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1 * $NetBSD: binstr.sa,v 1.3 1994/10/26 07:48:53 cgd Exp $
3 * MOTOROLA MICROPROCESSOR & MEMORY TECHNOLOGY GROUP
4 * M68000 Hi-Performance Microprocessor Division
5 * M68040 Software Package
6 *
7 * M68040 Software Package Copyright (c) 1993, 1994 Motorola Inc.
8 * All rights reserved.
9 *
10 * THE SOFTWARE is provided on an "AS IS" basis and without warranty.
11 * To the maximum extent permitted by applicable law,
12 * MOTOROLA DISCLAIMS ALL WARRANTIES WHETHER EXPRESS OR IMPLIED,
13 * INCLUDING IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A
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15 * regard to the SOFTWARE (INCLUDING ANY MODIFIED VERSIONS THEREOF)
16 * and any accompanying written materials.
17 *
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19 * IN NO EVENT SHALL MOTOROLA BE LIABLE FOR ANY DAMAGES WHATSOEVER
20 * (INCLUDING WITHOUT LIMITATION, DAMAGES FOR LOSS OF BUSINESS
21 * PROFITS, BUSINESS INTERRUPTION, LOSS OF BUSINESS INFORMATION, OR
22 * OTHER PECUNIARY LOSS) ARISING OF THE USE OR INABILITY TO USE THE
23 * SOFTWARE. Motorola assumes no responsibility for the maintenance
24 * and support of the SOFTWARE.
25 *
26 * You are hereby granted a copyright license to use, modify, and
27 * distribute the SOFTWARE so long as this entire notice is retained
28 * without alteration in any modified and/or redistributed versions,
29 * and that such modified versions are clearly identified as such.
30 * No licenses are granted by implication, estoppel or otherwise
31 * under any patents or trademarks of Motorola, Inc.
33 *
34 * binstr.sa 3.3 12/19/90
35 *
36 *
37 * Description: Converts a 64-bit binary integer to bcd.
38 *
39 * Input: 64-bit binary integer in d2:d3, desired length (LEN) in
40 * d0, and a pointer to start in memory for bcd characters
41 * in d0. (This pointer must point to byte 4 of the first
42 * lword of the packed decimal memory string.)
43 *
44 * Output: LEN bcd digits representing the 64-bit integer.
45 *
46 * Algorithm:
47 * The 64-bit binary is assumed to have a decimal point before
48 * bit 63. The fraction is multiplied by 10 using a mul by 2
49 * shift and a mul by 8 shift. The bits shifted out of the
50 * msb form a decimal digit. This process is iterated until
51 * LEN digits are formed.
52 *
53 * A1. Init d7 to 1. D7 is the byte digit counter, and if 1, the
54 * digit formed will be assumed the least significant. This is
55 * to force the first byte formed to have a 0 in the upper 4 bits.
56 *
57 * A2. Beginning of the loop:
58 * Copy the fraction in d2:d3 to d4:d5.
59 *
60 * A3. Multiply the fraction in d2:d3 by 8 using bit-field
61 * extracts and shifts. The three msbs from d2 will go into
62 * d1.
63 *
64 * A4. Multiply the fraction in d4:d5 by 2 using shifts. The msb
65 * will be collected by the carry.
66 *
67 * A5. Add using the carry the 64-bit quantities in d2:d3 and d4:d5
68 * into d2:d3. D1 will contain the bcd digit formed.
69 *
70 * A6. Test d7. If zero, the digit formed is the ms digit. If non-
71 * zero, it is the ls digit. Put the digit in its place in the
72 * upper word of d0. If it is the ls digit, write the word
73 * from d0 to memory.
74 *
75 * A7. Decrement d6 (LEN counter) and repeat the loop until zero.
76 *
77 * Implementation Notes:
78 *
79 * The registers are used as follows:
80 *
81 * d0: LEN counter
82 * d1: temp used to form the digit
83 * d2: upper 32-bits of fraction for mul by 8
84 * d3: lower 32-bits of fraction for mul by 8
85 * d4: upper 32-bits of fraction for mul by 2
86 * d5: lower 32-bits of fraction for mul by 2
87 * d6: temp for bit-field extracts
88 * d7: byte digit formation word;digit count {0,1}
89 * a0: pointer into memory for packed bcd string formation
90 *
92 BINSTR IDNT 2,1 Motorola 040 Floating Point Software Package
94 section 8
96 include fpsp.h
98 xdef binstr
99 binstr:
100 movem.l d0-d7,-(a7)
101 *
102 * A1: Init d7
103 *
104 moveq.l #1,d7 ;init d7 for second digit
105 subq.l #1,d0 ;for dbf d0 would have LEN+1 passes
106 *
107 * A2. Copy d2:d3 to d4:d5. Start loop.
108 *
109 loop:
110 move.l d2,d4 ;copy the fraction before muls
111 move.l d3,d5 ;to d4:d5
112 *
113 * A3. Multiply d2:d3 by 8; extract msbs into d1.
114 *
115 bfextu d2{0:3},d1 ;copy 3 msbs of d2 into d1
116 asl.l #3,d2 ;shift d2 left by 3 places
117 bfextu d3{0:3},d6 ;copy 3 msbs of d3 into d6
118 asl.l #3,d3 ;shift d3 left by 3 places
119 or.l d6,d2 ;or in msbs from d3 into d2
120 *
121 * A4. Multiply d4:d5 by 2; add carry out to d1.
122 *
123 add.l d5,d5 ;mul d5 by 2
124 addx.l d4,d4 ;mul d4 by 2
125 swap d6 ;put 0 in d6 lower word
126 addx.w d6,d1 ;add in extend from mul by 2
127 *
128 * A5. Add mul by 8 to mul by 2. D1 contains the digit formed.
129 *
130 add.l d5,d3 ;add lower 32 bits
131 nop ;ERRATA FIX #13 (Rev. 1.2 6/6/90)
132 addx.l d4,d2 ;add with extend upper 32 bits
133 nop ;ERRATA FIX #13 (Rev. 1.2 6/6/90)
134 addx.w d6,d1 ;add in extend from add to d1
135 swap d6 ;with d6 = 0; put 0 in upper word
136 *
137 * A6. Test d7 and branch.
138 *
139 tst.w d7 ;if zero, store digit & to loop
140 beq.b first_d ;if non-zero, form byte & write
141 sec_d:
142 swap d7 ;bring first digit to word d7b
143 asl.w #4,d7 ;first digit in upper 4 bits d7b
144 add.w d1,d7 ;add in ls digit to d7b
145 move.b d7,(a0)+ ;store d7b byte in memory
146 swap d7 ;put LEN counter in word d7a
147 clr.w d7 ;set d7a to signal no digits done
148 dbf.w d0,loop ;do loop some more!
149 bra.b end_bstr ;finished, so exit
150 first_d:
151 swap d7 ;put digit word in d7b
152 move.w d1,d7 ;put new digit in d7b
153 swap d7 ;put LEN counter in word d7a
154 addq.w #1,d7 ;set d7a to signal first digit done
155 dbf.w d0,loop ;do loop some more!
156 swap d7 ;put last digit in string
157 lsl.w #4,d7 ;move it to upper 4 bits
158 move.b d7,(a0)+ ;store it in memory string
159 *
160 * Clean up and return with result in fp0.
161 *
162 end_bstr:
163 movem.l (a7)+,d0-d7
164 rts
165 end