1* $NetBSD: binstr.sa,v 1.3 1994/10/26 07:48:53 cgd Exp $ 2 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 14* PARTICULAR PURPOSE and any warranty against infringement with 15* regard to the SOFTWARE (INCLUDING ANY MODIFIED VERSIONS THEREOF) 16* and any accompanying written materials. 17* 18* To the maximum extent permitted by applicable law, 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. 32 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* 91 92BINSTR IDNT 2,1 Motorola 040 Floating Point Software Package 93 94 section 8 95 96 include fpsp.h 97 98 xdef binstr 99binstr: 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* 109loop: 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 141sec_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 150first_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* 162end_bstr: 163 movem.l (a7)+,d0-d7 164 rts 165 end 166