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1 | | | ||
2 | | binstr.sa 3.3 12/19/90 | ||
3 | | | ||
4 | | | ||
5 | | Description: Converts a 64-bit binary integer to bcd. | ||
6 | | | ||
7 | | Input: 64-bit binary integer in d2:d3, desired length (LEN) in | ||
8 | | d0, and a pointer to start in memory for bcd characters | ||
9 | | in d0. (This pointer must point to byte 4 of the first | ||
10 | | lword of the packed decimal memory string.) | ||
11 | | | ||
12 | | Output: LEN bcd digits representing the 64-bit integer. | ||
13 | | | ||
14 | | Algorithm: | ||
15 | | The 64-bit binary is assumed to have a decimal point before | ||
16 | | bit 63. The fraction is multiplied by 10 using a mul by 2 | ||
17 | | shift and a mul by 8 shift. The bits shifted out of the | ||
18 | | msb form a decimal digit. This process is iterated until | ||
19 | | LEN digits are formed. | ||
20 | | | ||
21 | | A1. Init d7 to 1. D7 is the byte digit counter, and if 1, the | ||
22 | | digit formed will be assumed the least significant. This is | ||
23 | | to force the first byte formed to have a 0 in the upper 4 bits. | ||
24 | | | ||
25 | | A2. Beginning of the loop: | ||
26 | | Copy the fraction in d2:d3 to d4:d5. | ||
27 | | | ||
28 | | A3. Multiply the fraction in d2:d3 by 8 using bit-field | ||
29 | | extracts and shifts. The three msbs from d2 will go into | ||
30 | | d1. | ||
31 | | | ||
32 | | A4. Multiply the fraction in d4:d5 by 2 using shifts. The msb | ||
33 | | will be collected by the carry. | ||
34 | | | ||
35 | | A5. Add using the carry the 64-bit quantities in d2:d3 and d4:d5 | ||
36 | | into d2:d3. D1 will contain the bcd digit formed. | ||
37 | | | ||
38 | | A6. Test d7. If zero, the digit formed is the ms digit. If non- | ||
39 | | zero, it is the ls digit. Put the digit in its place in the | ||
40 | | upper word of d0. If it is the ls digit, write the word | ||
41 | | from d0 to memory. | ||
42 | | | ||
43 | | A7. Decrement d6 (LEN counter) and repeat the loop until zero. | ||
44 | | | ||
45 | | Implementation Notes: | ||
46 | | | ||
47 | | The registers are used as follows: | ||
48 | | | ||
49 | | d0: LEN counter | ||
50 | | d1: temp used to form the digit | ||
51 | | d2: upper 32-bits of fraction for mul by 8 | ||
52 | | d3: lower 32-bits of fraction for mul by 8 | ||
53 | | d4: upper 32-bits of fraction for mul by 2 | ||
54 | | d5: lower 32-bits of fraction for mul by 2 | ||
55 | | d6: temp for bit-field extracts | ||
56 | | d7: byte digit formation word;digit count {0,1} | ||
57 | | a0: pointer into memory for packed bcd string formation | ||
58 | | | ||
59 | |||
60 | | Copyright (C) Motorola, Inc. 1990 | ||
61 | | All Rights Reserved | ||
62 | | | ||
63 | | THIS IS UNPUBLISHED PROPRIETARY SOURCE CODE OF MOTOROLA | ||
64 | | The copyright notice above does not evidence any | ||
65 | | actual or intended publication of such source code. | ||
66 | |||
67 | |BINSTR idnt 2,1 | Motorola 040 Floating Point Software Package | ||
68 | |||
69 | |section 8 | ||
70 | |||
71 | #include "fpsp.h" | ||
72 | |||
73 | .global binstr | ||
74 | binstr: | ||
75 | moveml %d0-%d7,-(%a7) | ||
76 | | | ||
77 | | A1: Init d7 | ||
78 | | | ||
79 | moveql #1,%d7 |init d7 for second digit | ||
80 | subql #1,%d0 |for dbf d0 would have LEN+1 passes | ||
81 | | | ||
82 | | A2. Copy d2:d3 to d4:d5. Start loop. | ||
83 | | | ||
84 | loop: | ||
85 | movel %d2,%d4 |copy the fraction before muls | ||
86 | movel %d3,%d5 |to d4:d5 | ||
87 | | | ||
88 | | A3. Multiply d2:d3 by 8; extract msbs into d1. | ||
89 | | | ||
90 | bfextu %d2{#0:#3},%d1 |copy 3 msbs of d2 into d1 | ||
91 | asll #3,%d2 |shift d2 left by 3 places | ||
92 | bfextu %d3{#0:#3},%d6 |copy 3 msbs of d3 into d6 | ||
93 | asll #3,%d3 |shift d3 left by 3 places | ||
94 | orl %d6,%d2 |or in msbs from d3 into d2 | ||
95 | | | ||
96 | | A4. Multiply d4:d5 by 2; add carry out to d1. | ||
97 | | | ||
98 | asll #1,%d5 |mul d5 by 2 | ||
99 | roxll #1,%d4 |mul d4 by 2 | ||
100 | swap %d6 |put 0 in d6 lower word | ||
101 | addxw %d6,%d1 |add in extend from mul by 2 | ||
102 | | | ||
103 | | A5. Add mul by 8 to mul by 2. D1 contains the digit formed. | ||
104 | | | ||
105 | addl %d5,%d3 |add lower 32 bits | ||
106 | nop |ERRATA ; FIX #13 (Rev. 1.2 6/6/90) | ||
107 | addxl %d4,%d2 |add with extend upper 32 bits | ||
108 | nop |ERRATA ; FIX #13 (Rev. 1.2 6/6/90) | ||
109 | addxw %d6,%d1 |add in extend from add to d1 | ||
110 | swap %d6 |with d6 = 0; put 0 in upper word | ||
111 | | | ||
112 | | A6. Test d7 and branch. | ||
113 | | | ||
114 | tstw %d7 |if zero, store digit & to loop | ||
115 | beqs first_d |if non-zero, form byte & write | ||
116 | sec_d: | ||
117 | swap %d7 |bring first digit to word d7b | ||
118 | aslw #4,%d7 |first digit in upper 4 bits d7b | ||
119 | addw %d1,%d7 |add in ls digit to d7b | ||
120 | moveb %d7,(%a0)+ |store d7b byte in memory | ||
121 | swap %d7 |put LEN counter in word d7a | ||
122 | clrw %d7 |set d7a to signal no digits done | ||
123 | dbf %d0,loop |do loop some more! | ||
124 | bras end_bstr |finished, so exit | ||
125 | first_d: | ||
126 | swap %d7 |put digit word in d7b | ||
127 | movew %d1,%d7 |put new digit in d7b | ||
128 | swap %d7 |put LEN counter in word d7a | ||
129 | addqw #1,%d7 |set d7a to signal first digit done | ||
130 | dbf %d0,loop |do loop some more! | ||
131 | swap %d7 |put last digit in string | ||
132 | lslw #4,%d7 |move it to upper 4 bits | ||
133 | moveb %d7,(%a0)+ |store it in memory string | ||
134 | | | ||
135 | | Clean up and return with result in fp0. | ||
136 | | | ||
137 | end_bstr: | ||
138 | moveml (%a7)+,%d0-%d7 | ||
139 | rts | ||
140 | |end | ||