 2nd quotient "digit"
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ComputeSecond:
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div esi ;Guess 2nd quotient "digit"
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mov ebp,eax ;Save quotient
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mov ecx,edx ;Save remainder
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mul edi ;Quotient * low half divisor
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neg eax ;Subtract from dividend extended with 0
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sbb ecx,edx ;Subtract from remainder
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jnc DivSecondDigit ;Was quotient OK?
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DivCorrect2:
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dec ebp ;Quotient was too big
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add eax,edi ;Add divisor back into remainder
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adc ecx,esi
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jnc DivCorrect2 ;Repeat if quotient is still too big
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DivSecondDigit:
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;ebx:ebp = quotient
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;ecx:eax = remainder
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;esi:edi = divisor
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;Now compare remainder*2 with divisor to compute round and sticky bits.
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mov edx,-1 ;Set round and sticky bits
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shld ecx,eax,1 ;Double remainder
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jc DDivEnd ;If too big, round & sticky set
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shl eax,1
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sub edi,eax
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sbb esi,ecx ;Subtract remainder*2 from divisor
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jb DDivEnd ;If <0, use round & sticky bits set
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;Observe, oh wondering one, how you can assume the result of this last
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;compare is not equality. Use the following notation: n=numerator,
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;d=denominator,q=quotient,r=remainder,b=base(2^64 here). If
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;initially we had n < d then there was no shift and we will find q and r
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;so that q*d+r=n*b, if initially we had n >= d then there was a shift and
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;we will find q and r so that q*d+r=n*b/2. If we have equality here
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;then r=d/2 ==> n={possibly 2*}(2*q+1)*d/(2*b), since this can only
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;be integral if d is a multiple of b, but by definition b/2 <= d < b, we
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;have a contradiction. Equality is thus impossible at this point.
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;No round bit, but set sticky bit if remainder != 0.
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or eax,ecx ;Is remainder zero?
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add eax,-1 ;Set CY if non-zero
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adc edx,1 ;edx==0 if NC, ==1 if CY (was -1)
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DDivEnd:
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mov esi,ebp ;Result in ebx:esi
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mov eax,edx ;Round/sticky bits to eax
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pop ecx ;Recover sign/exponent
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jmp EMSEG:[RoundMode]
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MaxQuo:
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;ebx = first quotient "digit"
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;ecx:eax = remainder
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;esi:edi = divisor
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;On exit, ebp = second quotient "digit"
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;Come here if divide instruction would overflow. This must mean that ecx == esi,
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;i.e., the high halves of the dividend and divisor are equal. Assume a result
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;of 2^32-1, thus remainder = dividend - ( divisor * (2^32-1) )
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; = dividend - divisor * 2^32 + divisor. Since the high halves of the dividend
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;and divisor are equal, dividend - divisor * 2^32 can be computed by
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;subtracting only the low halves. When adding divisor (in esi) to this, note
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;that ecx == esi, and we want the result in ecx anyway.
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;Note also that since the dividend is a previous remainder, the
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;dividend - divisor * 2^32 calculation must always be negative. Thus the
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;addition of divisor back to it should generate a carry if it goes positive.
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mov ebp,-1 ;Max quotient digit
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sub eax,edi ;Calculate correct remainder
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add ecx,eax ;Should set CY if quotient fit
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mov eax,edi ;ecx:eax has new remainder
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jc DivSecondDigit ;Remainder was positive
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jmp DivCorrect2
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