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20-02-88/T\OSB14
This website contains an archive of files for the Acorn Electron, BBC Micro, Acorn Archimedes, Commodore 16 and Commodore 64 computers, which Dominic Ford has rescued from his private collection of floppy disks and cassettes.
Some of these files were originally commercial releases in the 1980s and 1990s, but they are now widely available online. I assume that copyright over them is no longer being asserted. If you own the copyright and would like files to be removed, please contact me.
| Tape/disk: | Home » CEEFAX disks » telesoftware5.adl |
| Filename: | 20-02-88/T\OSB14 |
| Read OK: | ✔ |
| File size: | 1A61 bytes |
| Load address: | 0000 |
| Exec address: | FFFFFFFF |
File contents
OSBITS - An Exploration of the BBC Micro at Machine Level
By Programmer
..........................................................
Part 14: Binary to ASCII Conversion
This module brings us around to division and I'll introduce
it by way of a routine to take a 4 byte number in memory and
print out it's value as a decimal number in ASCII.
Unlike input we have no error conditions to worry about
since there are only four bytes (the routine can be expanded
as required) and the number can be either negative or
positive.
If it is negative we will first note its sign and then
convert to positive for processing.
The processing goes like this:
Take the number and divide it by the base, in this case 10
Take the remainder (which represents the number of units)
and it will be a binary value between 0 and 9. Add &30 to
it, which converts it to the ASCII code for its number 0-9
and put it into a buffer.
Divide the answer to the last division by 10, which leaves
us the number of 10s in the remainder .... and so on around
the loop.
We do this until the answer on division by 10 is zero.
But how do we divide in machine code?
For this exercise I'll go for the simplest algorithm, which
is essentially long division. So let's divide 10101101 by
101 (173 by 5). This should give us the result 34 remainder
3. We use conventional long division but remembering that we
only need to remember the 'one times table'! This gives us
the following:
100010
---------
101 | 10101101
101
---
00110
101
---
011
We could continue this division beyond the 'binary' point
which will give you an indication of how we will produce
real (as opposed to integer) numbers later on.
Effectively we are shifting (or rotating in 6502 terms)
along the dividend until we have shifted out a number
greater than the divisor. We then subtract the divisor from
that part of the dividend. We put a zero into the result
for each time we shift unless we can subtract in which case
we put in a one.
Starting again with the remainder from that last subtraction
we add bits onto it by shifting further along the dividend
until we have again produced a number greater than the
divisor .... and so on. Because we are dividing by a number
less than 256 we can use the accumulator for some of this
work.
So here is an algorithm based on that long division.
Set accumulator and result to zero
Loop starts here
Rotate result left
Rotate dividend left into accumulator
If accumulator is < divisor repeat loop
If accumulator is >= divisor then:
Add one to result
Subtract divisor from accumulator
Repeat loop until finished
You know when you are finished because you are counting the
number of bits in the dividend. When you finish the result
contains the DIV part and the accumulator contains the
remainder or MOD part.
This division fits into the whole output routine as follows:
Check the sign of the number to be output
If -ve set the sign flag and subtract number from zero
Output loop starts here
If dividend is now zero then print out number from
buffer and finish
Divide number by 10 (again) and put remainder into
buffer as ASCII
Repeat the output loop
This essentially is how the output routine works. The
assembly program is annotated and should explain itself.
The number is entered using BASIC indirection operators
again but you could add on the input routine from module 12.
The subroutine at 'div_by_base' divides the number remaining
in 'dividend' by 10 and puts it back in 'dividend' with the
remainder in the accumulator ready for &30 to be added and
the ASCII value stored in the buffer. As we proceed the Y
register points to the right place in the buffer. Note that
we work backwards from the end to the beginning of the
number.
The buffer is finally printed out at 'print_characters'
which moves forwards along the buffer from the point where
the Y register finished moving backwards while loading
digits into the buffer. This is unlikely to be the front of
the buffer unless the number is very large. Note also that
there could be a minus sign at the first occupied position
in the buffer. I put a zero byte (a null) at the end of the
buffer space early in the program so you can use a BEQ to
find the end of your number and stop printing.
There is a simple method employed to see if a multi-byte
number is zero. If you logically OR all the bytes together
you will get a zero result only if each byte individually
was zero (since A OR B is only zero if both A and B are
zero). This is used at the start of 'output_character_loop'
to see if there is anything left in the source number in
'dividend'. This checking avoids any leading zeros
appearing in the number.
There's another similar trick, which I have not used here,
but it can be used to see if the product or quotient of two
numbers is negative or positive. If you EOR the most
significant bytes together then the top bit of the result is
clear if they have the same sign (i.e. the product or
quotient is positive) and set if the opposite is true. This
means a BPL or BMI (which effectively looks at the top bit)
can be used to control the action of the program at this
point.
There is a simple change you can make to the division
routine. As it is at the moment the result and the dividend
are separate words but you will notice that as the dividend
is rotated left into the accumulator, bit by bit, empty bits
are added at the right. At the same time the result is
being rotated left and meaningful bits are added at the
right.
The change to make is to combine the two. This way you will
not have to clear the result space at the beginning, only
one word will be rotated instead of two, and you will not
have to transfer the contents of the workspaces at the end
of the division routine.
This leads to an algorithm like this:
Set accumulator to zero
Loop starts here
Rotate dividend/result left into accumulator
If accumulator is < divisor repeat loop
If accumulator is >= divisor then:
Add one to dividend/result
Subtract divisor from accumulator
Repeat loop until finished
It is simple to make this change but it speeds things up and
also removes the need for one of the workspace words
('result'). 'dividend' now will start holding the dividend
but will gradually, bit by bit, become the result. This is
the method I will use in a general routine next time, but
you could try modifying this module for yourself.
00000000 4f 53 42 49 54 53 20 2d 20 41 6e 20 45 78 70 6c |OSBITS - An Expl|
00000010 6f 72 61 74 69 6f 6e 20 6f 66 20 74 68 65 20 42 |oration of the B|
00000020 42 43 20 4d 69 63 72 6f 20 61 74 20 4d 61 63 68 |BC Micro at Mach|
00000030 69 6e 65 20 4c 65 76 65 6c 0d 0d 42 79 20 50 72 |ine Level..By Pr|
00000040 6f 67 72 61 6d 6d 65 72 0d 0d 2e 2e 2e 2e 2e 2e |ogrammer........|
00000050 2e 2e 2e 2e 2e 2e 2e 2e 2e 2e 2e 2e 2e 2e 2e 2e |................|
*
00000080 2e 2e 2e 2e 0d 0d 0d 50 61 72 74 20 31 34 3a 20 |.......Part 14: |
00000090 42 69 6e 61 72 79 20 74 6f 20 41 53 43 49 49 20 |Binary to ASCII |
000000a0 43 6f 6e 76 65 72 73 69 6f 6e 0d 0d 54 68 69 73 |Conversion..This|
000000b0 20 6d 6f 64 75 6c 65 20 62 72 69 6e 67 73 20 75 | module brings u|
000000c0 73 20 61 72 6f 75 6e 64 20 74 6f 20 64 69 76 69 |s around to divi|
000000d0 73 69 6f 6e 20 61 6e 64 20 49 27 6c 6c 20 69 6e |sion and I'll in|
000000e0 74 72 6f 64 75 63 65 0d 69 74 20 62 79 20 77 61 |troduce.it by wa|
000000f0 79 20 6f 66 20 61 20 72 6f 75 74 69 6e 65 20 74 |y of a routine t|
00000100 6f 20 74 61 6b 65 20 61 20 34 20 62 79 74 65 20 |o take a 4 byte |
00000110 6e 75 6d 62 65 72 20 69 6e 20 6d 65 6d 6f 72 79 |number in memory|
00000120 20 61 6e 64 0d 70 72 69 6e 74 20 6f 75 74 20 69 | and.print out i|
00000130 74 27 73 20 76 61 6c 75 65 20 61 73 20 61 20 64 |t's value as a d|
00000140 65 63 69 6d 61 6c 20 6e 75 6d 62 65 72 20 69 6e |ecimal number in|
00000150 20 41 53 43 49 49 2e 0d 0d 55 6e 6c 69 6b 65 20 | ASCII...Unlike |
00000160 69 6e 70 75 74 20 77 65 20 68 61 76 65 20 6e 6f |input we have no|
00000170 20 65 72 72 6f 72 20 63 6f 6e 64 69 74 69 6f 6e | error condition|
00000180 73 20 74 6f 20 77 6f 72 72 79 20 61 62 6f 75 74 |s to worry about|
00000190 0d 73 69 6e 63 65 20 74 68 65 72 65 20 61 72 65 |.since there are|
000001a0 20 6f 6e 6c 79 20 66 6f 75 72 20 62 79 74 65 73 | only four bytes|
000001b0 20 28 74 68 65 20 72 6f 75 74 69 6e 65 20 63 61 | (the routine ca|
000001c0 6e 20 62 65 20 65 78 70 61 6e 64 65 64 0d 61 73 |n be expanded.as|
000001d0 20 72 65 71 75 69 72 65 64 29 20 61 6e 64 20 74 | required) and t|
000001e0 68 65 20 6e 75 6d 62 65 72 20 63 61 6e 20 62 65 |he number can be|
000001f0 20 65 69 74 68 65 72 20 6e 65 67 61 74 69 76 65 | either negative|
00000200 20 6f 72 0d 70 6f 73 69 74 69 76 65 2e 0d 0d 49 | or.positive...I|
00000210 66 20 69 74 20 69 73 20 6e 65 67 61 74 69 76 65 |f it is negative|
00000220 20 77 65 20 77 69 6c 6c 20 66 69 72 73 74 20 6e | we will first n|
00000230 6f 74 65 20 69 74 73 20 73 69 67 6e 20 61 6e 64 |ote its sign and|
00000240 20 74 68 65 6e 0d 63 6f 6e 76 65 72 74 20 74 6f | then.convert to|
00000250 20 70 6f 73 69 74 69 76 65 20 66 6f 72 20 70 72 | positive for pr|
00000260 6f 63 65 73 73 69 6e 67 2e 0d 0d 54 68 65 20 70 |ocessing...The p|
00000270 72 6f 63 65 73 73 69 6e 67 20 67 6f 65 73 20 6c |rocessing goes l|
00000280 69 6b 65 20 74 68 69 73 3a 0d 0d 54 61 6b 65 20 |ike this:..Take |
00000290 74 68 65 20 6e 75 6d 62 65 72 20 61 6e 64 20 64 |the number and d|
000002a0 69 76 69 64 65 20 69 74 20 62 79 20 74 68 65 20 |ivide it by the |
000002b0 62 61 73 65 2c 20 69 6e 20 74 68 69 73 20 63 61 |base, in this ca|
000002c0 73 65 20 31 30 0d 0d 54 61 6b 65 20 74 68 65 20 |se 10..Take the |
000002d0 72 65 6d 61 69 6e 64 65 72 20 28 77 68 69 63 68 |remainder (which|
000002e0 20 72 65 70 72 65 73 65 6e 74 73 20 74 68 65 20 | represents the |
000002f0 6e 75 6d 62 65 72 20 6f 66 20 75 6e 69 74 73 29 |number of units)|
00000300 0d 61 6e 64 20 69 74 20 77 69 6c 6c 20 62 65 20 |.and it will be |
00000310 61 20 62 69 6e 61 72 79 20 76 61 6c 75 65 20 62 |a binary value b|
00000320 65 74 77 65 65 6e 20 30 20 61 6e 64 20 39 2e 20 |etween 0 and 9. |
00000330 20 41 64 64 20 26 33 30 20 74 6f 0d 69 74 2c 20 | Add &30 to.it, |
00000340 77 68 69 63 68 20 63 6f 6e 76 65 72 74 73 20 69 |which converts i|
00000350 74 20 74 6f 20 74 68 65 20 41 53 43 49 49 20 63 |t to the ASCII c|
00000360 6f 64 65 20 66 6f 72 20 69 74 73 20 6e 75 6d 62 |ode for its numb|
00000370 65 72 20 30 2d 39 0d 61 6e 64 20 70 75 74 20 69 |er 0-9.and put i|
00000380 74 20 69 6e 74 6f 20 61 20 62 75 66 66 65 72 2e |t into a buffer.|
00000390 0d 0d 44 69 76 69 64 65 20 74 68 65 20 61 6e 73 |..Divide the ans|
000003a0 77 65 72 20 74 6f 20 74 68 65 20 6c 61 73 74 20 |wer to the last |
000003b0 64 69 76 69 73 69 6f 6e 20 62 79 20 31 30 2c 20 |division by 10, |
000003c0 77 68 69 63 68 20 6c 65 61 76 65 73 0d 75 73 20 |which leaves.us |
000003d0 74 68 65 20 6e 75 6d 62 65 72 20 6f 66 20 31 30 |the number of 10|
000003e0 73 20 69 6e 20 74 68 65 20 72 65 6d 61 69 6e 64 |s in the remaind|
000003f0 65 72 20 2e 2e 2e 2e 20 61 6e 64 20 73 6f 20 6f |er .... and so o|
00000400 6e 20 61 72 6f 75 6e 64 0d 74 68 65 20 6c 6f 6f |n around.the loo|
00000410 70 2e 0d 0d 57 65 20 64 6f 20 74 68 69 73 20 75 |p...We do this u|
00000420 6e 74 69 6c 20 74 68 65 20 61 6e 73 77 65 72 20 |ntil the answer |
00000430 6f 6e 20 64 69 76 69 73 69 6f 6e 20 62 79 20 31 |on division by 1|
00000440 30 20 69 73 20 7a 65 72 6f 2e 0d 0d 42 75 74 20 |0 is zero...But |
00000450 68 6f 77 20 64 6f 20 77 65 20 64 69 76 69 64 65 |how do we divide|
00000460 20 69 6e 20 6d 61 63 68 69 6e 65 20 63 6f 64 65 | in machine code|
00000470 3f 0d 0d 46 6f 72 20 74 68 69 73 20 65 78 65 72 |?..For this exer|
00000480 63 69 73 65 20 49 27 6c 6c 20 67 6f 20 66 6f 72 |cise I'll go for|
00000490 20 74 68 65 20 73 69 6d 70 6c 65 73 74 20 61 6c | the simplest al|
000004a0 67 6f 72 69 74 68 6d 2c 20 77 68 69 63 68 0d 69 |gorithm, which.i|
000004b0 73 20 65 73 73 65 6e 74 69 61 6c 6c 79 20 6c 6f |s essentially lo|
000004c0 6e 67 20 64 69 76 69 73 69 6f 6e 2e 20 20 53 6f |ng division. So|
000004d0 20 6c 65 74 27 73 20 64 69 76 69 64 65 20 31 30 | let's divide 10|
000004e0 31 30 31 31 30 31 20 62 79 0d 31 30 31 20 28 31 |101101 by.101 (1|
000004f0 37 33 20 62 79 20 35 29 2e 20 20 54 68 69 73 20 |73 by 5). This |
00000500 73 68 6f 75 6c 64 20 67 69 76 65 20 75 73 20 74 |should give us t|
00000510 68 65 20 72 65 73 75 6c 74 20 33 34 20 72 65 6d |he result 34 rem|
00000520 61 69 6e 64 65 72 0d 33 2e 20 57 65 20 75 73 65 |ainder.3. We use|
00000530 20 63 6f 6e 76 65 6e 74 69 6f 6e 61 6c 20 6c 6f | conventional lo|
00000540 6e 67 20 64 69 76 69 73 69 6f 6e 20 62 75 74 20 |ng division but |
00000550 72 65 6d 65 6d 62 65 72 69 6e 67 20 74 68 61 74 |remembering that|
00000560 20 77 65 0d 6f 6e 6c 79 20 6e 65 65 64 20 74 6f | we.only need to|
00000570 20 72 65 6d 65 6d 62 65 72 20 74 68 65 20 27 6f | remember the 'o|
00000580 6e 65 20 74 69 6d 65 73 20 74 61 62 6c 65 27 21 |ne times table'!|
00000590 20 20 54 68 69 73 20 67 69 76 65 73 20 75 73 0d | This gives us.|
000005a0 74 68 65 20 66 6f 6c 6c 6f 77 69 6e 67 3a 0d 0d |the following:..|
000005b0 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | |
000005c0 20 20 20 20 20 20 20 31 30 30 30 31 30 0d 20 20 | 100010. |
000005d0 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | |
000005e0 20 20 2d 2d 2d 2d 2d 2d 2d 2d 2d 0d 20 20 20 20 | ---------. |
000005f0 20 20 20 20 20 20 20 20 20 20 20 31 30 31 20 7c | 101 ||
00000600 20 31 30 31 30 31 31 30 31 0d 20 20 20 20 20 20 | 10101101. |
00000610 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 31 | 1|
00000620 30 31 0d 20 20 20 20 20 20 20 20 20 20 20 20 20 |01. |
00000630 20 20 20 20 20 20 20 20 2d 2d 2d 0d 20 20 20 20 | ---. |
00000640 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | |
00000650 20 20 20 30 30 31 31 30 20 0d 20 20 20 20 20 20 | 00110 . |
00000660 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | |
00000670 20 20 20 31 30 31 20 0d 20 20 20 20 20 20 20 20 | 101 . |
00000680 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | |
00000690 20 2d 2d 2d 20 0d 20 20 20 20 20 20 20 20 20 20 | --- . |
000006a0 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | |
000006b0 30 31 31 0d 0d 57 65 20 63 6f 75 6c 64 20 63 6f |011..We could co|
000006c0 6e 74 69 6e 75 65 20 74 68 69 73 20 64 69 76 69 |ntinue this divi|
000006d0 73 69 6f 6e 20 62 65 79 6f 6e 64 20 74 68 65 20 |sion beyond the |
000006e0 27 62 69 6e 61 72 79 27 20 70 6f 69 6e 74 0d 77 |'binary' point.w|
000006f0 68 69 63 68 20 77 69 6c 6c 20 67 69 76 65 20 79 |hich will give y|
00000700 6f 75 20 61 6e 20 69 6e 64 69 63 61 74 69 6f 6e |ou an indication|
00000710 20 6f 66 20 68 6f 77 20 77 65 20 77 69 6c 6c 20 | of how we will |
00000720 70 72 6f 64 75 63 65 0d 72 65 61 6c 20 28 61 73 |produce.real (as|
00000730 20 6f 70 70 6f 73 65 64 20 74 6f 20 69 6e 74 65 | opposed to inte|
00000740 67 65 72 29 20 6e 75 6d 62 65 72 73 20 6c 61 74 |ger) numbers lat|
00000750 65 72 20 6f 6e 2e 0d 0d 45 66 66 65 63 74 69 76 |er on...Effectiv|
00000760 65 6c 79 20 77 65 20 61 72 65 20 73 68 69 66 74 |ely we are shift|
00000770 69 6e 67 20 28 6f 72 20 72 6f 74 61 74 69 6e 67 |ing (or rotating|
00000780 20 69 6e 20 36 35 30 32 20 74 65 72 6d 73 29 0d | in 6502 terms).|
00000790 61 6c 6f 6e 67 20 74 68 65 20 64 69 76 69 64 65 |along the divide|
000007a0 6e 64 20 75 6e 74 69 6c 20 77 65 20 68 61 76 65 |nd until we have|
000007b0 20 73 68 69 66 74 65 64 20 6f 75 74 20 61 20 6e | shifted out a n|
000007c0 75 6d 62 65 72 0d 67 72 65 61 74 65 72 20 74 68 |umber.greater th|
000007d0 61 6e 20 74 68 65 20 64 69 76 69 73 6f 72 2e 20 |an the divisor. |
000007e0 20 57 65 20 74 68 65 6e 20 73 75 62 74 72 61 63 | We then subtrac|
000007f0 74 20 74 68 65 20 64 69 76 69 73 6f 72 20 66 72 |t the divisor fr|
00000800 6f 6d 0d 74 68 61 74 20 70 61 72 74 20 6f 66 20 |om.that part of |
00000810 74 68 65 20 64 69 76 69 64 65 6e 64 2e 20 20 57 |the dividend. W|
00000820 65 20 70 75 74 20 61 20 7a 65 72 6f 20 69 6e 74 |e put a zero int|
00000830 6f 20 74 68 65 20 72 65 73 75 6c 74 0d 66 6f 72 |o the result.for|
00000840 20 65 61 63 68 20 74 69 6d 65 20 77 65 20 73 68 | each time we sh|
00000850 69 66 74 20 75 6e 6c 65 73 73 20 77 65 20 63 61 |ift unless we ca|
00000860 6e 20 73 75 62 74 72 61 63 74 20 69 6e 20 77 68 |n subtract in wh|
00000870 69 63 68 20 63 61 73 65 0d 77 65 20 70 75 74 20 |ich case.we put |
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000008a0 68 65 20 72 65 6d 61 69 6e 64 65 72 20 66 72 6f |he remainder fro|
000008b0 6d 20 74 68 61 74 20 6c 61 73 74 20 73 75 62 74 |m that last subt|
000008c0 72 61 63 74 69 6f 6e 0d 77 65 20 61 64 64 20 62 |raction.we add b|
000008d0 69 74 73 20 6f 6e 74 6f 20 69 74 20 62 79 20 73 |its onto it by s|
000008e0 68 69 66 74 69 6e 67 20 66 75 72 74 68 65 72 20 |hifting further |
000008f0 61 6c 6f 6e 67 20 74 68 65 20 64 69 76 69 64 65 |along the divide|
00000900 6e 64 0d 75 6e 74 69 6c 20 77 65 20 68 61 76 65 |nd.until we have|
00000910 20 61 67 61 69 6e 20 70 72 6f 64 75 63 65 64 20 | again produced |
00000920 61 20 6e 75 6d 62 65 72 20 67 72 65 61 74 65 72 |a number greater|
00000930 20 74 68 61 6e 20 74 68 65 0d 64 69 76 69 73 6f | than the.diviso|
00000940 72 20 2e 2e 2e 2e 20 61 6e 64 20 73 6f 20 6f 6e |r .... and so on|
00000950 2e 20 20 42 65 63 61 75 73 65 20 77 65 20 61 72 |. Because we ar|
00000960 65 20 64 69 76 69 64 69 6e 67 20 62 79 20 61 20 |e dividing by a |
00000970 6e 75 6d 62 65 72 0d 6c 65 73 73 20 74 68 61 6e |number.less than|
00000980 20 32 35 36 20 77 65 20 63 61 6e 20 75 73 65 20 | 256 we can use |
00000990 74 68 65 20 61 63 63 75 6d 75 6c 61 74 6f 72 20 |the accumulator |
000009a0 66 6f 72 20 73 6f 6d 65 20 6f 66 20 74 68 69 73 |for some of this|
000009b0 0d 77 6f 72 6b 2e 0d 0d 53 6f 20 68 65 72 65 20 |.work...So here |
000009c0 69 73 20 61 6e 20 61 6c 67 6f 72 69 74 68 6d 20 |is an algorithm |
000009d0 62 61 73 65 64 20 6f 6e 20 74 68 61 74 20 6c 6f |based on that lo|
000009e0 6e 67 20 64 69 76 69 73 69 6f 6e 2e 0d 0d 20 20 |ng division... |
000009f0 20 20 53 65 74 20 61 63 63 75 6d 75 6c 61 74 6f | Set accumulato|
00000a00 72 20 61 6e 64 20 72 65 73 75 6c 74 20 74 6f 20 |r and result to |
00000a10 7a 65 72 6f 0d 20 20 20 20 4c 6f 6f 70 20 73 74 |zero. Loop st|
00000a20 61 72 74 73 20 68 65 72 65 0d 20 20 20 20 20 20 |arts here. |
00000a30 52 6f 74 61 74 65 20 72 65 73 75 6c 74 20 6c 65 |Rotate result le|
00000a40 66 74 0d 20 20 20 20 20 20 52 6f 74 61 74 65 20 |ft. Rotate |
00000a50 64 69 76 69 64 65 6e 64 20 6c 65 66 74 20 69 6e |dividend left in|
00000a60 74 6f 20 61 63 63 75 6d 75 6c 61 74 6f 72 0d 20 |to accumulator. |
00000a70 20 20 20 20 20 49 66 20 61 63 63 75 6d 75 6c 61 | If accumula|
00000a80 74 6f 72 20 69 73 20 3c 20 64 69 76 69 73 6f 72 |tor is < divisor|
00000a90 20 72 65 70 65 61 74 20 6c 6f 6f 70 0d 20 20 20 | repeat loop. |
00000aa0 20 20 20 49 66 20 61 63 63 75 6d 75 6c 61 74 6f | If accumulato|
00000ab0 72 20 69 73 20 3e 3d 20 64 69 76 69 73 6f 72 20 |r is >= divisor |
00000ac0 74 68 65 6e 3a 0d 20 20 20 20 20 20 20 20 20 20 |then:. |
00000ad0 41 64 64 20 6f 6e 65 20 74 6f 20 72 65 73 75 6c |Add one to resul|
00000ae0 74 0d 20 20 20 20 20 20 20 20 20 20 53 75 62 74 |t. Subt|
00000af0 72 61 63 74 20 64 69 76 69 73 6f 72 20 66 72 6f |ract divisor fro|
00000b00 6d 20 61 63 63 75 6d 75 6c 61 74 6f 72 0d 20 20 |m accumulator. |
00000b10 20 20 52 65 70 65 61 74 20 6c 6f 6f 70 20 75 6e | Repeat loop un|
00000b20 74 69 6c 20 66 69 6e 69 73 68 65 64 0d 0d 59 6f |til finished..Yo|
00000b30 75 20 6b 6e 6f 77 20 77 68 65 6e 20 79 6f 75 20 |u know when you |
00000b40 61 72 65 20 66 69 6e 69 73 68 65 64 20 62 65 63 |are finished bec|
00000b50 61 75 73 65 20 79 6f 75 20 61 72 65 20 63 6f 75 |ause you are cou|
00000b60 6e 74 69 6e 67 20 74 68 65 0d 6e 75 6d 62 65 72 |nting the.number|
00000b70 20 6f 66 20 62 69 74 73 20 69 6e 20 74 68 65 20 | of bits in the |
00000b80 64 69 76 69 64 65 6e 64 2e 20 20 57 68 65 6e 20 |dividend. When |
00000b90 79 6f 75 20 66 69 6e 69 73 68 20 74 68 65 20 72 |you finish the r|
00000ba0 65 73 75 6c 74 0d 63 6f 6e 74 61 69 6e 73 20 74 |esult.contains t|
00000bb0 68 65 20 44 49 56 20 70 61 72 74 20 61 6e 64 20 |he DIV part and |
00000bc0 74 68 65 20 61 63 63 75 6d 75 6c 61 74 6f 72 20 |the accumulator |
00000bd0 63 6f 6e 74 61 69 6e 73 20 74 68 65 0d 72 65 6d |contains the.rem|
00000be0 61 69 6e 64 65 72 20 6f 72 20 4d 4f 44 20 70 61 |ainder or MOD pa|
00000bf0 72 74 2e 0d 0d 54 68 69 73 20 64 69 76 69 73 69 |rt...This divisi|
00000c00 6f 6e 20 66 69 74 73 20 69 6e 74 6f 20 74 68 65 |on fits into the|
00000c10 20 77 68 6f 6c 65 20 6f 75 74 70 75 74 20 72 6f | whole output ro|
00000c20 75 74 69 6e 65 20 61 73 20 66 6f 6c 6c 6f 77 73 |utine as follows|
00000c30 3a 0d 0d 20 20 20 20 43 68 65 63 6b 20 74 68 65 |:.. Check the|
00000c40 20 73 69 67 6e 20 6f 66 20 74 68 65 20 6e 75 6d | sign of the num|
00000c50 62 65 72 20 74 6f 20 62 65 20 6f 75 74 70 75 74 |ber to be output|
00000c60 0d 20 20 20 20 49 66 20 2d 76 65 20 73 65 74 20 |. If -ve set |
00000c70 74 68 65 20 73 69 67 6e 20 66 6c 61 67 20 61 6e |the sign flag an|
00000c80 64 20 73 75 62 74 72 61 63 74 20 6e 75 6d 62 65 |d subtract numbe|
00000c90 72 20 66 72 6f 6d 20 7a 65 72 6f 0d 20 20 20 20 |r from zero. |
00000ca0 4f 75 74 70 75 74 20 6c 6f 6f 70 20 73 74 61 72 |Output loop star|
00000cb0 74 73 20 68 65 72 65 0d 20 20 20 20 20 20 49 66 |ts here. If|
00000cc0 20 64 69 76 69 64 65 6e 64 20 69 73 20 6e 6f 77 | dividend is now|
00000cd0 20 7a 65 72 6f 20 74 68 65 6e 20 70 72 69 6e 74 | zero then print|
00000ce0 20 6f 75 74 20 6e 75 6d 62 65 72 20 66 72 6f 6d | out number from|
00000cf0 0d 20 20 20 20 20 20 20 20 62 75 66 66 65 72 20 |. buffer |
00000d00 61 6e 64 20 66 69 6e 69 73 68 0d 20 20 20 20 20 |and finish. |
00000d10 20 44 69 76 69 64 65 20 6e 75 6d 62 65 72 20 62 | Divide number b|
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20-02-88/T\OSB14.m0
20-02-88/T\OSB14.m1
20-02-88/T\OSB14.m2
20-02-88/T\OSB14.m4
20-02-88/T\OSB14.m5
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