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* $NetBSD: binstr.sa,v 1.3 1994/10/26 07:48:53 cgd Exp $
* MOTOROLA MICROPROCESSOR & MEMORY TECHNOLOGY GROUP
* M68000 Hi-Performance Microprocessor Division
* M68040 Software Package
*
* M68040 Software Package Copyright (c) 1993, 1994 Motorola Inc.
* All rights reserved.
*
* THE SOFTWARE is provided on an "AS IS" basis and without warranty.
* To the maximum extent permitted by applicable law,
* MOTOROLA DISCLAIMS ALL WARRANTIES WHETHER EXPRESS OR IMPLIED,
* INCLUDING IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A
* PARTICULAR PURPOSE and any warranty against infringement with
* regard to the SOFTWARE (INCLUDING ANY MODIFIED VERSIONS THEREOF)
* and any accompanying written materials.
*
* To the maximum extent permitted by applicable law,
* IN NO EVENT SHALL MOTOROLA BE LIABLE FOR ANY DAMAGES WHATSOEVER
* (INCLUDING WITHOUT LIMITATION, DAMAGES FOR LOSS OF BUSINESS
* PROFITS, BUSINESS INTERRUPTION, LOSS OF BUSINESS INFORMATION, OR
* OTHER PECUNIARY LOSS) ARISING OF THE USE OR INABILITY TO USE THE
* SOFTWARE. Motorola assumes no responsibility for the maintenance
* and support of the SOFTWARE.
*
* You are hereby granted a copyright license to use, modify, and
* distribute the SOFTWARE so long as this entire notice is retained
* without alteration in any modified and/or redistributed versions,
* and that such modified versions are clearly identified as such.
* No licenses are granted by implication, estoppel or otherwise
* under any patents or trademarks of Motorola, Inc.
*
* binstr.sa 3.3 12/19/90
*
*
* Description: Converts a 64-bit binary integer to bcd.
*
* Input: 64-bit binary integer in d2:d3, desired length (LEN) in
* d0, and a pointer to start in memory for bcd characters
* in d0. (This pointer must point to byte 4 of the first
* lword of the packed decimal memory string.)
*
* Output: LEN bcd digits representing the 64-bit integer.
*
* Algorithm:
* The 64-bit binary is assumed to have a decimal point before
* bit 63. The fraction is multiplied by 10 using a mul by 2
* shift and a mul by 8 shift. The bits shifted out of the
* msb form a decimal digit. This process is iterated until
* LEN digits are formed.
*
* A1. Init d7 to 1. D7 is the byte digit counter, and if 1, the
* digit formed will be assumed the least significant. This is
* to force the first byte formed to have a 0 in the upper 4 bits.
*
* A2. Beginning of the loop:
* Copy the fraction in d2:d3 to d4:d5.
*
* A3. Multiply the fraction in d2:d3 by 8 using bit-field
* extracts and shifts. The three msbs from d2 will go into
* d1.
*
* A4. Multiply the fraction in d4:d5 by 2 using shifts. The msb
* will be collected by the carry.
*
* A5. Add using the carry the 64-bit quantities in d2:d3 and d4:d5
* into d2:d3. D1 will contain the bcd digit formed.
*
* A6. Test d7. If zero, the digit formed is the ms digit. If non-
* zero, it is the ls digit. Put the digit in its place in the
* upper word of d0. If it is the ls digit, write the word
* from d0 to memory.
*
* A7. Decrement d6 (LEN counter) and repeat the loop until zero.
*
* Implementation Notes:
*
* The registers are used as follows:
*
* d0: LEN counter
* d1: temp used to form the digit
* d2: upper 32-bits of fraction for mul by 8
* d3: lower 32-bits of fraction for mul by 8
* d4: upper 32-bits of fraction for mul by 2
* d5: lower 32-bits of fraction for mul by 2
* d6: temp for bit-field extracts
* d7: byte digit formation word;digit count {0,1}
* a0: pointer into memory for packed bcd string formation
*
BINSTR IDNT 2,1 Motorola 040 Floating Point Software Package
section 8
include fpsp.h
xdef binstr
binstr:
movem.l d0-d7,-(a7)
*
* A1: Init d7
*
moveq.l #1,d7 ;init d7 for second digit
subq.l #1,d0 ;for dbf d0 would have LEN+1 passes
*
* A2. Copy d2:d3 to d4:d5. Start loop.
*
loop:
move.l d2,d4 ;copy the fraction before muls
move.l d3,d5 ;to d4:d5
*
* A3. Multiply d2:d3 by 8; extract msbs into d1.
*
bfextu d2{0:3},d1 ;copy 3 msbs of d2 into d1
asl.l #3,d2 ;shift d2 left by 3 places
bfextu d3{0:3},d6 ;copy 3 msbs of d3 into d6
asl.l #3,d3 ;shift d3 left by 3 places
or.l d6,d2 ;or in msbs from d3 into d2
*
* A4. Multiply d4:d5 by 2; add carry out to d1.
*
add.l d5,d5 ;mul d5 by 2
addx.l d4,d4 ;mul d4 by 2
swap d6 ;put 0 in d6 lower word
addx.w d6,d1 ;add in extend from mul by 2
*
* A5. Add mul by 8 to mul by 2. D1 contains the digit formed.
*
add.l d5,d3 ;add lower 32 bits
nop ;ERRATA FIX #13 (Rev. 1.2 6/6/90)
addx.l d4,d2 ;add with extend upper 32 bits
nop ;ERRATA FIX #13 (Rev. 1.2 6/6/90)
addx.w d6,d1 ;add in extend from add to d1
swap d6 ;with d6 = 0; put 0 in upper word
*
* A6. Test d7 and branch.
*
tst.w d7 ;if zero, store digit & to loop
beq.b first_d ;if non-zero, form byte & write
sec_d:
swap d7 ;bring first digit to word d7b
asl.w #4,d7 ;first digit in upper 4 bits d7b
add.w d1,d7 ;add in ls digit to d7b
move.b d7,(a0)+ ;store d7b byte in memory
swap d7 ;put LEN counter in word d7a
clr.w d7 ;set d7a to signal no digits done
dbf.w d0,loop ;do loop some more!
bra.b end_bstr ;finished, so exit
first_d:
swap d7 ;put digit word in d7b
move.w d1,d7 ;put new digit in d7b
swap d7 ;put LEN counter in word d7a
addq.w #1,d7 ;set d7a to signal first digit done
dbf.w d0,loop ;do loop some more!
swap d7 ;put last digit in string
lsl.w #4,d7 ;move it to upper 4 bits
move.b d7,(a0)+ ;store it in memory string
*
* Clean up and return with result in fp0.
*
end_bstr:
movem.l (a7)+,d0-d7
rts
end
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