* $NetBSD: setox.sa,v 1.5 2014/09/01 08:21:26 matt Exp $
* MOTOROLA MICROPROCESSOR & MEMORY TECHNOLOGY GROUP
* M68000 Hi-Performance Microprocessor Division
* M68040 Software Package
*
* M68040 Software Package Copyright (c) 1993, 1994 Motorola Inc.
* All rights reserved.
*
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*
* setox.sa 3.1 12/10/90
*
* The entry point setox computes the exponential of a value.
* setoxd does the same except the input value is a denormalized
* number. setoxm1 computes exp(X)-1, and setoxm1d computes
* exp(X)-1 for denormalized X.
*
* INPUT
* -----
* Double-extended value in memory location pointed to by address
* register a0.
*
* OUTPUT
* ------
* exp(X) or exp(X)-1 returned in floating-point register fp0.
*
* ACCURACY and MONOTONICITY
* -------------------------
* The returned result is within 0.85 ulps in 64 significant bit, i.e.
* within 0.5001 ulp to 53 bits if the result is subsequently rounded
* to double precision. The result is provably monotonic in double
* precision.
*
* SPEED
* -----
* Two timings are measured, both in the copy-back mode. The
* first one is measured when the function is invoked the first time
* (so the instructions and data are not in cache), and the
* second one is measured when the function is reinvoked at the same
* input argument.
*
* The program setox takes approximately 210/190 cycles for input
* argument X whose magnitude is less than 16380 log2, which
* is the usual situation. For the less common arguments,
* depending on their values, the program may run faster or slower --
* but no worse than 10% slower even in the extreme cases.
*
* The program setoxm1 takes approximately ??? / ??? cycles for input
* argument X, 0.25 <= |X| < 70log2. For |X| < 0.25, it takes
* approximately ??? / ??? cycles. For the less common arguments,
* depending on their values, the program may run faster or slower --
* but no worse than 10% slower even in the extreme cases.
*
* ALGORITHM and IMPLEMENTATION NOTES
* ----------------------------------
*
* setoxd
* ------
* Step 1. Set ans := 1.0
*
* Step 2. Return ans := ans + sign(X)*2^(-126). Exit.
* Notes: This will always generate one exception -- inexact.
*
*
* setox
* -----
*
* Step 1. Filter out extreme cases of input argument.
* 1.1 If |X| >= 2^(-65), go to Step 1.3.
* 1.2 Go to Step 7.
* 1.3 If |X| < 16380 log(2), go to Step 2.
* 1.4 Go to Step 8.
* Notes: The usual case should take the branches 1.1 -> 1.3 -> 2.
* To avoid the use of floating-point comparisons, a
* compact representation of |X| is used. This format is a
* 32-bit integer, the upper (more significant) 16 bits are
* the sign and biased exponent field of |X|; the lower 16
* bits are the 16 most significant fraction (including the
* explicit bit) bits of |X|. Consequently, the comparisons
* in Steps 1.1 and 1.3 can be performed by integer comparison.
* Note also that the constant 16380 log(2) used in Step 1.3
* is also in the compact form. Thus taking the branch
* to Step 2 guarantees |X| < 16380 log(2). There is no harm
* to have a small number of cases where |X| is less than,
* but close to, 16380 log(2) and the branch to Step 9 is
* taken.
*
* Step 2. Calculate N = round-to-nearest-int( X * 64/log2 ).
* 2.1 Set AdjFlag := 0 (indicates the branch 1.3 -> 2 was taken)
* 2.2 N := round-to-nearest-integer( X * 64/log2 ).
* 2.3 Calculate J = N mod 64; so J = 0,1,2,..., or 63.
* 2.4 Calculate M = (N - J)/64; so N = 64M + J.
* 2.5 Calculate the address of the stored value of 2^(J/64).
* 2.6 Create the value Scale = 2^M.
* Notes: The calculation in 2.2 is really performed by
*
* Z := X * constant
* N := round-to-nearest-integer(Z)
*
* where
*
* constant := single-precision( 64/log 2 ).
*
* Using a single-precision constant avoids memory access.
* Another effect of using a single-precision "constant" is
* that the calculated value Z is
*
* Z = X*(64/log2)*(1+eps), |eps| <= 2^(-24).
*
* This error has to be considered later in Steps 3 and 4.
*
* Step 3. Calculate X - N*log2/64.
* 3.1 R := X + N*L1, where L1 := single-precision(-log2/64).
* 3.2 R := R + N*L2, L2 := extended-precision(-log2/64 - L1).
* Notes: a) The way L1 and L2 are chosen ensures L1+L2 approximate
* the value -log2/64 to 88 bits of accuracy.
* b) N*L1 is exact because N is no longer than 22 bits and
* L1 is no longer than 24 bits.
* c) The calculation X+N*L1 is also exact due to cancellation.
* Thus, R is practically X+N(L1+L2) to full 64 bits.
* d) It is important to estimate how large can |R| be after
* Step 3.2.
*
* N = rnd-to-int( X*64/log2 (1+eps) ), |eps|<=2^(-24)
* X*64/log2 (1+eps) = N + f, |f| <= 0.5
* X*64/log2 - N = f - eps*X 64/log2
* X - N*log2/64 = f*log2/64 - eps*X
*
*
* Now |X| <= 16446 log2, thus
*
* |X - N*log2/64| <= (0.5 + 16446/2^(18))*log2/64
* <= 0.57 log2/64.
* This bound will be used in Step 4.
*
* Step 4. Approximate exp(R)-1 by a polynomial
* p = R + R*R*(A1 + R*(A2 + R*(A3 + R*(A4 + R*A5))))
* Notes: a) In order to reduce memory access, the coefficients are
* made as "short" as possible: A1 (which is 1/2), A4 and A5
* are single precision; A2 and A3 are double precision.
* b) Even with the restrictions above,
* |p - (exp(R)-1)| < 2^(-68.8) for all |R| <= 0.0062.
* Note that 0.0062 is slightly bigger than 0.57 log2/64.
* c) To fully use the pipeline, p is separated into
* two independent pieces of roughly equal complexities
* p = [ R + R*S*(A2 + S*A4) ] +
* [ S*(A1 + S*(A3 + S*A5)) ]
* where S = R*R.
*
* Step 5. Compute 2^(J/64)*exp(R) = 2^(J/64)*(1+p) by
* ans := T + ( T*p + t)
* where T and t are the stored values for 2^(J/64).
* Notes: 2^(J/64) is stored as T and t where T+t approximates
* 2^(J/64) to roughly 85 bits; T is in extended precision
* and t is in single precision. Note also that T is rounded
* to 62 bits so that the last two bits of T are zero. The
* reason for such a special form is that T-1, T-2, and T-8
* will all be exact --- a property that will give much
* more accurate computation of the function EXPM1.
*
* Step 6. Reconstruction of exp(X)
* exp(X) = 2^M * 2^(J/64) * exp(R).
* 6.1 If AdjFlag = 0, go to 6.3
* 6.2 ans := ans * AdjScale
* 6.3 Restore the user FPCR
* 6.4 Return ans := ans * Scale. Exit.
* Notes: If AdjFlag = 0, we have X = Mlog2 + Jlog2/64 + R,
* |M| <= 16380, and Scale = 2^M. Moreover, exp(X) will
* neither overflow nor underflow. If AdjFlag = 1, that
* means that
* X = (M1+M)log2 + Jlog2/64 + R, |M1+M| >= 16380.
* Hence, exp(X) may overflow or underflow or neither.
* When that is the case, AdjScale = 2^(M1) where M1 is
* approximately M. Thus 6.2 will never cause over/underflow.
* Possible exception in 6.4 is overflow or underflow.
* The inexact exception is not generated in 6.4. Although
* one can argue that the inexact flag should always be
* raised, to simulate that exception cost to much than the
* flag is worth in practical uses.
*
* Step 7. Return 1 + X.
* 7.1 ans := X
* 7.2 Restore user FPCR.
* 7.3 Return ans := 1 + ans. Exit
* Notes: For non-zero X, the inexact exception will always be
* raised by 7.3. That is the only exception raised by 7.3.
* Note also that we use the FMOVEM instruction to move X
* in Step 7.1 to avoid unnecessary trapping. (Although
* the FMOVEM may not seem relevant since X is normalized,
* the precaution will be useful in the library version of
* this code where the separate entry for denormalized inputs
* will be done away with.)
*
* Step 8. Handle exp(X) where |X| >= 16380log2.
* 8.1 If |X| > 16480 log2, go to Step 9.
* (mimic 2.2 - 2.6)
* 8.2 N := round-to-integer( X * 64/log2 )
* 8.3 Calculate J = N mod 64, J = 0,1,...,63
* 8.4 K := (N-J)/64, M1 := truncate(K/2), M = K-M1, AdjFlag := 1.
* 8.5 Calculate the address of the stored value 2^(J/64).
* 8.6 Create the values Scale = 2^M, AdjScale = 2^M1.
* 8.7 Go to Step 3.
* Notes: Refer to notes for 2.2 - 2.6.
*
* Step 9. Handle exp(X), |X| > 16480 log2.
* 9.1 If X < 0, go to 9.3
* 9.2 ans := Huge, go to 9.4
* 9.3 ans := Tiny.
* 9.4 Restore user FPCR.
* 9.5 Return ans := ans * ans. Exit.
* Notes: Exp(X) will surely overflow or underflow, depending on
* X's sign. "Huge" and "Tiny" are respectively large/tiny
* extended-precision numbers whose square over/underflow
* with an inexact result. Thus, 9.5 always raises the
* inexact together with either overflow or underflow.
*
*
* setoxm1d
* --------
*
* Step 1. Set ans := 0
*
* Step 2. Return ans := X + ans. Exit.
* Notes: This will return X with the appropriate rounding
* precision prescribed by the user FPCR.
*
* setoxm1
* -------
*
* Step 1. Check |X|
* 1.1 If |X| >= 1/4, go to Step 1.3.
* 1.2 Go to Step 7.
* 1.3 If |X| < 70 log(2), go to Step 2.
* 1.4 Go to Step 10.
* Notes: The usual case should take the branches 1.1 -> 1.3 -> 2.
* However, it is conceivable |X| can be small very often
* because EXPM1 is intended to evaluate exp(X)-1 accurately
* when |X| is small. For further details on the comparisons,
* see the notes on Step 1 of setox.
*
* Step 2. Calculate N = round-to-nearest-int( X * 64/log2 ).
* 2.1 N := round-to-nearest-integer( X * 64/log2 ).
* 2.2 Calculate J = N mod 64; so J = 0,1,2,..., or 63.
* 2.3 Calculate M = (N - J)/64; so N = 64M + J.
* 2.4 Calculate the address of the stored value of 2^(J/64).
* 2.5 Create the values Sc = 2^M and OnebySc := -2^(-M).
* Notes: See the notes on Step 2 of setox.
*
* Step 3. Calculate X - N*log2/64.
* 3.1 R := X + N*L1, where L1 := single-precision(-log2/64).
* 3.2 R := R + N*L2, L2 := extended-precision(-log2/64 - L1).
* Notes: Applying the analysis of Step 3 of setox in this case
* shows that |R| <= 0.0055 (note that |X| <= 70 log2 in
* this case).
*
* Step 4. Approximate exp(R)-1 by a polynomial
* p = R+R*R*(A1+R*(A2+R*(A3+R*(A4+R*(A5+R*A6)))))
* Notes: a) In order to reduce memory access, the coefficients are
* made as "short" as possible: A1 (which is 1/2), A5 and A6
* are single precision; A2, A3 and A4 are double precision.
* b) Even with the restriction above,
* |p - (exp(R)-1)| < |R| * 2^(-72.7)
* for all |R| <= 0.0055.
* c) To fully use the pipeline, p is separated into
* two independent pieces of roughly equal complexity
* p = [ R*S*(A2 + S*(A4 + S*A6)) ] +
* [ R + S*(A1 + S*(A3 + S*A5)) ]
* where S = R*R.
*
* Step 5. Compute 2^(J/64)*p by
* p := T*p
* where T and t are the stored values for 2^(J/64).
* Notes: 2^(J/64) is stored as T and t where T+t approximates
* 2^(J/64) to roughly 85 bits; T is in extended precision
* and t is in single precision. Note also that T is rounded
* to 62 bits so that the last two bits of T are zero. The
* reason for such a special form is that T-1, T-2, and T-8
* will all be exact --- a property that will be exploited
* in Step 6 below. The total relative error in p is no
* bigger than 2^(-67.7) compared to the final result.
*
* Step 6. Reconstruction of exp(X)-1
* exp(X)-1 = 2^M * ( 2^(J/64) + p - 2^(-M) ).
* 6.1 If M <= 63, go to Step 6.3.
* 6.2 ans := T + (p + (t + OnebySc)). Go to 6.6
* 6.3 If M >= -3, go to 6.5.
* 6.4 ans := (T + (p + t)) + OnebySc. Go to 6.6
* 6.5 ans := (T + OnebySc) + (p + t).
* 6.6 Restore user FPCR.
* 6.7 Return ans := Sc * ans. Exit.
* Notes: The various arrangements of the expressions give accurate
* evaluations.
*
* Step 7. exp(X)-1 for |X| < 1/4.
* 7.1 If |X| >= 2^(-65), go to Step 9.
* 7.2 Go to Step 8.
*
* Step 8. Calculate exp(X)-1, |X| < 2^(-65).
* 8.1 If |X| < 2^(-16312), goto 8.3
* 8.2 Restore FPCR; return ans := X - 2^(-16382). Exit.
* 8.3 X := X * 2^(140).
* 8.4 Restore FPCR; ans := ans - 2^(-16382).
* Return ans := ans*2^(140). Exit
* Notes: The idea is to return "X - tiny" under the user
* precision and rounding modes. To avoid unnecessary
* inefficiency, we stay away from denormalized numbers the
* best we can. For |X| >= 2^(-16312), the straightforward
* 8.2 generates the inexact exception as the case warrants.
*
* Step 9. Calculate exp(X)-1, |X| < 1/4, by a polynomial
* p = X + X*X*(B1 + X*(B2 + ... + X*B12))
* Notes: a) In order to reduce memory access, the coefficients are
* made as "short" as possible: B1 (which is 1/2), B9 to B12
* are single precision; B3 to B8 are double precision; and
* B2 is double extended.
* b) Even with the restriction above,
* |p - (exp(X)-1)| < |X| 2^(-70.6)
* for all |X| <= 0.251.
* Note that 0.251 is slightly bigger than 1/4.
* c) To fully preserve accuracy, the polynomial is computed
* as X + ( S*B1 + Q ) where S = X*X and
* Q = X*S*(B2 + X*(B3 + ... + X*B12))
* d) To fully use the pipeline, Q is separated into
* two independent pieces of roughly equal complexity
* Q = [ X*S*(B2 + S*(B4 + ... + S*B12)) ] +
* [ S*S*(B3 + S*(B5 + ... + S*B11)) ]
*
* Step 10. Calculate exp(X)-1 for |X| >= 70 log 2.
* 10.1 If X >= 70log2 , exp(X) - 1 = exp(X) for all practical
* purposes. Therefore, go to Step 1 of setox.
* 10.2 If X <= -70log2, exp(X) - 1 = -1 for all practical purposes.
* ans := -1
* Restore user FPCR
* Return ans := ans + 2^(-126). Exit.
* Notes: 10.2 will always create an inexact and return -1 + tiny
* in the user rounding precision and mode.
*
setox IDNT 2,1 Motorola 040 Floating Point Software Package
xdef setoxd
setoxd:
*--entry point for EXP(X), X is denormalized
MOVE.L (a0),d0
ANDI.L #$80000000,d0
ORI.L #$00800000,d0 ...sign(X)*2^(-126)
MOVE.L d0,-(sp)
FMOVE.S #:3F800000,fp0
fmove.l d1,fpcr
FADD.S (sp)+,fp0
bra t_frcinx
xdef setox
setox:
*--entry point for EXP(X), here X is finite, non-zero, and not NaN's
*--Step 1.
MOVE.L (a0),d0 ...load part of input X
ANDI.L #$7FFF0000,d0 ...biased expo. of X
CMPI.L #$3FBE0000,d0 ...2^(-65)
BGE.B EXPC1 ...normal case
BRA.W EXPSM
EXPC1:
*--The case |X| >= 2^(-65)
MOVE.W 4(a0),d0 ...expo. and partial sig. of |X|
CMPI.L #$400CB167,d0 ...16380 log2 trunc. 16 bits
BLT.B EXPMAIN ...normal case
BRA.W EXPBIG
EXPMAIN:
*--Step 2.
*--This is the normal branch: 2^(-65) <= |X| < 16380 log2.
FMOVE.X (a0),fp0 ...load input from (a0)
FMOVE.X fp0,fp1
FMUL.S #:42B8AA3B,fp0 ...64/log2 * X
fmovem.x fp2/fp3,-(a7) ...save fp2
CLR.L ADJFLAG(a6)
FMOVE.L fp0,d0 ...N = int( X * 64/log2 )
LEA EXPTBL,a1
FMOVE.L d0,fp0 ...convert to floating-format
MOVE.L d0,L_SCR1(a6) ...save N temporarily
ANDI.L #$3F,d0 ...D0 is J = N mod 64
LSL.L #4,d0
ADDA.L d0,a1 ...address of 2^(J/64)
MOVE.L L_SCR1(a6),d0
ASR.L #6,d0 ...D0 is M
ADDI.W #$3FFF,d0 ...biased expo. of 2^(M)
MOVE.W L2,L_SCR1(a6) ...prefetch L2, no need in CB
EXPCONT1:
*--Step 3.
*--fp1,fp2 saved on the stack. fp0 is N, fp1 is X,
*--a0 points to 2^(J/64), D0 is biased expo. of 2^(M)
FMOVE.X fp0,fp2
FMUL.S #:BC317218,fp0 ...N * L1, L1 = lead(-log2/64)
FMUL.X L2,fp2 ...N * L2, L1+L2 = -log2/64
FADD.X fp1,fp0 ...X + N*L1
FADD.X fp2,fp0 ...fp0 is R, reduced arg.
* MOVE.W #$3FA5,EXPA3 ...load EXPA3 in cache
*--Step 4.
*--WE NOW COMPUTE EXP(R)-1 BY A POLYNOMIAL
*-- R + R*R*(A1 + R*(A2 + R*(A3 + R*(A4 + R*A5))))
*--TO FULLY USE THE PIPELINE, WE COMPUTE S = R*R
*--[R+R*S*(A2+S*A4)] + [S*(A1+S*(A3+S*A5))]
FMOVE.X fp0,fp1
FMUL.X fp1,fp1 ...fp1 IS S = R*R
FMOVE.S #:3AB60B70,fp2 ...fp2 IS A5
* CLR.W 2(a1) ...load 2^(J/64) in cache
FMUL.X fp1,fp2 ...fp2 IS S*A5
FMOVE.X fp1,fp3
FMUL.S #:3C088895,fp3 ...fp3 IS S*A4
FADD.D EXPA3,fp2 ...fp2 IS A3+S*A5
FADD.D EXPA2,fp3 ...fp3 IS A2+S*A4
FMUL.X fp1,fp2 ...fp2 IS S*(A3+S*A5)
MOVE.W d0,SCALE(a6) ...SCALE is 2^(M) in extended
clr.w SCALE+2(a6)
move.l #$80000000,SCALE+4(a6)
clr.l SCALE+8(a6)
FMUL.X fp1,fp3 ...fp3 IS S*(A2+S*A4)
FADD.S #:3F000000,fp2 ...fp2 IS A1+S*(A3+S*A5)
FMUL.X fp0,fp3 ...fp3 IS R*S*(A2+S*A4)
FMUL.X fp1,fp2 ...fp2 IS S*(A1+S*(A3+S*A5))
FADD.X fp3,fp0 ...fp0 IS R+R*S*(A2+S*A4),
* ...fp3 released
FMOVE.X (a1)+,fp1 ...fp1 is lead. pt. of 2^(J/64)
FADD.X fp2,fp0 ...fp0 is EXP(R) - 1
* ...fp2 released
xdef setoxm1d
setoxm1d:
*--entry point for EXPM1(X), here X is denormalized
*--Step 0.
bra t_extdnrm
xdef setoxm1
setoxm1:
*--entry point for EXPM1(X), here X is finite, non-zero, non-NaN
*--Step 1.
*--Step 1.1
MOVE.L (a0),d0 ...load part of input X
ANDI.L #$7FFF0000,d0 ...biased expo. of X
CMPI.L #$3FFD0000,d0 ...1/4
BGE.B EM1CON1 ...|X| >= 1/4
BRA.W EM1SM
EM1CON1:
*--Step 1.3
*--The case |X| >= 1/4
MOVE.W 4(a0),d0 ...expo. and partial sig. of |X|
CMPI.L #$4004C215,d0 ...70log2 rounded up to 16 bits
BLE.B EM1MAIN ...1/4 <= |X| <= 70log2
BRA.W EM1BIG
EM1MAIN:
*--Step 2.
*--This is the case: 1/4 <= |X| <= 70 log2.
FMOVE.X (a0),fp0 ...load input from (a0)
FMOVE.X fp0,fp1
FMUL.S #:42B8AA3B,fp0 ...64/log2 * X
fmovem.x fp2/fp3,-(a7) ...save fp2
* MOVE.W #$3F81,EM1A4 ...prefetch in CB mode
FMOVE.L fp0,d0 ...N = int( X * 64/log2 )
LEA EXPTBL,a1
FMOVE.L d0,fp0 ...convert to floating-format
MOVE.L d0,L_SCR1(a6) ...save N temporarily
ANDI.L #$3F,d0 ...D0 is J = N mod 64
LSL.L #4,d0
ADDA.L d0,a1 ...address of 2^(J/64)
MOVE.L L_SCR1(a6),d0
ASR.L #6,d0 ...D0 is M
MOVE.L d0,L_SCR1(a6) ...save a copy of M
* MOVE.W #$3FDC,L2 ...prefetch L2 in CB mode
*--Step 3.
*--fp1,fp2 saved on the stack. fp0 is N, fp1 is X,
*--a0 points to 2^(J/64), D0 and a1 both contain M
FMOVE.X fp0,fp2
FMUL.S #:BC317218,fp0 ...N * L1, L1 = lead(-log2/64)
FMUL.X L2,fp2 ...N * L2, L1+L2 = -log2/64
FADD.X fp1,fp0 ...X + N*L1
FADD.X fp2,fp0 ...fp0 is R, reduced arg.
* MOVE.W #$3FC5,EM1A2 ...load EM1A2 in cache
ADDI.W #$3FFF,d0 ...D0 is biased expo. of 2^M
*--Step 4.
*--WE NOW COMPUTE EXP(R)-1 BY A POLYNOMIAL
*-- R + R*R*(A1 + R*(A2 + R*(A3 + R*(A4 + R*(A5 + R*A6)))))
*--TO FULLY USE THE PIPELINE, WE COMPUTE S = R*R
*--[R*S*(A2+S*(A4+S*A6))] + [R+S*(A1+S*(A3+S*A5))]
FMOVE.X fp0,fp1
FMUL.X fp1,fp1 ...fp1 IS S = R*R
FMOVE.S #:3950097B,fp2 ...fp2 IS a6
* CLR.W 2(a1) ...load 2^(J/64) in cache
FMUL.X fp1,fp2 ...fp2 IS S*A6
FMOVE.X fp1,fp3
FMUL.S #:3AB60B6A,fp3 ...fp3 IS S*A5
FADD.D EM1A4,fp2 ...fp2 IS A4+S*A6
FADD.D EM1A3,fp3 ...fp3 IS A3+S*A5
MOVE.W d0,SC(a6) ...SC is 2^(M) in extended
clr.w SC+2(a6)
move.l #$80000000,SC+4(a6)
clr.l SC+8(a6)
FMUL.X fp1,fp2 ...fp2 IS S*(A4+S*A6)
MOVE.L L_SCR1(a6),d0 ...D0 is M
NEG.W D0 ...D0 is -M
FMUL.X fp1,fp3 ...fp3 IS S*(A3+S*A5)
ADDI.W #$3FFF,d0 ...biased expo. of 2^(-M)
FADD.D EM1A2,fp2 ...fp2 IS A2+S*(A4+S*A6)
FADD.S #:3F000000,fp3 ...fp3 IS A1+S*(A3+S*A5)
FMUL.X fp1,fp2 ...fp2 IS S*(A2+S*(A4+S*A6))
ORI.W #$8000,d0 ...signed/expo. of -2^(-M)
MOVE.W d0,ONEBYSC(a6) ...OnebySc is -2^(-M)
clr.w ONEBYSC+2(a6)
move.l #$80000000,ONEBYSC+4(a6)
clr.l ONEBYSC+8(a6)
FMUL.X fp3,fp1 ...fp1 IS S*(A1+S*(A3+S*A5))
* ...fp3 released
FMUL.X fp0,fp2 ...fp2 IS R*S*(A2+S*(A4+S*A6))
FADD.X fp1,fp0 ...fp0 IS R+S*(A1+S*(A3+S*A5))
* ...fp1 released
FADD.X fp2,fp0 ...fp0 IS EXP(R)-1
* ...fp2 released
fmovem.x (a7)+,fp2/fp3 ...fp2 restored
*--Step 5
*--Compute 2^(J/64)*p
FMUL.X (a1),fp0 ...2^(J/64)*(Exp(R)-1)
*--Step 6
*--Step 6.1
MOVE.L L_SCR1(a6),d0 ...retrieve M
CMPI.L #63,d0
BLE.B MLE63
*--Step 6.2 M >= 64
FMOVE.S 12(a1),fp1 ...fp1 is t
FADD.X ONEBYSC(a6),fp1 ...fp1 is t+OnebySc
FADD.X fp1,fp0 ...p+(t+OnebySc), fp1 released
FADD.X (a1),fp0 ...T+(p+(t+OnebySc))
BRA.B EM1SCALE
MLE63:
*--Step 6.3 M <= 63
CMPI.L #-3,d0
BGE.B MGEN3
MLTN3:
*--Step 6.4 M <= -4
FADD.S 12(a1),fp0 ...p+t
FADD.X (a1),fp0 ...T+(p+t)
FADD.X ONEBYSC(a6),fp0 ...OnebySc + (T+(p+t))
BRA.B EM1SCALE
MGEN3:
*--Step 6.5 -3 <= M <= 63
FMOVE.X (a1)+,fp1 ...fp1 is T
FADD.S (a1),fp0 ...fp0 is p+t
FADD.X ONEBYSC(a6),fp1 ...fp1 is T+OnebySc
FADD.X fp1,fp0 ...(T+OnebySc)+(p+t)
EM1POLY:
*--Step 9 exp(X)-1 by a simple polynomial
FMOVE.X (a0),fp0 ...fp0 is X
FMUL.X fp0,fp0 ...fp0 is S := X*X
fmovem.x fp2/fp3,-(a7) ...save fp2
FMOVE.S #:2F30CAA8,fp1 ...fp1 is B12
FMUL.X fp0,fp1 ...fp1 is S*B12
FMOVE.S #:310F8290,fp2 ...fp2 is B11
FADD.S #:32D73220,fp1 ...fp1 is B10+S*B12
FMUL.X fp0,fp2 ...fp2 is S*B11
FMUL.X fp0,fp1 ...fp1 is S*(B10 + ...
FADD.S #:3493F281,fp2 ...fp2 is B9+S*...
FADD.D EM1B8,fp1 ...fp1 is B8+S*...
FMUL.X fp0,fp2 ...fp2 is S*(B9+...
FMUL.X fp0,fp1 ...fp1 is S*(B8+...
FADD.D EM1B7,fp2 ...fp2 is B7+S*...
FADD.D EM1B6,fp1 ...fp1 is B6+S*...
FMUL.X fp0,fp2 ...fp2 is S*(B7+...
FMUL.X fp0,fp1 ...fp1 is S*(B6+...
FADD.D EM1B5,fp2 ...fp2 is B5+S*...
FADD.D EM1B4,fp1 ...fp1 is B4+S*...
FMUL.X fp0,fp2 ...fp2 is S*(B5+...
FMUL.X fp0,fp1 ...fp1 is S*(B4+...
FADD.D EM1B3,fp2 ...fp2 is B3+S*...
FADD.X EM1B2,fp1 ...fp1 is B2+S*...
FMUL.X fp0,fp2 ...fp2 is S*(B3+...
FMUL.X fp0,fp1 ...fp1 is S*(B2+...
FMUL.X fp0,fp2 ...fp2 is S*S*(B3+...)
FMUL.X (a0),fp1 ...fp1 is X*S*(B2...
FMUL.S #:3F000000,fp0 ...fp0 is S*B1
FADD.X fp2,fp1 ...fp1 is Q
* ...fp2 released
