2005-04-17 05:20:36 +07:00
|
|
|
|
|
|
|
|
| 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
|
|
|
|
|
|
|
|
|
|
|
|
|
| Copyright (C) Motorola, Inc. 1990
|
|
|
|
| All Rights Reserved
|
|
|
|
|
|
2006-02-12 08:55:48 +07:00
|
|
|
| For details on the license for this file, please see the
|
|
|
|
| file, README, in this same directory.
|
2005-04-17 05:20:36 +07:00
|
|
|
|
|
|
|
|BINSTR idnt 2,1 | Motorola 040 Floating Point Software Package
|
|
|
|
|
|
|
|
|section 8
|
|
|
|
|
|
|
|
#include "fpsp.h"
|
|
|
|
|
|
|
|
.global binstr
|
|
|
|
binstr:
|
|
|
|
moveml %d0-%d7,-(%a7)
|
|
|
|
|
|
|
|
|
| A1: Init d7
|
|
|
|
|
|
|
|
|
moveql #1,%d7 |init d7 for second digit
|
|
|
|
subql #1,%d0 |for dbf d0 would have LEN+1 passes
|
|
|
|
|
|
|
|
|
| A2. Copy d2:d3 to d4:d5. Start loop.
|
|
|
|
|
|
|
|
|
loop:
|
|
|
|
movel %d2,%d4 |copy the fraction before muls
|
|
|
|
movel %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
|
|
|
|
asll #3,%d2 |shift d2 left by 3 places
|
|
|
|
bfextu %d3{#0:#3},%d6 |copy 3 msbs of d3 into d6
|
|
|
|
asll #3,%d3 |shift d3 left by 3 places
|
|
|
|
orl %d6,%d2 |or in msbs from d3 into d2
|
|
|
|
|
|
|
|
|
| A4. Multiply d4:d5 by 2; add carry out to d1.
|
|
|
|
|
|
|
|
|
asll #1,%d5 |mul d5 by 2
|
|
|
|
roxll #1,%d4 |mul d4 by 2
|
|
|
|
swap %d6 |put 0 in d6 lower word
|
|
|
|
addxw %d6,%d1 |add in extend from mul by 2
|
|
|
|
|
|
|
|
|
| A5. Add mul by 8 to mul by 2. D1 contains the digit formed.
|
|
|
|
|
|
|
|
|
addl %d5,%d3 |add lower 32 bits
|
|
|
|
nop |ERRATA ; FIX #13 (Rev. 1.2 6/6/90)
|
|
|
|
addxl %d4,%d2 |add with extend upper 32 bits
|
|
|
|
nop |ERRATA ; FIX #13 (Rev. 1.2 6/6/90)
|
|
|
|
addxw %d6,%d1 |add in extend from add to d1
|
|
|
|
swap %d6 |with d6 = 0; put 0 in upper word
|
|
|
|
|
|
|
|
|
| A6. Test d7 and branch.
|
|
|
|
|
|
|
|
|
tstw %d7 |if zero, store digit & to loop
|
|
|
|
beqs first_d |if non-zero, form byte & write
|
|
|
|
sec_d:
|
|
|
|
swap %d7 |bring first digit to word d7b
|
|
|
|
aslw #4,%d7 |first digit in upper 4 bits d7b
|
|
|
|
addw %d1,%d7 |add in ls digit to d7b
|
|
|
|
moveb %d7,(%a0)+ |store d7b byte in memory
|
|
|
|
swap %d7 |put LEN counter in word d7a
|
|
|
|
clrw %d7 |set d7a to signal no digits done
|
|
|
|
dbf %d0,loop |do loop some more!
|
|
|
|
bras end_bstr |finished, so exit
|
|
|
|
first_d:
|
|
|
|
swap %d7 |put digit word in d7b
|
|
|
|
movew %d1,%d7 |put new digit in d7b
|
|
|
|
swap %d7 |put LEN counter in word d7a
|
|
|
|
addqw #1,%d7 |set d7a to signal first digit done
|
|
|
|
dbf %d0,loop |do loop some more!
|
|
|
|
swap %d7 |put last digit in string
|
|
|
|
lslw #4,%d7 |move it to upper 4 bits
|
|
|
|
moveb %d7,(%a0)+ |store it in memory string
|
|
|
|
|
|
|
|
|
| Clean up and return with result in fp0.
|
|
|
|
|
|
|
|
|
end_bstr:
|
|
|
|
moveml (%a7)+,%d0-%d7
|
|
|
|
rts
|
|
|
|
|end
|