* 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.
*
* THE SOFTWARE is provided on an "AS IS" basis and without warranty.
* To the maximum extent permitted by applicable law,
* MOTOROLA DISCLAIMS ALL WARRANTIES WHETHER EXPRESS OR IMPLIED,
* INCLUDING IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A
* PARTICULAR PURPOSE and any warranty against infringement with
* regard to the SOFTWARE (INCLUDING ANY MODIFIED VERSIONS THEREOF)
* and any accompanying written materials.
*
* To the maximum extent permitted by applicable law,
* IN NO EVENT SHALL MOTOROLA BE LIABLE FOR ANY DAMAGES WHATSOEVER
* (INCLUDING WITHOUT LIMITATION, DAMAGES FOR LOSS OF BUSINESS
* PROFITS, BUSINESS INTERRUPTION, LOSS OF BUSINESS INFORMATION, OR
* OTHER PECUNIARY LOSS) ARISING OF THE USE OR INABILITY TO USE THE
* SOFTWARE. Motorola assumes no responsibility for the maintenance
* and support of the SOFTWARE.
*
* You are hereby granted a copyright license to use, modify, and
* distribute the SOFTWARE so long as this entire notice is retained
* without alteration in any modified and/or redistributed versions,
* and that such modified versions are clearly identified as such.
* No licenses are granted by implication, estoppel or otherwise
* under any patents or trademarks of Motorola, Inc.
*
* ssin.sa 3.3 7/29/91
*
* The entry point sSIN computes the sine of an input argument
* sCOS computes the cosine, and sSINCOS computes both. The
* corresponding entry points with a "d" computes the same
* corresponding function values for denormalized inputs.
*
* Input: Double-extended number X in location pointed to
* by address register a0.
*
* Output: The function value sin(X) or cos(X) returned in Fp0 if SIN or
* COS is requested. Otherwise, for SINCOS, sin(X) is returned
* in Fp0, and cos(X) is returned in Fp1.
*
* Modifies: Fp0 for SIN or COS; both Fp0 and Fp1 for SINCOS.
*
* Accuracy and Monotonicity: The returned result is within 1 ulp 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: The programs sSIN and sCOS take approximately 150 cycles for
* input argument X such that |X| < 15Pi, which is the usual
* situation. The speed for sSINCOS is approximately 190 cycles.
*
* Algorithm:
*
* SIN and COS:
* 1. If SIN is invoked, set AdjN := 0; otherwise, set AdjN := 1.
*
* 2. If |X| >= 15Pi or |X| < 2**(-40), go to 7.
*
* 3. Decompose X as X = N(Pi/2) + r where |r| <= Pi/4. Let
* k = N mod 4, so in particular, k = 0,1,2,or 3. Overwrite
* k by k := k + AdjN.
*
* 4. If k is even, go to 6.
*
* 5. (k is odd) Set j := (k-1)/2, sgn := (-1)**j. Return sgn*cos(r)
* where cos(r) is approximated by an even polynomial in r,
* 1 + r*r*(B1+s*(B2+ ... + s*B8)), s = r*r.
* Exit.
*
* 6. (k is even) Set j := k/2, sgn := (-1)**j. Return sgn*sin(r)
* where sin(r) is approximated by an odd polynomial in r
* r + r*s*(A1+s*(A2+ ... + s*A7)), s = r*r.
* Exit.
*
* 7. If |X| > 1, go to 9.
*
* 8. (|X|<2**(-40)) If SIN is invoked, return X; otherwise return 1.
*
* 9. Overwrite X by X := X rem 2Pi. Now that |X| <= Pi, go back to 3.
*
* SINCOS:
* 1. If |X| >= 15Pi or |X| < 2**(-40), go to 6.
*
* 2. Decompose X as X = N(Pi/2) + r where |r| <= Pi/4. Let
* k = N mod 4, so in particular, k = 0,1,2,or 3.
*
* 3. If k is even, go to 5.
*
* 4. (k is odd) Set j1 := (k-1)/2, j2 := j1 (EOR) (k mod 2), i.e.
* j1 exclusive or with the l.s.b. of k.
* sgn1 := (-1)**j1, sgn2 := (-1)**j2.
* SIN(X) = sgn1 * cos(r) and COS(X) = sgn2*sin(r) where
* sin(r) and cos(r) are computed as odd and even polynomials
* in r, respectively. Exit
*
* 5. (k is even) Set j1 := k/2, sgn1 := (-1)**j1.
* SIN(X) = sgn1 * sin(r) and COS(X) = sgn1*cos(r) where
* sin(r) and cos(r) are computed as odd and even polynomials
* in r, respectively. Exit
*
* 6. If |X| > 1, go to 8.
*
* 7. (|X|<2**(-40)) SIN(X) = X and COS(X) = 1. Exit.
*
* 8. Overwrite X by X := X rem 2Pi. Now that |X| <= Pi, go back to 2.
*
SSIN IDNT 2,1 Motorola 040 Floating Point Software Package
SINMAIN:
*--THIS IS THE USUAL CASE, |X| <= 15 PI.
*--THE ARGUMENT REDUCTION IS DONE BY TABLE LOOK UP.
FMOVE.X FP0,FP1
FMUL.D TWOBYPI,FP1 ...X*2/PI
*--HIDE THE NEXT THREE INSTRUCTIONS
LEA PITBL+$200,A1 ...TABLE OF N*PI/2, N = -32,...,32
*--FP1 IS NOW READY
FMOVE.L FP1,N(a6) ...CONVERT TO INTEGER
MOVE.L N(a6),D0
ASL.L #4,D0
ADDA.L D0,A1 ...A1 IS THE ADDRESS OF N*PIBY2
* ...WHICH IS IN TWO PIECES Y1 & Y2
FSUB.X (A1)+,FP0 ...X-Y1
*--HIDE THE NEXT ONE
FSUB.S (A1),FP0 ...FP0 IS R = (X-Y1)-Y2
SINCONT:
*--continuation from REDUCEX
*--GET N+ADJN AND SEE IF SIN(R) OR COS(R) IS NEEDED
MOVE.L N(a6),D0
ADD.L ADJN(a6),D0 ...SEE IF D0 IS ODD OR EVEN
ROR.L #1,D0 ...D0 WAS ODD IFF D0 IS NEGATIVE
TST.L D0
BLT.W COSPOLY
SINPOLY:
*--LET J BE THE LEAST SIG. BIT OF D0, LET SGN := (-1)**J.
*--THEN WE RETURN SGN*SIN(R). SGN*SIN(R) IS COMPUTED BY
*--R' + R'*S*(A1 + S(A2 + S(A3 + S(A4 + ... + SA7)))), WHERE
*--R' = SGN*R, S=R*R. THIS CAN BE REWRITTEN AS
*--R' + R'*S*( [A1+T(A3+T(A5+TA7))] + [S(A2+T(A4+TA6))])
*--WHERE T=S*S.
*--NOTE THAT A3 THROUGH A7 ARE STORED IN DOUBLE PRECISION
*--WHILE A1 AND A2 ARE IN DOUBLE-EXTENDED FORMAT.
FMOVE.X FP0,X(a6) ...X IS R
FMUL.X FP0,FP0 ...FP0 IS S
*---HIDE THE NEXT TWO WHILE WAITING FOR FP0
FMOVE.D SINA7,FP3
FMOVE.D SINA6,FP2
*--FP0 IS NOW READY
FMOVE.X FP0,FP1
FMUL.X FP1,FP1 ...FP1 IS T
*--HIDE THE NEXT TWO WHILE WAITING FOR FP1
ROR.L #1,D0
ANDI.L #$80000000,D0
* ...LEAST SIG. BIT OF D0 IN SIGN POSITION
EOR.L D0,X(a6) ...X IS NOW R'= SGN*R
FADD.X FP2,FP1 ...[A1+T(A3+T(A5+TA7))]+[S(A2+T(A4+TA6))]
*--FP3 RELEASED, RESTORE NOW AND TAKE SOME ADVANTAGE OF HIDING
*--FP2 RELEASED, RESTORE NOW AND TAKE FULL ADVANTAGE OF HIDING
FMUL.X FP1,FP0 ...SIN(R')-R'
*--FP1 RELEASED.
FMOVE.L d1,FPCR ;restore users exceptions
FADD.X X(a6),FP0 ;last inst - possible exception set
bra t_frcinx
COSPOLY:
*--LET J BE THE LEAST SIG. BIT OF D0, LET SGN := (-1)**J.
*--THEN WE RETURN SGN*COS(R). SGN*COS(R) IS COMPUTED BY
*--SGN + S'*(B1 + S(B2 + S(B3 + S(B4 + ... + SB8)))), WHERE
*--S=R*R AND S'=SGN*S. THIS CAN BE REWRITTEN AS
*--SGN + S'*([B1+T(B3+T(B5+TB7))] + [S(B2+T(B4+T(B6+TB8)))])
*--WHERE T=S*S.
*--NOTE THAT B4 THROUGH B8 ARE STORED IN DOUBLE PRECISION
*--WHILE B2 AND B3 ARE IN DOUBLE-EXTENDED FORMAT, B1 IS -1/2
*--AND IS THEREFORE STORED AS SINGLE PRECISION.
FMUL.X FP0,FP0 ...FP0 IS S
*---HIDE THE NEXT TWO WHILE WAITING FOR FP0
FMOVE.D COSB8,FP2
FMOVE.D COSB7,FP3
*--FP0 IS NOW READY
FMOVE.X FP0,FP1
FMUL.X FP1,FP1 ...FP1 IS T
*--HIDE THE NEXT TWO WHILE WAITING FOR FP1
FMOVE.X FP0,X(a6) ...X IS S
ROR.L #1,D0
ANDI.L #$80000000,D0
* ...LEAST SIG. BIT OF D0 IN SIGN POSITION
FMUL.X FP1,FP2 ...TB8
*--HIDE THE NEXT TWO WHILE WAITING FOR THE XU
EOR.L D0,X(a6) ...X IS NOW S'= SGN*S
ANDI.L #$80000000,D0
FMUL.X FP1,FP3 ...TB7
*--HIDE THE NEXT TWO WHILE WAITING FOR THE XU
ORI.L #$3F800000,D0 ...D0 IS SGN IN SINGLE
MOVE.L D0,POSNEG1(a6)
FMUL.X FP2,FP0 ...S(B2+T(B4+T(B6+TB8)))
*--FP3 RELEASED, RESTORE NOW AND TAKE SOME ADVANTAGE OF HIDING
*--FP2 RELEASED.
FADD.X FP1,FP0
*--FP1 RELEASED
FMUL.X X(a6),FP0
FMOVE.L d1,FPCR ;restore users exceptions
FADD.S POSNEG1(a6),FP0 ;last inst - possible exception set
bra t_frcinx
SINBORS:
*--IF |X| > 15PI, WE USE THE GENERAL ARGUMENT REDUCTION.
*--IF |X| < 2**(-40), RETURN X OR 1.
CMPI.L #$3FFF8000,D0
BGT.B REDUCEX
SINSM:
MOVE.L ADJN(a6),D0
TST.L D0
BGT.B COSTINY
SINTINY:
CLR.W XDCARE(a6) ...JUST IN CASE
FMOVE.L d1,FPCR ;restore users exceptions
FMOVE.X X(a6),FP0 ;last inst - possible exception set
bra t_frcinx
COSTINY:
FMOVE.S #:3F800000,FP0
FMOVE.L d1,FPCR ;restore users exceptions
FSUB.S #:00800000,FP0 ;last inst - possible exception set
bra t_frcinx
REDUCEX:
*--WHEN REDUCEX IS USED, THE CODE WILL INEVITABLY BE SLOW.
*--THIS REDUCTION METHOD, HOWEVER, IS MUCH FASTER THAN USING
*--THE REMAINDER INSTRUCTION WHICH IS NOW IN SOFTWARE.
FMOVEM.X FP2-FP5,-(A7) ...save FP2 through FP5
MOVE.L D2,-(A7)
FMOVE.S #:00000000,FP1
*--If compact form of abs(arg) in d0=$7ffeffff, argument is so large that
*--there is a danger of unwanted overflow in first LOOP iteration. In this
*--case, reduce argument by one remainder step to make subsequent reduction
*--safe.
cmpi.l #$7ffeffff,d0 ;is argument dangerously large?
bne.b LOOP
move.l #$7ffe0000,FP_SCR2(a6) ;yes
* ;create 2**16383*PI/2
move.l #$c90fdaa2,FP_SCR2+4(a6)
clr.l FP_SCR2+8(a6)
ftst.x fp0 ;test sign of argument
move.l #$7fdc0000,FP_SCR3(a6) ;create low half of 2**16383*
* ;PI/2 at FP_SCR3
move.l #$85a308d3,FP_SCR3+4(a6)
clr.l FP_SCR3+8(a6)
fblt.w red_neg
or.w #$8000,FP_SCR2(a6) ;positive arg
or.w #$8000,FP_SCR3(a6)
red_neg:
fadd.x FP_SCR2(a6),fp0 ;high part of reduction is exact
fmove.x fp0,fp1 ;save high result in fp1
fadd.x FP_SCR3(a6),fp0 ;low part of reduction
fsub.x fp0,fp1 ;determine low component of result
fadd.x FP_SCR3(a6),fp1 ;fp0/fp1 are reduced argument.
*--ON ENTRY, FP0 IS X, ON RETURN, FP0 IS X REM PI/2, |X| <= PI/4.
*--integer quotient will be stored in N
*--Intermeditate remainder is 66-bit long; (R,r) in (FP0,FP1)
LOOP:
FMOVE.X FP0,INARG(a6) ...+-2**K * F, 1 <= F < 2
MOVE.W INARG(a6),D0
MOVE.L D0,A1 ...save a copy of D0
ANDI.L #$00007FFF,D0
SUBI.L #$00003FFF,D0 ...D0 IS K
CMPI.L #28,D0
BLE.B LASTLOOP
CONTLOOP:
SUBI.L #27,D0 ...D0 IS L := K-27
CLR.L ENDFLAG(a6)
BRA.B WORK
LASTLOOP:
CLR.L D0 ...D0 IS L := 0
MOVE.L #1,ENDFLAG(a6)
WORK:
*--FIND THE REMAINDER OF (R,r) W.R.T. 2**L * (PI/2). L IS SO CHOSEN
*--THAT INT( X * (2/PI) / 2**(L) ) < 2**29.
MOVE.L #$00003FFE,D2 ...BIASED EXPO OF 2/PI
SUB.L D0,D2 ...BIASED EXPO OF 2**(-L)*(2/PI)
MOVE.L #$A2F9836E,FP_SCR1+4(a6)
MOVE.L #$4E44152A,FP_SCR1+8(a6)
MOVE.W D2,FP_SCR1(a6) ...FP_SCR1 is 2**(-L)*(2/PI)
FMOVE.X FP0,FP2
FMUL.X FP_SCR1(a6),FP2
*--WE MUST NOW FIND INT(FP2). SINCE WE NEED THIS VALUE IN
*--FLOATING POINT FORMAT, THE TWO FMOVE'S FMOVE.L FP <--> N
*--WILL BE TOO INEFFICIENT. THE WAY AROUND IT IS THAT
*--(SIGN(INARG)*2**63 + FP2) - SIGN(INARG)*2**63 WILL GIVE
*--US THE DESIRED VALUE IN FLOATING POINT.
*--HIDE SIX CYCLES OF INSTRUCTION
MOVE.L A1,D2
SWAP D2
ANDI.L #$80000000,D2
ORI.L #$5F000000,D2 ...D2 IS SIGN(INARG)*2**63 IN SGL
MOVE.L D2,TWOTO63(a6)
MOVE.L D0,D2
ADDI.L #$00003FFF,D2 ...BIASED EXPO OF 2**L * (PI/2)
*--FP2 IS READY
FADD.S TWOTO63(a6),FP2 ...THE FRACTIONAL PART OF FP1 IS ROUNDED
*--HIDE 4 CYCLES OF INSTRUCTION; creating 2**(L)*Piby2_1 and 2**(L)*Piby2_2
MOVE.W D2,FP_SCR2(a6)
CLR.W FP_SCR2+2(a6)
MOVE.L #$C90FDAA2,FP_SCR2+4(a6)
CLR.L FP_SCR2+8(a6) ...FP_SCR2 is 2**(L) * Piby2_1
*--FP2 IS READY
FSUB.S TWOTO63(a6),FP2 ...FP2 is N
*--We are now ready to perform (R+r) - N*P1 - N*P2, P1 = 2**(L) * Piby2_1 and
*--P2 = 2**(L) * Piby2_2
FMOVE.X FP2,FP4
FMul.X FP_SCR2(a6),FP4 ...W = N*P1
FMove.X FP2,FP5
FMul.X FP_SCR3(a6),FP5 ...w = N*P2
FMove.X FP4,FP3
*--we want P+p = W+w but |p| <= half ulp of P
*--Then, we need to compute A := R-P and a := r-p
FAdd.X FP5,FP3 ...FP3 is P
FSub.X FP3,FP4 ...W-P
FSub.X FP3,FP0 ...FP0 is A := R - P
FAdd.X FP5,FP4 ...FP4 is p = (W-P)+w
FMove.X FP0,FP3 ...FP3 A
FSub.X FP4,FP1 ...FP1 is a := r - p
*--Now we need to normalize (A,a) to "new (R,r)" where R+r = A+a but
*--|r| <= half ulp of R.
FAdd.X FP1,FP0 ...FP0 is R := A+a
*--No need to calculate r if this is the last loop
TST.L D0
BGT.W RESTORE
*--Need to calculate r
FSub.X FP0,FP3 ...A-R
FAdd.X FP3,FP1 ...FP1 is r := (A-R)+a
BRA.W LOOP
