ifpsp060/src/fpsp.S

17 IN NO EVENT SHALL MOTOROLA BE LIABLE FOR ANY DAMAGES WHATSOEVER
5166 # ...WHICH IS IN TWO PIECES Y1 & Y2
5186 #--NOTE THAT A3 THROUGH A7 ARE STORED IN DOUBLE PRECISION
5187 #--WHILE A1 AND A2 ARE IN DOUBLE-EXTENDED FORMAT.
5202 # ...LEAST SIG. BIT OF D0 IN SIGN POSITION
5239 #--NOTE THAT B4 THROUGH B8 ARE STORED IN DOUBLE PRECISION
5240 #--WHILE B2 AND B3 ARE IN DOUBLE-EXTENDED FORMAT, B1 IS -1/2
5256 # ...LEAST SIG. BIT OF D0 IN SIGN POSITION
5265 	or.l		&0x3F800000,%d1		# D0 IS SGN IN SINGLE
5312 #	mov.w		&0x0000,XDCARE(%a6)	# JUST IN CASE
5375 	add.l		%d1,%a1			# ADDRESS OF N*PIBY2, IN Y1, Y2
5570 #--THE REMAINDER INSTRUCTION WHICH IS NOW IN SOFTWARE.
5641 #--WE MUST NOW FIND INT(FP2). SINCE WE NEED THIS VALUE IN
5645 #--US THE DESIRED VALUE IN FLOATING POINT.
5649 	or.l		&0x5F000000,%d2		# d2 = SIGN(INARG)*2**63 IN SGL
5787 #--N*PI/2, -32 <= N <= 32, IN A LEADING TERM IN EXT. AND TRAILING
5788 #--TERM IN SGL. NOTE THAT PI IS 64-BIT LONG, THUS N*PI/2 IS AT
5891 	add.l		%d1,%a1			# ADDRESS N*PIBY2 IN Y1, Y2
6001 #--THE REMAINDER INSTRUCTION WHICH IS NOW IN SOFTWARE.
6072 #--WE MUST NOW FIND INT(FP2). SINCE WE NEED THIS VALUE IN
6076 #--US THE DESIRED VALUE IN FLOATING POINT.
6080 	or.l		&0x5F000000,%d2		# d2 = SIGN(INARG)*2**63 IN SGL
6364 #--THE MOST LIKELY CASE, |X| IN [1/16, 16). WE USE TABLE TECHNIQUE
6366 #--SO IF F IS CHOSEN TO BE CLOSE TO X AND ATAN(F) IS STORED IN
6372 #--DIVIDE. IN THE END THIS METHOD IS MUCH FASTER THAN A TRADITIONAL
6398 #--CREATE ATAN(F) AND STORE IT IN ATANF, AND
6422 #--U IN FP0, WE ARE NOW READY TO COMPUTE ATAN(U) AS
6450 #--|X| IS IN d0 IN COMPACT FORM. FP1, d0 SAVED.
6675 	or.l		&0x3F800000,%d1		# +-1 IN SGL FORMAT
6676 	mov.l		%d1,-(%sp)		# push SIGN(X) IN SGL-FMT
8030 	fmov.s		%d1,%fp1		# -SIGN(X)*2 IN SGL FMT
8035 	fmov.s		%d1,%fp0		# SGN IN SGL FMT
8345 #--X = 2^(K) * Y, 1 <= Y < 2. THUS, Y = 1.XXXXXXXX....XX IN BINARY.
8351 #--ALSO NOTE THAT THE VALUE 1/F IS STORED IN A TABLE SO THAT NO
8388 	fmov.x		%fp1,KLOG2(%a6)		# PUT K*LOG2 IN MEMEORY, FREE FP1
8439 #--IN U, U = 2(X-1)/(X+1) = FP1/FP0
8451 	fmov.x		%fp1,SAVEU(%a6)		# STORE U IN MEMORY, FREE FP1
8572 #--CONTAINS AT LEAST 63 BITS OF INFORMATION OF Z. IN THAT CASE,
8593 #--PRESERVE ALL THE INFORMATION. BECAUSE 1+Z IS IN [1/2,3/2],
8597 #--ON RETURNING TO LP1CONT1, WE MUST HAVE K IN FP1, ADDRESS OF
8598 #--(1/F) IN A0, Y-F IN FP0, AND FP2 SAVED.
8642 #--FPCR SAVED. D0 IS X IN COMPACT FORM.
9131 #--REGISTERS SAVED SO FAR ARE (IN ORDER) FPCR, D0, FP1, a1, AND FP2.
9228 #--REGISTERS SAVED SO FAR ARE (IN ORDER) FPCR, D0, FP1, a1, AND FP2.
9254 #--FPCR, FP2, FP3 ARE SAVED IN ORDER AS SHOWN.
12070 # FP MOVE IN: NORMs and DENORMs ONLY!
18349 # MOVE IN: #