Linus Torvalds | 1da177e | 2005-04-16 15:20:36 -0700 | [diff] [blame] | 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 | | |
Matt Waddel | e00d82d | 2006-02-11 17:55:48 -0800 | [diff] [blame] | 63 | | For details on the license for this file, please see the |
| 64 | | file, README, in this same directory. |
Linus Torvalds | 1da177e | 2005-04-16 15:20:36 -0700 | [diff] [blame] | 65 | |
| 66 | |BINSTR idnt 2,1 | Motorola 040 Floating Point Software Package |
| 67 | |
| 68 | |section 8 |
| 69 | |
| 70 | #include "fpsp.h" |
| 71 | |
| 72 | .global binstr |
| 73 | binstr: |
| 74 | moveml %d0-%d7,-(%a7) |
| 75 | | |
| 76 | | A1: Init d7 |
| 77 | | |
| 78 | moveql #1,%d7 |init d7 for second digit |
| 79 | subql #1,%d0 |for dbf d0 would have LEN+1 passes |
| 80 | | |
| 81 | | A2. Copy d2:d3 to d4:d5. Start loop. |
| 82 | | |
| 83 | loop: |
| 84 | movel %d2,%d4 |copy the fraction before muls |
| 85 | movel %d3,%d5 |to d4:d5 |
| 86 | | |
| 87 | | A3. Multiply d2:d3 by 8; extract msbs into d1. |
| 88 | | |
| 89 | bfextu %d2{#0:#3},%d1 |copy 3 msbs of d2 into d1 |
| 90 | asll #3,%d2 |shift d2 left by 3 places |
| 91 | bfextu %d3{#0:#3},%d6 |copy 3 msbs of d3 into d6 |
| 92 | asll #3,%d3 |shift d3 left by 3 places |
| 93 | orl %d6,%d2 |or in msbs from d3 into d2 |
| 94 | | |
| 95 | | A4. Multiply d4:d5 by 2; add carry out to d1. |
| 96 | | |
| 97 | asll #1,%d5 |mul d5 by 2 |
| 98 | roxll #1,%d4 |mul d4 by 2 |
| 99 | swap %d6 |put 0 in d6 lower word |
| 100 | addxw %d6,%d1 |add in extend from mul by 2 |
| 101 | | |
| 102 | | A5. Add mul by 8 to mul by 2. D1 contains the digit formed. |
| 103 | | |
| 104 | addl %d5,%d3 |add lower 32 bits |
| 105 | nop |ERRATA ; FIX #13 (Rev. 1.2 6/6/90) |
| 106 | addxl %d4,%d2 |add with extend upper 32 bits |
| 107 | nop |ERRATA ; FIX #13 (Rev. 1.2 6/6/90) |
| 108 | addxw %d6,%d1 |add in extend from add to d1 |
| 109 | swap %d6 |with d6 = 0; put 0 in upper word |
| 110 | | |
| 111 | | A6. Test d7 and branch. |
| 112 | | |
| 113 | tstw %d7 |if zero, store digit & to loop |
| 114 | beqs first_d |if non-zero, form byte & write |
| 115 | sec_d: |
| 116 | swap %d7 |bring first digit to word d7b |
| 117 | aslw #4,%d7 |first digit in upper 4 bits d7b |
| 118 | addw %d1,%d7 |add in ls digit to d7b |
| 119 | moveb %d7,(%a0)+ |store d7b byte in memory |
| 120 | swap %d7 |put LEN counter in word d7a |
| 121 | clrw %d7 |set d7a to signal no digits done |
| 122 | dbf %d0,loop |do loop some more! |
| 123 | bras end_bstr |finished, so exit |
| 124 | first_d: |
| 125 | swap %d7 |put digit word in d7b |
| 126 | movew %d1,%d7 |put new digit in d7b |
| 127 | swap %d7 |put LEN counter in word d7a |
| 128 | addqw #1,%d7 |set d7a to signal first digit done |
| 129 | dbf %d0,loop |do loop some more! |
| 130 | swap %d7 |put last digit in string |
| 131 | lslw #4,%d7 |move it to upper 4 bits |
| 132 | moveb %d7,(%a0)+ |store it in memory string |
| 133 | | |
| 134 | | Clean up and return with result in fp0. |
| 135 | | |
| 136 | end_bstr: |
| 137 | moveml (%a7)+,%d0-%d7 |
| 138 | rts |
| 139 | |end |