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30-01-88/T\OSB12
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: | 30-01-88/T\OSB12 |
| Read OK: | ✔ |
| File size: | 32DF bytes |
| Load address: | 0000 |
| Exec address: | FFFFFFFF |
File contents
OSBITS - An Exploration of the BBC Micro at Machine Level
By Programmer
...........................................................
Part 12: ASCII to Binary Conversion
I think we get so used to inputting numbers into a computer
that we can forget that they are not in the right format.
The keys we press put the ASCII value of the particular
digit we type into the micro rather than the number itself.
Some sort of conversion program is needed to take a string
of ASCII characters that represent a number and convert them
into a binary number in memory on which the program can
work.
In the previous module we produced a routine that took
digits typed in at the keyboard, refusing to accept invalid
characters, and put the result into a small chunk of memory
called the input buffer. That program returns in this
module as a subroutine in a larger program which will leave
us with a 4 byte number in work space in the computer
memory.
Once that number is in memory we can perform arithmetical
operations on it: add, subtract, multiply and divide. Once
we have a result, that will be a binary number in memory and
to print it out we have to convert in the opposite
direction, binary to ASCII.
B/osb11 left us with a string in a buffer. The first digit
of that string could be a number, or a minus sign. I'll
leave the sign for a moment. First let's look at the
problem of conversion.
As an example let's take a string that contains the
following characters:
1234
Remember that is a string of four characters, not a number
with the value 1234. If we start at the left hand end we
find a '1'. Now I know that under the Arabic numeral system
the actual value of that '1' depends on it's position, but
for ASCII to binary conversion we do not know, indeed do not
need to know, it's position directly. If we look at the
next character in the string we find the '2' but more
important we now know that the '1' is in fact worth 10 so we
have a 'running total' of 12. As we look at the next
position we find a '3' and we know that our '12' to the left
is in fact now worth 120 giving us a total of 123. Finally
we find a '4', the figures to our left are now worth 1230
and so we have a total of 1234, which is what was blindingly
obvious in the first place!
Let's think of this in terms of a computer algorithm (a
recipe if you like) and that goes like this:
Multiply the number 'so far' by 10
Add the value of the digit currently being read to the
number 'so far'
Repeat loop until there are no more digits
When you start the number 'so far' is of course zero. I
should perhaps call it the partial result or the accumulated
result.
This algorithm has an advantage in that it can readily be
adapted to read in numbers in bases other than 10. In that
case it would read:
Multiply the number 'so far' by the base
Add the value of the digit currently being read to the
number 'so far'
Repeat until there are no more digits
There are a couple of unanswered questions here. Just how
do you determine that there are no more digits? The easiest
way is to continue until you hit a character that is not in
your character set. In our case, for decimal input, we
continue until we hit a character that is not between zero
and nine.
What is the value of a digit? That is easy for the numbers
0 to 9 because the ASCII codes for them are &30 to &39 or 48
to 57. We just subtract 48 from the ASCII.
So we need to be able to multiply a number by the base in
order to carry out this input procedure. The routine in
this module is in fact general enough to cope with any base
up to 255 (although sorting out the values of the characters
in that base might cause some problems) but for the time
being we'll use it to deal with decimal numbers. The bases
you might want to use eventually are binary (2) octal (8)
decimal (10) and hexadecimal (16).
The largest number that can be held in 4 bytes using 2's
complement negative is 2147483647 positive or - 2147483648
negative which correspond to &7FFFFFFF and &80000000. In
fact in this routine I have sacrificed the ability to enter
- 2147483648 itself in order to simplify the coding a
little. The largest negative number is - 2147483647, but I
doubt if that will cause much wailing and gnashing of teeth.
So just how do we multiply in machine code? Well
fortunately we can fall back on good old fashioned long
multiplication, which is absurdly simple in binary since you
have a very rudimentary multiplication table.
Let's multiply 11011 by 1011 (i.e. 27 by 11)
11011
x
1011
---------
11011000
110110
11011
---------
100101001 which is 297
I hope you can see that for each set bit of the multiplier
(1101) there is a row in the space between the dotted lines.
Just like decimal long multiplication the rows are offset to
the left corresponding to the 'power' of the bit the row
belongs to. You should note also that you essentially only
need to know the one times table and that the adding up of
the partial products is more difficult for a mere human than
the multiplying that produces each partial product.
In machine code the 'difficult' adding is taken care of
easily using ADC. So what about those partial products
being offset left .... well that's rotating isn't it? So
how does it look as an algorithm?
We'll actually do this backwards so it looks more like
11011
x
1011
---------
11011
110110
11011000
---------
100101001
but the result is the same.
Take the rightmost (least significant bit) of the multiplier
and if it is set add the multiplicand to the contents of the
partial product workspace (ppws) which is empty to start
with, if it is clear do nothing.
Rotate the multiplier right and the multiplicand left.
Repeat the two stages until the multiplier is zero.
In practice you rotate the multiplier first because you use
the carry flag to check whether a bit is set, so it looks
like this if both numbers and their product are less than a
byte in size:
.loop
LSR multiplier
BCC no_add
CLC
LDA multiplicand
ADC ppws
STA ppws
.no_add
BEQ exit
ASL multiplicand
JMP loop
If your product is larger than a byte (as it was in our long
multiplication example above) then you simply have to add
zero (with carry) to the high byte, after you have added the
low bytes together. Let's say that the partial product was
in ppws and ppws+1 and that you were adding in a value from
the accumulator. It would look like this:
CLC
ADC ppws
STA ppws
LDA #0
ADC ppws+1
STA ppws+1
Note that we don't clear the carry before the second
addition.
It is possible to hold the multiplier in the accumulator
(this is done in the program module) but if either
multiplier or multiplicand are greater than 255 then
multi-byte rotating and multi-byte addition must be used.
Multi-byte addition I have already covered. Multi-byte
rotating is straightforward enough. Basically your first
rotation is either an ASL or an LSR because you do not want
to bring in the carry flag at first, but other rotations are
ROL or ROR so that you can use the carry flag to carry
between bytes. Here is an example with some memory called
mem to mem+3 being rotated left, which multiplies its value
by two.
ASL mem
ROL mem+1
ROL mem+2
ROL mem+3
The multiplication in B/osb12 is a compromise in that the
multiplicand is 4 bytes long but the multiplier is a single
byte (usually 10). I hope this will make the technique
clear, next time I will break from input and output of
numbers to give you a general multi-byte multiplication
routine which we can use later when we come on to numbers
with a decimal point.
Having discussed the techniques let's run through the
program in this module. It is copiously annotated, which
makes it a bit long. If you have a micro without second
processor or shadow screen you will have to go into mode 7
to run it although it is best displayed as a listing in mode
3.
In setting up B/osb12 I have decided not to enable the
assembly listing which is why line 190 goes to 2 instead of
3. The final code takes up over 500 bytes and I have
reserved &250 bytes of memory in line 170. To help testing
function key 2 inputs the largest positive and negative
numbers.
The changes to the input routine from last time are as
follows:
The buffer size is increased as 11 digit numbers, including
a minus sign, can be accepted. A plus sign has been added
to the valid character list. The zero flag is set on exit
if the string is of zero length.
The main body of the code, from line 260 to line 400, calls
two subroutines, subject to trapping ESCAPEs and zero length
strings (lines 370 and 380). The subroutine starting at
'input_character' is essentially that from the last module.
The subroutine at 'convert_ASCII_to_binary' takes the string
in the input buffer and converts it into a 4 byte binary
number and leaves the result in 4 bytes of memory starting
at 'digit_ws'.
The conversion subroutine is at line 1550. Note that the
workspace must be cleared before use. I have been known to
forget to do this with silly results! The sign flag is
cleared also.
We go through the string in the input buffer a character at
a time using X as a counter/index. If a character is not
between 0 and 9 we branch to 'not_number'. There, if we are
looking at the first character in the string we check for a
+ or a - sign and set the sign flag if it is a minus. A +
is just ignored. Any other character would cause an
'invalid number' error but this should not occur here as the
input routine filters other characters out. A non numeric
character anywhere else in the string is counted as
terminating the string. This terminator will usually be a
carriage return.
There were two possible ways of dealing with a negative
number. I could turn the number negative after it
was entered or I could actually build up a negative number
during the number entry routine. In the end I decided to
convert after entry, since this makes the code simpler and
has only the slight side effect of invalidating the largest
negative number (&80000000) as I mentioned before. After
the string has been terminated, and the code reaches the
address 'exit_number_conversion' we check to see if the sign
flag is set. If it is the number is negative and we simply
subtract it from zero (lines 2160 - 2280).
I have put the numerical base in memory as a variable to
make conversion of this routine to other bases more
straightforward. You should try to convert to base 8, or
even 16, to check if you understand how it works and that I
have explained it clearly enough. In each case, of course,
you will need different characters. 0-7 for octal and 0-9
A-F for hexadecimal.
To process a digit we first subtract the ASCII value of '0'
from it (48) and then call a subroutine at 'process_number'.
As the comment says we multiply the number held in digit_ws
(et al) by the base and add the new digit (held in
'accustore' temporarily) to produce a new partial result in
digit_ws.
Note line 2630. Here is an error trap that catches any
number that sets the top bit (becomes negative) during this
addition. If that happens the number has overflowed 4
bytes. This is also done at line 3050. I have given the
add and rotate error traps different messages so you can see
which operates when if you deliberately enter huge numbers.
You might try deleting the traps and seeing what happens.
The trap at line 3050 is a bit belt and braces since I am
not convinced it catches anything.
We save X to the stack because X is holding the current
offset down the input buffer and so we must not lose it.
The 'mult_by_base' subroutine uses 4 bytes at 'ppws' as
partial product workspace and holds the multiplier in the
accumulator most of the time. If a number rotates to set
the top bit then we have another overflow situation and that
is trapped in line 3160. Otherwise this section of code is
a straightforward multiplication where the multiplier is
only a byte in size and the multiplicand is 4 bytes in size.
I wanted to put the final result back in digit_ws so lines
3270 to 3340 copy it across there. The BASIC routine
between line 3430 and 3460 enables you to check the results.
The next module will deal with multi-byte multiplication,
also called multiple precision multiplication, in a more
general way.
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 2e 0d 0d 0d 50 61 72 74 20 31 32 3a |........Part 12:| 00000090 20 41 53 43 49 49 20 74 6f 20 42 69 6e 61 72 79 | ASCII to Binary| 000000a0 20 43 6f 6e 76 65 72 73 69 6f 6e 0d 0d 49 20 74 | Conversion..I t| 000000b0 68 69 6e 6b 20 77 65 20 67 65 74 20 73 6f 20 75 |hink we get so u| 000000c0 73 65 64 20 74 6f 20 69 6e 70 75 74 74 69 6e 67 |sed to inputting| 000000d0 20 6e 75 6d 62 65 72 73 20 69 6e 74 6f 20 61 20 | numbers into a | 000000e0 63 6f 6d 70 75 74 65 72 0d 74 68 61 74 20 77 65 |computer.that we| 000000f0 20 63 61 6e 20 66 6f 72 67 65 74 20 74 68 61 74 | can forget that| 00000100 20 74 68 65 79 20 61 72 65 20 6e 6f 74 20 69 6e | they are not in| 00000110 20 74 68 65 20 72 69 67 68 74 20 66 6f 72 6d 61 | the right forma| 00000120 74 2e 20 0d 54 68 65 20 6b 65 79 73 20 77 65 20 |t. .The keys we | 00000130 70 72 65 73 73 20 70 75 74 20 74 68 65 20 41 53 |press put the AS| 00000140 43 49 49 20 76 61 6c 75 65 20 6f 66 20 74 68 65 |CII value of the| 00000150 20 70 61 72 74 69 63 75 6c 61 72 0d 64 69 67 69 | particular.digi| 00000160 74 20 77 65 20 74 79 70 65 20 69 6e 74 6f 20 74 |t we type into t| 00000170 68 65 20 6d 69 63 72 6f 20 72 61 74 68 65 72 20 |he micro rather | 00000180 74 68 61 6e 20 74 68 65 20 6e 75 6d 62 65 72 20 |than the number | 00000190 69 74 73 65 6c 66 2e 0d 0d 53 6f 6d 65 20 73 6f |itself...Some so| 000001a0 72 74 20 6f 66 20 63 6f 6e 76 65 72 73 69 6f 6e |rt of conversion| 000001b0 20 70 72 6f 67 72 61 6d 20 69 73 20 6e 65 65 64 | program is need| 000001c0 65 64 20 74 6f 20 74 61 6b 65 20 61 20 73 74 72 |ed to take a str| 000001d0 69 6e 67 0d 6f 66 20 41 53 43 49 49 20 63 68 61 |ing.of ASCII cha| 000001e0 72 61 63 74 65 72 73 20 74 68 61 74 20 72 65 70 |racters that rep| 000001f0 72 65 73 65 6e 74 20 61 20 6e 75 6d 62 65 72 20 |resent a number | 00000200 61 6e 64 20 63 6f 6e 76 65 72 74 20 74 68 65 6d |and convert them| 00000210 0d 69 6e 74 6f 20 61 20 62 69 6e 61 72 79 20 6e |.into a binary n| 00000220 75 6d 62 65 72 20 69 6e 20 6d 65 6d 6f 72 79 20 |umber in memory | 00000230 6f 6e 20 77 68 69 63 68 20 74 68 65 20 70 72 6f |on which the pro| 00000240 67 72 61 6d 20 63 61 6e 0d 77 6f 72 6b 2e 0d 0d |gram can.work...| 00000250 49 6e 20 74 68 65 20 70 72 65 76 69 6f 75 73 20 |In the previous | 00000260 6d 6f 64 75 6c 65 20 77 65 20 70 72 6f 64 75 63 |module we produc| 00000270 65 64 20 61 20 72 6f 75 74 69 6e 65 20 74 68 61 |ed a routine tha| 00000280 74 20 74 6f 6f 6b 0d 64 69 67 69 74 73 20 74 79 |t took.digits ty| 00000290 70 65 64 20 69 6e 20 61 74 20 74 68 65 20 6b 65 |ped in at the ke| 000002a0 79 62 6f 61 72 64 2c 20 72 65 66 75 73 69 6e 67 |yboard, refusing| 000002b0 20 74 6f 20 61 63 63 65 70 74 20 69 6e 76 61 6c | to accept inval| 000002c0 69 64 0d 63 68 61 72 61 63 74 65 72 73 2c 20 61 |id.characters, a| 000002d0 6e 64 20 70 75 74 20 74 68 65 20 72 65 73 75 6c |nd put the resul| 000002e0 74 20 69 6e 74 6f 20 61 20 73 6d 61 6c 6c 20 63 |t into a small c| 000002f0 68 75 6e 6b 20 6f 66 20 6d 65 6d 6f 72 79 0d 63 |hunk of memory.c| 00000300 61 6c 6c 65 64 20 74 68 65 20 69 6e 70 75 74 20 |alled the input | 00000310 62 75 66 66 65 72 2e 20 20 54 68 61 74 20 70 72 |buffer. That pr| 00000320 6f 67 72 61 6d 20 72 65 74 75 72 6e 73 20 69 6e |ogram returns in| 00000330 20 74 68 69 73 0d 6d 6f 64 75 6c 65 20 61 73 20 | this.module as | 00000340 61 20 73 75 62 72 6f 75 74 69 6e 65 20 69 6e 20 |a subroutine in | 00000350 61 20 6c 61 72 67 65 72 20 70 72 6f 67 72 61 6d |a larger program| 00000360 20 77 68 69 63 68 20 77 69 6c 6c 20 6c 65 61 76 | which will leav| 00000370 65 0d 75 73 20 77 69 74 68 20 61 20 34 20 62 79 |e.us with a 4 by| 00000380 74 65 20 6e 75 6d 62 65 72 20 69 6e 20 77 6f 72 |te number in wor| 00000390 6b 20 73 70 61 63 65 20 69 6e 20 74 68 65 20 63 |k space in the c| 000003a0 6f 6d 70 75 74 65 72 0d 6d 65 6d 6f 72 79 2e 0d |omputer.memory..| 000003b0 0d 4f 6e 63 65 20 74 68 61 74 20 6e 75 6d 62 65 |.Once that numbe| 000003c0 72 20 69 73 20 69 6e 20 6d 65 6d 6f 72 79 20 77 |r is in memory w| 000003d0 65 20 63 61 6e 20 70 65 72 66 6f 72 6d 20 61 72 |e can perform ar| 000003e0 69 74 68 6d 65 74 69 63 61 6c 0d 6f 70 65 72 61 |ithmetical.opera| 000003f0 74 69 6f 6e 73 20 6f 6e 20 69 74 3a 20 61 64 64 |tions on it: add| 00000400 2c 20 73 75 62 74 72 61 63 74 2c 20 6d 75 6c 74 |, subtract, mult| 00000410 69 70 6c 79 20 61 6e 64 20 64 69 76 69 64 65 2e |iply and divide.| 00000420 20 20 4f 6e 63 65 0d 77 65 20 68 61 76 65 20 61 | Once.we have a| 00000430 20 72 65 73 75 6c 74 2c 20 74 68 61 74 20 77 69 | result, that wi| 00000440 6c 6c 20 62 65 20 61 20 62 69 6e 61 72 79 20 6e |ll be a binary n| 00000450 75 6d 62 65 72 20 69 6e 20 6d 65 6d 6f 72 79 20 |umber in memory | 00000460 61 6e 64 0d 74 6f 20 70 72 69 6e 74 20 69 74 20 |and.to print it | 00000470 6f 75 74 20 77 65 20 68 61 76 65 20 74 6f 20 63 |out we have to c| 00000480 6f 6e 76 65 72 74 20 69 6e 20 74 68 65 20 6f 70 |onvert in the op| 00000490 70 6f 73 69 74 65 0d 64 69 72 65 63 74 69 6f 6e |posite.direction| 000004a0 2c 20 62 69 6e 61 72 79 20 74 6f 20 41 53 43 49 |, binary to ASCI| 000004b0 49 2e 0d 0d 42 2f 6f 73 62 31 31 20 6c 65 66 74 |I...B/osb11 left| 000004c0 20 75 73 20 77 69 74 68 20 61 20 73 74 72 69 6e | us with a strin| 000004d0 67 20 69 6e 20 61 20 62 75 66 66 65 72 2e 20 20 |g in a buffer. | 000004e0 54 68 65 20 66 69 72 73 74 20 64 69 67 69 74 0d |The first digit.| 000004f0 6f 66 20 74 68 61 74 20 73 74 72 69 6e 67 20 63 |of that string c| 00000500 6f 75 6c 64 20 62 65 20 61 20 6e 75 6d 62 65 72 |ould be a number| 00000510 2c 20 6f 72 20 61 20 6d 69 6e 75 73 20 73 69 67 |, or a minus sig| 00000520 6e 2e 20 20 49 27 6c 6c 0d 6c 65 61 76 65 20 74 |n. I'll.leave t| 00000530 68 65 20 73 69 67 6e 20 66 6f 72 20 61 20 6d 6f |he sign for a mo| 00000540 6d 65 6e 74 2e 20 20 46 69 72 73 74 20 6c 65 74 |ment. First let| 00000550 27 73 20 6c 6f 6f 6b 20 61 74 20 74 68 65 0d 70 |'s look at the.p| 00000560 72 6f 62 6c 65 6d 20 6f 66 20 63 6f 6e 76 65 72 |roblem of conver| 00000570 73 69 6f 6e 2e 0d 0d 41 73 20 61 6e 20 65 78 61 |sion...As an exa| 00000580 6d 70 6c 65 20 6c 65 74 27 73 20 74 61 6b 65 20 |mple let's take | 00000590 61 20 73 74 72 69 6e 67 20 74 68 61 74 20 63 6f |a string that co| 000005a0 6e 74 61 69 6e 73 20 74 68 65 0d 66 6f 6c 6c 6f |ntains the.follo| 000005b0 77 69 6e 67 20 63 68 61 72 61 63 74 65 72 73 3a |wing characters:| 000005c0 0d 0d 20 20 20 20 20 20 20 20 20 20 20 20 20 20 |.. | 000005d0 20 20 20 20 20 20 20 31 32 33 34 0d 0d 52 65 6d | 1234..Rem| 000005e0 65 6d 62 65 72 20 74 68 61 74 20 69 73 20 61 20 |ember that is a | 000005f0 73 74 72 69 6e 67 20 6f 66 20 66 6f 75 72 20 63 |string of four c| 00000600 68 61 72 61 63 74 65 72 73 2c 20 6e 6f 74 20 61 |haracters, not a| 00000610 20 6e 75 6d 62 65 72 0d 77 69 74 68 20 74 68 65 | number.with the| 00000620 20 76 61 6c 75 65 20 31 32 33 34 2e 20 20 49 66 | value 1234. If| 00000630 20 77 65 20 73 74 61 72 74 20 61 74 20 74 68 65 | we start at the| 00000640 20 6c 65 66 74 20 68 61 6e 64 20 65 6e 64 20 77 | left hand end w| 00000650 65 0d 66 69 6e 64 20 61 20 27 31 27 2e 20 20 4e |e.find a '1'. N| 00000660 6f 77 20 49 20 6b 6e 6f 77 20 74 68 61 74 20 75 |ow I know that u| 00000670 6e 64 65 72 20 74 68 65 20 41 72 61 62 69 63 20 |nder the Arabic | 00000680 6e 75 6d 65 72 61 6c 20 73 79 73 74 65 6d 0d 74 |numeral system.t| 00000690 68 65 20 61 63 74 75 61 6c 20 76 61 6c 75 65 20 |he actual value | 000006a0 6f 66 20 74 68 61 74 20 27 31 27 20 64 65 70 65 |of that '1' depe| 000006b0 6e 64 73 20 6f 6e 20 69 74 27 73 20 70 6f 73 69 |nds on it's posi| 000006c0 74 69 6f 6e 2c 20 62 75 74 0d 66 6f 72 20 41 53 |tion, but.for AS| 000006d0 43 49 49 20 74 6f 20 62 69 6e 61 72 79 20 63 6f |CII to binary co| 000006e0 6e 76 65 72 73 69 6f 6e 20 77 65 20 64 6f 20 6e |nversion we do n| 000006f0 6f 74 20 6b 6e 6f 77 2c 20 69 6e 64 65 65 64 20 |ot know, indeed | 00000700 64 6f 20 6e 6f 74 0d 6e 65 65 64 20 74 6f 20 6b |do not.need to k| 00000710 6e 6f 77 2c 20 69 74 27 73 20 70 6f 73 69 74 69 |now, it's positi| 00000720 6f 6e 20 64 69 72 65 63 74 6c 79 2e 20 20 49 66 |on directly. If| 00000730 20 77 65 20 6c 6f 6f 6b 20 61 74 20 74 68 65 0d | we look at the.| 00000740 6e 65 78 74 20 63 68 61 72 61 63 74 65 72 20 69 |next character i| 00000750 6e 20 74 68 65 20 73 74 72 69 6e 67 20 77 65 20 |n the string we | 00000760 66 69 6e 64 20 74 68 65 20 27 32 27 20 62 75 74 |find the '2' but| 00000770 20 6d 6f 72 65 0d 69 6d 70 6f 72 74 61 6e 74 20 | more.important | 00000780 77 65 20 6e 6f 77 20 6b 6e 6f 77 20 74 68 61 74 |we now know that| 00000790 20 74 68 65 20 27 31 27 20 69 73 20 69 6e 20 66 | the '1' is in f| 000007a0 61 63 74 20 77 6f 72 74 68 20 31 30 20 73 6f 20 |act worth 10 so | 000007b0 77 65 0d 68 61 76 65 20 61 20 27 72 75 6e 6e 69 |we.have a 'runni| 000007c0 6e 67 20 74 6f 74 61 6c 27 20 6f 66 20 31 32 2e |ng total' of 12.| 000007d0 20 20 41 73 20 77 65 20 6c 6f 6f 6b 20 61 74 20 | As we look at | 000007e0 74 68 65 20 6e 65 78 74 0d 70 6f 73 69 74 69 6f |the next.positio| 000007f0 6e 20 77 65 20 66 69 6e 64 20 61 20 27 33 27 20 |n we find a '3' | 00000800 61 6e 64 20 77 65 20 6b 6e 6f 77 20 74 68 61 74 |and we know that| 00000810 20 6f 75 72 20 27 31 32 27 20 74 6f 20 74 68 65 | our '12' to the| 00000820 20 6c 65 66 74 0d 69 73 20 69 6e 20 66 61 63 74 | left.is in fact| 00000830 20 6e 6f 77 20 77 6f 72 74 68 20 31 32 30 20 67 | now worth 120 g| 00000840 69 76 69 6e 67 20 75 73 20 61 20 74 6f 74 61 6c |iving us a total| 00000850 20 6f 66 20 31 32 33 2e 20 20 46 69 6e 61 6c 6c | of 123. Finall| 00000860 79 0d 77 65 20 66 69 6e 64 20 61 20 27 34 27 2c |y.we find a '4',| 00000870 20 74 68 65 20 66 69 67 75 72 65 73 20 74 6f 20 | the figures to | 00000880 6f 75 72 20 6c 65 66 74 20 61 72 65 20 6e 6f 77 |our left are now| 00000890 20 77 6f 72 74 68 20 31 32 33 30 0d 61 6e 64 20 | worth 1230.and | 000008a0 73 6f 20 77 65 20 68 61 76 65 20 61 20 74 6f 74 |so we have a tot| 000008b0 61 6c 20 6f 66 20 31 32 33 34 2c 20 77 68 69 63 |al of 1234, whic| 000008c0 68 20 69 73 20 77 68 61 74 20 77 61 73 20 62 6c |h is what was bl| 000008d0 69 6e 64 69 6e 67 6c 79 0d 6f 62 76 69 6f 75 73 |indingly.obvious| 000008e0 20 69 6e 20 74 68 65 20 66 69 72 73 74 20 70 6c | in the first pl| 000008f0 61 63 65 21 0d 0d 4c 65 74 27 73 20 74 68 69 6e |ace!..Let's thin| 00000900 6b 20 6f 66 20 74 68 69 73 20 69 6e 20 74 65 72 |k of this in ter| 00000910 6d 73 20 6f 66 20 61 20 63 6f 6d 70 75 74 65 72 |ms of a computer| 00000920 20 61 6c 67 6f 72 69 74 68 6d 20 28 61 0d 72 65 | algorithm (a.re| 00000930 63 69 70 65 20 69 66 20 79 6f 75 20 6c 69 6b 65 |cipe if you like| 00000940 29 20 61 6e 64 20 74 68 61 74 20 67 6f 65 73 20 |) and that goes | 00000950 6c 69 6b 65 20 74 68 69 73 3a 0d 0d 20 20 20 20 |like this:.. | 00000960 4d 75 6c 74 69 70 6c 79 20 74 68 65 20 6e 75 6d |Multiply the num| 00000970 62 65 72 20 27 73 6f 20 66 61 72 27 20 62 79 20 |ber 'so far' by | 00000980 31 30 0d 20 20 20 20 41 64 64 20 74 68 65 20 76 |10. Add the v| 00000990 61 6c 75 65 20 6f 66 20 74 68 65 20 64 69 67 69 |alue of the digi| 000009a0 74 20 63 75 72 72 65 6e 74 6c 79 20 62 65 69 6e |t currently bein| 000009b0 67 20 72 65 61 64 20 74 6f 20 74 68 65 0d 20 20 |g read to the. | 000009c0 20 20 20 20 20 20 6e 75 6d 62 65 72 20 27 73 6f | number 'so| 000009d0 20 66 61 72 27 0d 20 20 20 20 52 65 70 65 61 74 | far'. Repeat| 000009e0 20 6c 6f 6f 70 20 75 6e 74 69 6c 20 74 68 65 72 | loop until ther| 000009f0 65 20 61 72 65 20 6e 6f 20 6d 6f 72 65 20 64 69 |e are no more di| 00000a00 67 69 74 73 0d 0d 57 68 65 6e 20 79 6f 75 20 73 |gits..When you s| 00000a10 74 61 72 74 20 74 68 65 20 6e 75 6d 62 65 72 20 |tart the number | 00000a20 27 73 6f 20 66 61 72 27 20 69 73 20 6f 66 20 63 |'so far' is of c| 00000a30 6f 75 72 73 65 20 7a 65 72 6f 2e 20 20 49 0d 73 |ourse zero. I.s| 00000a40 68 6f 75 6c 64 20 70 65 72 68 61 70 73 20 63 61 |hould perhaps ca| 00000a50 6c 6c 20 69 74 20 74 68 65 20 70 61 72 74 69 61 |ll it the partia| 00000a60 6c 20 72 65 73 75 6c 74 20 6f 72 20 74 68 65 20 |l result or the | 00000a70 61 63 63 75 6d 75 6c 61 74 65 64 0d 72 65 73 75 |accumulated.resu| 00000a80 6c 74 2e 0d 0d 54 68 69 73 20 61 6c 67 6f 72 69 |lt...This algori| 00000a90 74 68 6d 20 68 61 73 20 61 6e 20 61 64 76 61 6e |thm has an advan| 00000aa0 74 61 67 65 20 69 6e 20 74 68 61 74 20 69 74 20 |tage in that it | 00000ab0 63 61 6e 20 72 65 61 64 69 6c 79 20 62 65 0d 61 |can readily be.a| 00000ac0 64 61 70 74 65 64 20 74 6f 20 72 65 61 64 20 69 |dapted to read i| 00000ad0 6e 20 6e 75 6d 62 65 72 73 20 69 6e 20 62 61 73 |n numbers in bas| 00000ae0 65 73 20 6f 74 68 65 72 20 74 68 61 6e 20 31 30 |es other than 10| 00000af0 2e 20 20 49 6e 20 74 68 61 74 0d 63 61 73 65 20 |. In that.case | 00000b00 69 74 20 77 6f 75 6c 64 20 72 65 61 64 3a 0d 0d |it would read:..| 00000b10 20 20 20 20 4d 75 6c 74 69 70 6c 79 20 74 68 65 | Multiply the| 00000b20 20 6e 75 6d 62 65 72 20 27 73 6f 20 66 61 72 27 | number 'so far'| 00000b30 20 62 79 20 74 68 65 20 62 61 73 65 0d 20 20 20 | by the base. | 00000b40 20 41 64 64 20 74 68 65 20 76 61 6c 75 65 20 6f | Add the value o| 00000b50 66 20 74 68 65 20 64 69 67 69 74 20 63 75 72 72 |f the digit curr| 00000b60 65 6e 74 6c 79 20 62 65 69 6e 67 20 72 65 61 64 |ently being read| 00000b70 20 74 6f 20 74 68 65 0d 20 20 20 20 20 20 20 20 | to the. | 00000b80 6e 75 6d 62 65 72 20 27 73 6f 20 66 61 72 27 0d |number 'so far'.| 00000b90 20 20 20 20 52 65 70 65 61 74 20 75 6e 74 69 6c | Repeat until| 00000ba0 20 74 68 65 72 65 20 61 72 65 20 6e 6f 20 6d 6f | there are no mo| 00000bb0 72 65 20 64 69 67 69 74 73 0d 0d 54 68 65 72 65 |re digits..There| 00000bc0 20 61 72 65 20 61 20 63 6f 75 70 6c 65 20 6f 66 | are a couple of| 00000bd0 20 75 6e 61 6e 73 77 65 72 65 64 20 71 75 65 73 | unanswered ques| 00000be0 74 69 6f 6e 73 20 68 65 72 65 2e 20 20 4a 75 73 |tions here. Jus| 00000bf0 74 20 68 6f 77 0d 64 6f 20 79 6f 75 20 64 65 74 |t how.do you det| 00000c00 65 72 6d 69 6e 65 20 74 68 61 74 20 74 68 65 72 |ermine that ther| 00000c10 65 20 61 72 65 20 6e 6f 20 6d 6f 72 65 20 64 69 |e are no more di| 00000c20 67 69 74 73 3f 20 20 54 68 65 20 65 61 73 69 65 |gits? The easie| 00000c30 73 74 0d 77 61 79 20 69 73 20 74 6f 20 63 6f 6e |st.way is to con| 00000c40 74 69 6e 75 65 20 75 6e 74 69 6c 20 79 6f 75 20 |tinue until you | 00000c50 68 69 74 20 61 20 63 68 61 72 61 63 74 65 72 20 |hit a character | 00000c60 74 68 61 74 20 69 73 20 6e 6f 74 20 69 6e 0d 79 |that is not in.y| 00000c70 6f 75 72 20 63 68 61 72 61 63 74 65 72 20 73 65 |our character se| 00000c80 74 2e 20 20 49 6e 20 6f 75 72 20 63 61 73 65 2c |t. In our case,| 00000c90 20 66 6f 72 20 64 65 63 69 6d 61 6c 20 69 6e 70 | for decimal inp| 00000ca0 75 74 2c 20 77 65 0d 63 6f 6e 74 69 6e 75 65 20 |ut, we.continue | 00000cb0 75 6e 74 69 6c 20 77 65 20 68 69 74 20 61 20 63 |until we hit a c| 00000cc0 68 61 72 61 63 74 65 72 20 74 68 61 74 20 69 73 |haracter that is| 00000cd0 20 6e 6f 74 20 62 65 74 77 65 65 6e 20 7a 65 72 | not between zer| 00000ce0 6f 0d 61 6e 64 20 6e 69 6e 65 2e 0d 0d 57 68 61 |o.and nine...Wha| 00000cf0 74 20 69 73 20 74 68 65 20 76 61 6c 75 65 20 6f |t is the value o| 00000d00 66 20 61 20 64 69 67 69 74 3f 20 20 54 68 61 74 |f a digit? That| 00000d10 20 69 73 20 65 61 73 79 20 66 6f 72 20 74 68 65 | is easy for the| 00000d20 20 6e 75 6d 62 65 72 73 0d 30 20 74 6f 20 39 20 | numbers.0 to 9 | 00000d30 62 65 63 61 75 73 65 20 74 68 65 20 41 53 43 49 |because the ASCI| 00000d40 49 20 63 6f 64 65 73 20 66 6f 72 20 74 68 65 6d |I codes for them| 00000d50 20 61 72 65 20 26 33 30 20 74 6f 20 26 33 39 20 | are &30 to &39 | 00000d60 6f 72 20 34 38 0d 74 6f 20 35 37 2e 20 20 57 65 |or 48.to 57. We| 00000d70 20 6a 75 73 74 20 73 75 62 74 72 61 63 74 20 34 | just subtract 4| 00000d80 38 20 66 72 6f 6d 20 74 68 65 20 41 53 43 49 49 |8 from the ASCII| 00000d90 2e 0d 0d 53 6f 20 77 65 20 6e 65 65 64 20 74 6f |...So we need to| 00000da0 20 62 65 20 61 62 6c 65 20 74 6f 20 6d 75 6c 74 | be able to mult| 00000db0 69 70 6c 79 20 61 20 6e 75 6d 62 65 72 20 62 79 |iply a number by| 00000dc0 20 74 68 65 20 62 61 73 65 20 69 6e 0d 6f 72 64 | the base in.ord| 00000dd0 65 72 20 74 6f 20 63 61 72 72 79 20 6f 75 74 20 |er to carry out | 00000de0 74 68 69 73 20 69 6e 70 75 74 20 70 72 6f 63 65 |this input proce| 00000df0 64 75 72 65 2e 20 20 54 68 65 20 72 6f 75 74 69 |dure. The routi| 00000e00 6e 65 20 69 6e 0d 74 68 69 73 20 6d 6f 64 75 6c |ne in.this modul| 00000e10 65 20 69 73 20 69 6e 20 66 61 63 74 20 67 65 6e |e is in fact gen| 00000e20 65 72 61 6c 20 65 6e 6f 75 67 68 20 74 6f 20 63 |eral enough to c| 00000e30 6f 70 65 20 77 69 74 68 20 61 6e 79 20 62 61 73 |ope with any bas| 00000e40 65 0d 75 70 20 74 6f 20 32 35 35 20 28 61 6c 74 |e.up to 255 (alt| 00000e50 68 6f 75 67 68 20 73 6f 72 74 69 6e 67 20 6f 75 |hough sorting ou| 00000e60 74 20 74 68 65 20 76 61 6c 75 65 73 20 6f 66 20 |t the values of | 00000e70 74 68 65 20 63 68 61 72 61 63 74 65 72 73 0d 69 |the characters.i| 00000e80 6e 20 74 68 61 74 20 62 61 73 65 20 6d 69 67 68 |n that base migh| 00000e90 74 20 63 61 75 73 65 20 73 6f 6d 65 20 70 72 6f |t cause some pro| 00000ea0 62 6c 65 6d 73 29 20 62 75 74 20 66 6f 72 20 74 |blems) but for t| 00000eb0 68 65 20 74 69 6d 65 0d 62 65 69 6e 67 20 77 65 |he time.being we| 00000ec0 27 6c 6c 20 75 73 65 20 69 74 20 74 6f 20 64 65 |'ll use it to de| 00000ed0 61 6c 20 77 69 74 68 20 64 65 63 69 6d 61 6c 20 |al with decimal | 00000ee0 6e 75 6d 62 65 72 73 2e 20 20 54 68 65 20 62 61 |numbers. The ba| 00000ef0 73 65 73 0d 79 6f 75 20 6d 69 67 68 74 20 77 61 |ses.you might wa| 00000f00 6e 74 20 74 6f 20 75 73 65 20 65 76 65 6e 74 75 |nt to use eventu| 00000f10 61 6c 6c 79 20 61 72 65 20 62 69 6e 61 72 79 20 |ally are binary | 00000f20 28 32 29 20 6f 63 74 61 6c 20 28 38 29 0d 64 65 |(2) octal (8).de| 00000f30 63 69 6d 61 6c 20 28 31 30 29 20 61 6e 64 20 68 |cimal (10) and h| 00000f40 65 78 61 64 65 63 69 6d 61 6c 20 28 31 36 29 2e |exadecimal (16).| 00000f50 0d 0d 54 68 65 20 6c 61 72 67 65 73 74 20 6e 75 |..The largest nu| 00000f60 6d 62 65 72 20 74 68 61 74 20 63 61 6e 20 62 65 |mber that can be| 00000f70 20 68 65 6c 64 20 69 6e 20 34 20 62 79 74 65 73 | held in 4 bytes| 00000f80 20 75 73 69 6e 67 20 32 27 73 0d 63 6f 6d 70 6c | using 2's.compl| 00000f90 65 6d 65 6e 74 20 6e 65 67 61 74 69 76 65 20 69 |ement negative i| 00000fa0 73 20 32 31 34 37 34 38 33 36 34 37 20 70 6f 73 |s 2147483647 pos| 00000fb0 69 74 69 76 65 20 6f 72 20 2d 20 32 31 34 37 34 |itive or - 21474| 00000fc0 38 33 36 34 38 0d 6e 65 67 61 74 69 76 65 20 77 |83648.negative w| 00000fd0 68 69 63 68 20 63 6f 72 72 65 73 70 6f 6e 64 20 |hich correspond | 00000fe0 74 6f 20 26 37 46 46 46 46 46 46 46 20 61 6e 64 |to &7FFFFFFF and| 00000ff0 20 26 38 30 30 30 30 30 30 30 2e 20 20 49 6e 0d | &80000000. In.| 00001000 66 61 63 74 20 69 6e 20 74 68 69 73 20 72 6f 75 |fact in this rou| 00001010 74 69 6e 65 20 49 20 68 61 76 65 20 73 61 63 72 |tine I have sacr| 00001020 69 66 69 63 65 64 20 74 68 65 20 61 62 69 6c 69 |ificed the abili| 00001030 74 79 20 74 6f 20 65 6e 74 65 72 0d 2d 20 32 31 |ty to enter.- 21| 00001040 34 37 34 38 33 36 34 38 20 69 74 73 65 6c 66 20 |47483648 itself | 00001050 69 6e 20 6f 72 64 65 72 20 74 6f 20 73 69 6d 70 |in order to simp| 00001060 6c 69 66 79 20 74 68 65 20 63 6f 64 69 6e 67 20 |lify the coding | 00001070 61 0d 6c 69 74 74 6c 65 2e 20 20 54 68 65 20 6c |a.little. The l| 00001080 61 72 67 65 73 74 20 6e 65 67 61 74 69 76 65 20 |argest negative | 00001090 6e 75 6d 62 65 72 20 69 73 20 2d 20 32 31 34 37 |number is - 2147| 000010a0 34 38 33 36 34 37 2c 20 62 75 74 20 49 0d 64 6f |483647, but I.do| 000010b0 75 62 74 20 69 66 20 74 68 61 74 20 77 69 6c 6c |ubt if that will| 000010c0 20 63 61 75 73 65 20 6d 75 63 68 20 77 61 69 6c | cause much wail| 000010d0 69 6e 67 20 61 6e 64 20 67 6e 61 73 68 69 6e 67 |ing and gnashing| 000010e0 20 6f 66 20 74 65 65 74 68 2e 0d 0d 53 6f 20 6a | of teeth...So j| 000010f0 75 73 74 20 68 6f 77 20 64 6f 20 77 65 20 6d 75 |ust how do we mu| 00001100 6c 74 69 70 6c 79 20 69 6e 20 6d 61 63 68 69 6e |ltiply in machin| 00001110 65 20 63 6f 64 65 3f 20 20 57 65 6c 6c 0d 66 6f |e code? Well.fo| 00001120 72 74 75 6e 61 74 65 6c 79 20 77 65 20 63 61 6e |rtunately we can| 00001130 20 66 61 6c 6c 20 62 61 63 6b 20 6f 6e 20 67 6f | fall back on go| 00001140 6f 64 20 6f 6c 64 20 66 61 73 68 69 6f 6e 65 64 |od old fashioned| 00001150 20 6c 6f 6e 67 0d 6d 75 6c 74 69 70 6c 69 63 61 | long.multiplica| 00001160 74 69 6f 6e 2c 20 77 68 69 63 68 20 69 73 20 61 |tion, which is a| 00001170 62 73 75 72 64 6c 79 20 73 69 6d 70 6c 65 20 69 |bsurdly simple i| 00001180 6e 20 62 69 6e 61 72 79 20 73 69 6e 63 65 20 79 |n binary since y| 00001190 6f 75 0d 68 61 76 65 20 61 20 76 65 72 79 20 72 |ou.have a very r| 000011a0 75 64 69 6d 65 6e 74 61 72 79 20 6d 75 6c 74 69 |udimentary multi| 000011b0 70 6c 69 63 61 74 69 6f 6e 20 74 61 62 6c 65 2e |plication table.| 000011c0 0d 0d 4c 65 74 27 73 20 6d 75 6c 74 69 70 6c 79 |..Let's multiply| 000011d0 20 31 31 30 31 31 20 62 79 20 31 30 31 31 20 28 | 11011 by 1011 (| 000011e0 69 2e 65 2e 20 32 37 20 62 79 20 31 31 29 0d 0d |i.e. 27 by 11)..| 000011f0 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | | 00001200 31 31 30 31 31 0d 20 20 20 20 20 20 20 20 20 20 |11011. | 00001210 20 20 20 20 20 20 20 20 20 20 78 0d 20 20 20 20 | x. | 00001220 20 20 20 20 20 20 20 20 20 20 20 20 20 31 30 31 | 101| 00001230 31 0d 20 20 20 20 20 20 20 20 20 20 20 20 2d 2d |1. --| 00001240 2d 2d 2d 2d 2d 2d 2d 0d 20 20 20 20 20 20 20 20 |-------. | 00001250 20 20 20 20 20 31 31 30 31 31 30 30 30 0d 20 20 | 11011000. | 00001260 20 20 20 20 20 20 20 20 20 20 20 20 20 31 31 30 | 110| 00001270 31 31 30 0d 20 20 20 20 20 20 20 20 20 20 20 20 |110. | 00001280 20 20 20 20 31 31 30 31 31 0d 20 20 20 20 20 20 | 11011. | 00001290 20 20 20 20 20 20 2d 2d 2d 2d 2d 2d 2d 2d 2d 0d | ---------.| 000012a0 20 20 20 20 20 20 20 20 20 20 20 20 31 30 30 31 | 1001| 000012b0 30 31 30 30 31 20 20 20 20 20 77 68 69 63 68 20 |01001 which | 000012c0 69 73 20 20 32 39 37 0d 0d 49 20 68 6f 70 65 20 |is 297..I hope | 000012d0 79 6f 75 20 63 61 6e 20 73 65 65 20 74 68 61 74 |you can see that| 000012e0 20 66 6f 72 20 65 61 63 68 20 73 65 74 20 62 69 | for each set bi| 000012f0 74 20 6f 66 20 74 68 65 20 6d 75 6c 74 69 70 6c |t of the multipl| 00001300 69 65 72 0d 28 31 31 30 31 29 20 74 68 65 72 65 |ier.(1101) there| 00001310 20 69 73 20 61 20 72 6f 77 20 69 6e 20 74 68 65 | is a row in the| 00001320 20 73 70 61 63 65 20 62 65 74 77 65 65 6e 20 74 | space between t| 00001330 68 65 20 64 6f 74 74 65 64 20 6c 69 6e 65 73 2e |he dotted lines.| 00001340 20 0d 4a 75 73 74 20 6c 69 6b 65 20 64 65 63 69 | .Just like deci| 00001350 6d 61 6c 20 6c 6f 6e 67 20 6d 75 6c 74 69 70 6c |mal long multipl| 00001360 69 63 61 74 69 6f 6e 20 74 68 65 20 72 6f 77 73 |ication the rows| 00001370 20 61 72 65 20 6f 66 66 73 65 74 20 74 6f 0d 74 | are offset to.t| 00001380 68 65 20 6c 65 66 74 20 63 6f 72 72 65 73 70 6f |he left correspo| 00001390 6e 64 69 6e 67 20 74 6f 20 74 68 65 20 27 70 6f |nding to the 'po| 000013a0 77 65 72 27 20 6f 66 20 74 68 65 20 62 69 74 20 |wer' of the bit | 000013b0 74 68 65 20 72 6f 77 0d 62 65 6c 6f 6e 67 73 20 |the row.belongs | 000013c0 74 6f 2e 20 20 59 6f 75 20 73 68 6f 75 6c 64 20 |to. You should | 000013d0 6e 6f 74 65 20 61 6c 73 6f 20 74 68 61 74 20 79 |note also that y| 000013e0 6f 75 20 65 73 73 65 6e 74 69 61 6c 6c 79 20 6f |ou essentially o| 000013f0 6e 6c 79 0d 6e 65 65 64 20 74 6f 20 6b 6e 6f 77 |nly.need to know| 00001400 20 74 68 65 20 6f 6e 65 20 74 69 6d 65 73 20 74 | the one times t| 00001410 61 62 6c 65 20 61 6e 64 20 74 68 61 74 20 74 68 |able and that th| 00001420 65 20 61 64 64 69 6e 67 20 75 70 20 6f 66 0d 74 |e adding up of.t| 00001430 68 65 20 70 61 72 74 69 61 6c 20 70 72 6f 64 75 |he partial produ| 00001440 63 74 73 20 69 73 20 6d 6f 72 65 20 64 69 66 66 |cts is more diff| 00001450 69 63 75 6c 74 20 66 6f 72 20 61 20 6d 65 72 65 |icult for a mere| 00001460 20 68 75 6d 61 6e 20 74 68 61 6e 0d 74 68 65 20 | human than.the | 00001470 6d 75 6c 74 69 70 6c 79 69 6e 67 20 74 68 61 74 |multiplying that| 00001480 20 70 72 6f 64 75 63 65 73 20 65 61 63 68 20 70 | produces each p| 00001490 61 72 74 69 61 6c 20 70 72 6f 64 75 63 74 2e 0d |artial product..| 000014a0 0d 49 6e 20 6d 61 63 68 69 6e 65 20 63 6f 64 65 |.In machine code| 000014b0 20 74 68 65 20 27 64 69 66 66 69 63 75 6c 74 27 | the 'difficult'| 000014c0 20 61 64 64 69 6e 67 20 69 73 20 74 61 6b 65 6e | adding is taken| 000014d0 20 63 61 72 65 20 6f 66 0d 65 61 73 69 6c 79 20 | care of.easily | 000014e0 75 73 69 6e 67 20 41 44 43 2e 20 20 53 6f 20 77 |using ADC. So w| 000014f0 68 61 74 20 61 62 6f 75 74 20 74 68 6f 73 65 20 |hat about those | 00001500 70 61 72 74 69 61 6c 20 70 72 6f 64 75 63 74 73 |partial products| 00001510 0d 62 65 69 6e 67 20 6f 66 66 73 65 74 20 6c 65 |.being offset le| 00001520 66 74 20 2e 2e 2e 2e 20 77 65 6c 6c 20 74 68 61 |ft .... well tha| 00001530 74 27 73 20 72 6f 74 61 74 69 6e 67 20 69 73 6e |t's rotating isn| 00001540 27 74 20 69 74 3f 20 20 53 6f 0d 68 6f 77 20 64 |'t it? So.how d| 00001550 6f 65 73 20 69 74 20 6c 6f 6f 6b 20 61 73 20 61 |oes it look as a| 00001560 6e 20 61 6c 67 6f 72 69 74 68 6d 3f 0d 0d 57 65 |n algorithm?..We| 00001570 27 6c 6c 20 61 63 74 75 61 6c 6c 79 20 64 6f 20 |'ll actually do | 00001580 74 68 69 73 20 62 61 63 6b 77 61 72 64 73 20 73 |this backwards s| 00001590 6f 20 69 74 20 6c 6f 6f 6b 73 20 6d 6f 72 65 20 |o it looks more | 000015a0 6c 69 6b 65 0d 0d 20 20 20 20 20 20 20 20 20 20 |like.. | 000015b0 20 20 20 20 20 20 20 20 20 20 31 31 30 31 31 0d | 11011.| 000015c0 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | | 000015d0 20 20 20 20 20 20 20 20 78 0d 20 20 20 20 20 20 | x. | 000015e0 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 31 | 1| 000015f0 30 31 31 0d 20 20 20 20 20 20 20 20 20 20 20 20 |011. | 00001600 20 20 20 20 2d 2d 2d 2d 2d 2d 2d 2d 2d 0d 20 20 | ---------. | 00001610 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | | 00001620 20 20 31 31 30 31 31 0d 20 20 20 20 20 20 20 20 | 11011. | 00001630 20 20 20 20 20 20 20 20 20 20 20 31 31 30 31 31 | 11011| 00001640 30 0d 20 20 20 20 20 20 20 20 20 20 20 20 20 20 |0. | 00001650 20 20 20 31 31 30 31 31 30 30 30 0d 20 20 20 20 | 11011000. | 00001660 20 20 20 20 20 20 20 20 20 20 20 20 2d 2d 2d 2d | ----| 00001670 2d 2d 2d 2d 2d 0d 20 20 20 20 20 20 20 20 20 20 |-----. | 00001680 20 20 20 20 20 20 31 30 30 31 30 31 30 30 31 0d | 100101001.| 00001690 0d 62 75 74 20 74 68 65 20 72 65 73 75 6c 74 20 |.but the result | 000016a0 69 73 20 74 68 65 20 73 61 6d 65 2e 0d 0d 54 61 |is the same...Ta| 000016b0 6b 65 20 74 68 65 20 72 69 67 68 74 6d 6f 73 74 |ke the rightmost| 000016c0 20 28 6c 65 61 73 74 20 73 69 67 6e 69 66 69 63 | (least signific| 000016d0 61 6e 74 20 62 69 74 29 20 6f 66 20 74 68 65 20 |ant bit) of the | 000016e0 6d 75 6c 74 69 70 6c 69 65 72 0d 61 6e 64 20 69 |multiplier.and i| 000016f0 66 20 69 74 20 69 73 20 73 65 74 20 61 64 64 20 |f it is set add | 00001700 74 68 65 20 6d 75 6c 74 69 70 6c 69 63 61 6e 64 |the multiplicand| 00001710 20 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20 69 73 20 7a 65 72 6f 2e 0d 0d |plier is zero...| 000017f0 49 6e 20 70 72 61 63 74 69 63 65 20 79 6f 75 20 |In practice you | 00001800 72 6f 74 61 74 65 20 74 68 65 20 6d 75 6c 74 69 |rotate the multi| 00001810 70 6c 69 65 72 20 66 69 72 73 74 20 62 65 63 61 |plier first beca| 00001820 75 73 65 20 79 6f 75 20 75 73 65 0d 74 68 65 20 |use you use.the | 00001830 63 61 72 72 79 20 66 6c 61 67 20 74 6f 20 63 68 |carry flag to ch| 00001840 65 63 6b 20 77 68 65 74 68 65 72 20 61 20 62 69 |eck whether a bi| 00001850 74 20 69 73 20 73 65 74 2c 20 73 6f 20 69 74 20 |t is set, so it | 00001860 6c 6f 6f 6b 73 0d 6c 69 6b 65 20 74 68 69 73 20 |looks.like this | 00001870 69 66 20 62 6f 74 68 20 6e 75 6d 62 65 72 73 20 |if both numbers | 00001880 61 6e 64 20 74 68 65 69 72 20 70 72 6f 64 75 63 |and their produc| 00001890 74 20 61 72 65 20 6c 65 73 73 20 74 68 61 6e 20 |t are less than | 000018a0 61 0d 62 79 74 65 20 69 6e 20 73 69 7a 65 3a 0d |a.byte in size:.| 000018b0 0d 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 |. | 000018c0 20 2e 6c 6f 6f 70 0d 20 20 20 20 20 20 20 20 20 | .loop. | 000018d0 20 20 20 20 20 20 20 4c 53 52 20 6d 75 6c 74 69 | LSR multi| 000018e0 70 6c 69 65 72 0d 20 20 20 20 20 20 20 20 20 20 |plier. | 000018f0 20 20 20 20 20 20 42 43 43 20 6e 6f 5f 61 64 64 | BCC no_add| 00001900 0d 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 |. | 00001910 20 43 4c 43 0d 20 20 20 20 20 20 20 20 20 20 20 | CLC. | 00001920 20 20 20 20 20 4c 44 41 20 6d 75 6c 74 69 70 6c | LDA multipl| 00001930 69 63 61 6e 64 0d 20 20 20 20 20 20 20 20 20 20 |icand. | 00001940 20 20 20 20 20 20 41 44 43 20 70 70 77 73 0d 20 | ADC ppws. | 00001950 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 53 | S| 00001960 54 41 20 70 70 77 73 0d 20 20 20 20 20 20 20 20 |TA ppws. | 00001970 20 20 20 20 20 20 20 20 2e 6e 6f 5f 61 64 64 0d | .no_add.| 00001980 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | | 00001990 42 45 51 20 65 78 69 74 0d 20 20 20 20 20 20 20 |BEQ exit. | 000019a0 20 20 20 20 20 20 20 20 20 41 53 4c 20 6d 75 6c | ASL mul| 000019b0 74 69 70 6c 69 63 61 6e 64 0d 20 20 20 20 20 20 |tiplicand. | 000019c0 20 20 20 20 20 20 20 20 20 20 4a 4d 50 20 6c 6f | JMP lo| 000019d0 6f 70 0d 0d 0d 49 66 20 79 6f 75 72 20 70 72 6f |op...If your pro| 000019e0 64 75 63 74 20 69 73 20 6c 61 72 67 65 72 20 74 |duct is larger t| 000019f0 68 61 6e 20 61 20 62 79 74 65 20 28 61 73 20 69 |han a byte (as i| 00001a00 74 20 77 61 73 20 69 6e 20 6f 75 72 20 6c 6f 6e |t was in our lon| 00001a10 67 0d 6d 75 6c 74 69 70 6c 69 63 61 74 69 6f 6e |g.multiplication| 00001a20 20 65 78 61 6d 70 6c 65 20 61 62 6f 76 65 29 20 | example above) | 00001a30 74 68 65 6e 20 79 6f 75 20 73 69 6d 70 6c 79 20 |then you simply | 00001a40 68 61 76 65 20 74 6f 20 61 64 64 0d 7a 65 72 6f |have to add.zero| 00001a50 20 28 77 69 74 68 20 63 61 72 72 79 29 20 74 6f | (with carry) to| 00001a60 20 74 68 65 20 68 69 67 68 20 62 79 74 65 2c 20 | the high byte, | 00001a70 61 66 74 65 72 20 79 6f 75 20 68 61 76 65 20 61 |after you have a| 00001a80 64 64 65 64 20 74 68 65 0d 6c 6f 77 20 62 79 74 |dded the.low byt| 00001a90 65 73 20 74 6f 67 65 74 68 65 72 2e 20 20 4c 65 |es together. Le| 00001aa0 74 27 73 20 73 61 79 20 74 68 61 74 20 74 68 65 |t's say that the| 00001ab0 20 70 61 72 74 69 61 6c 20 70 72 6f 64 75 63 74 | partial product| 00001ac0 20 77 61 73 0d 69 6e 20 70 70 77 73 20 61 6e 64 | was.in ppws and| 00001ad0 20 70 70 77 73 2b 31 20 61 6e 64 20 74 68 61 74 | ppws+1 and that| 00001ae0 20 79 6f 75 20 77 65 72 65 20 61 64 64 69 6e 67 | you were adding| 00001af0 20 69 6e 20 61 20 76 61 6c 75 65 20 66 72 6f 6d | in a value from| 00001b00 0d 74 68 65 20 61 63 63 75 6d 75 6c 61 74 6f 72 |.the accumulator| 00001b10 2e 20 20 49 74 20 77 6f 75 6c 64 20 6c 6f 6f 6b |. It would look| 00001b20 20 6c 69 6b 65 20 74 68 69 73 3a 20 0d 0d 20 20 | like this: .. | 00001b30 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | | 00001b40 20 43 4c 43 0d 20 20 20 20 20 20 20 20 20 20 20 | CLC. | 00001b50 20 20 20 20 20 20 20 20 41 44 43 20 70 70 77 73 | ADC ppws| 00001b60 0d 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 |. | 00001b70 20 20 20 20 53 54 41 20 70 70 77 73 0d 20 20 20 | STA ppws. | 00001b80 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 | | 00001b90 4c 44 41 20 23 30 0d 20 20 20 20 20 20 20 20 20 |LDA #0. | 00001ba0 20 20 20 20 20 20 20 20 20 20 41 44 43 20 70 70 | ADC pp| 00001bb0 77 73 2b 31 0d 20 20 20 20 20 20 20 20 20 20 20 |ws+1. | 00001bc0 20 20 20 20 20 20 20 20 53 54 41 20 70 70 77 73 | STA ppws| 00001bd0 2b 31 0d 0d 4e 6f 74 65 20 74 68 61 74 20 77 65 |+1..Note that we| 00001be0 20 64 6f 6e 27 74 20 63 6c 65 61 72 20 74 68 65 | don't clear the| 00001bf0 20 63 61 72 72 79 20 62 65 66 6f 72 65 20 74 68 | carry before th| 00001c00 65 20 73 65 63 6f 6e 64 0d 61 64 64 69 74 69 6f |e second.additio| 00001c10 6e 2e 0d 0d 49 74 20 69 73 20 70 6f 73 73 69 62 |n...It is possib| 00001c20 6c 65 20 74 6f 20 68 6f 6c 64 20 74 68 65 20 6d |le to hold the m| 00001c30 75 6c 74 69 70 6c 69 65 72 20 69 6e 20 74 68 65 |ultiplier in the| 00001c40 20 61 63 63 75 6d 75 6c 61 74 6f 72 0d 28 74 68 | accumulator.(th| 00001c50 69 73 20 69 73 20 64 6f 6e 65 20 69 6e 20 74 68 |is is done in th| 00001c60 65 20 70 72 6f 67 72 61 6d 20 6d 6f 64 75 6c 65 |e program module| 00001c70 29 20 62 75 74 20 69 66 20 65 69 74 68 65 72 0d |) but if either.| 00001c80 6d 75 6c 74 69 70 6c 69 65 72 20 6f 72 20 6d 75 |multiplier or mu| 00001c90 6c 74 69 70 6c 69 63 61 6e 64 20 61 72 65 20 67 |ltiplicand are g| 00001ca0 72 65 61 74 65 72 20 74 68 61 6e 20 32 35 35 20 |reater than 255 | 00001cb0 74 68 65 6e 0d 6d 75 6c 74 69 2d 62 79 74 65 20 |then.multi-byte | 00001cc0 72 6f 74 61 74 69 6e 67 20 61 6e 64 20 6d 75 6c |rotating and mul| 00001cd0 74 69 2d 62 79 74 65 20 61 64 64 69 74 69 6f 6e |ti-byte addition| 00001ce0 20 6d 75 73 74 20 62 65 20 75 73 65 64 2e 20 0d | must be used. .| 00001cf0 4d 75 6c 74 69 2d 62 79 74 65 20 61 64 64 69 74 |Multi-byte addit| 00001d00 69 6f 6e 20 49 20 68 61 76 65 20 61 6c 72 65 61 |ion I have alrea| 00001d10 64 79 20 63 6f 76 65 72 65 64 2e 20 20 4d 75 6c |dy covered. Mul| 00001d20 74 69 2d 62 79 74 65 0d 72 6f 74 61 74 69 6e 67 |ti-byte.rotating| 00001d30 20 69 73 20 73 74 72 61 69 67 68 74 66 6f 72 77 | is straightforw| 00001d40 61 72 64 20 65 6e 6f 75 67 68 2e 20 20 42 61 73 |ard enough. 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T| 00002470 68 65 20 73 75 62 72 6f 75 74 69 6e 65 20 73 74 |he subroutine st| 00002480 61 72 74 69 6e 67 20 61 74 0d 27 69 6e 70 75 74 |arting at.'input| 00002490 5f 63 68 61 72 61 63 74 65 72 27 20 69 73 20 65 |_character' is e| 000024a0 73 73 65 6e 74 69 61 6c 6c 79 20 74 68 61 74 20 |ssentially that | 000024b0 66 72 6f 6d 20 74 68 65 20 6c 61 73 74 20 6d 6f |from the last mo| 000024c0 64 75 6c 65 2e 0d 54 68 65 20 73 75 62 72 6f 75 |dule..The subrou| 000024d0 74 69 6e 65 20 61 74 20 27 63 6f 6e 76 65 72 74 |tine at 'convert| 000024e0 5f 41 53 43 49 49 5f 74 6f 5f 62 69 6e 61 72 79 |_ASCII_to_binary| 000024f0 27 20 74 61 6b 65 73 20 74 68 65 20 73 74 72 69 |' takes the stri| 00002500 6e 67 0d 69 6e 20 74 68 65 20 69 6e 70 75 74 20 |ng.in the input | 00002510 62 75 66 66 65 72 20 61 6e 64 20 63 6f 6e 76 65 |buffer and conve| 00002520 72 74 73 20 69 74 20 69 6e 74 6f 20 61 20 34 20 |rts it into a 4 | 00002530 62 79 74 65 20 62 69 6e 61 72 79 0d 6e 75 6d 62 |byte binary.numb| 00002540 65 72 20 61 6e 64 20 6c 65 61 76 65 73 20 74 68 |er and leaves th| 00002550 65 20 72 65 73 75 6c 74 20 69 6e 20 34 20 62 79 |e result in 4 by| 00002560 74 65 73 20 6f 66 20 6d 65 6d 6f 72 79 20 73 74 |tes of memory st| 00002570 61 72 74 69 6e 67 0d 61 74 20 27 64 69 67 69 74 |arting.at 'digit| 00002580 5f 77 73 27 2e 0d 0d 54 68 65 20 63 6f 6e 76 65 |_ws'...The conve| 00002590 72 73 69 6f 6e 20 73 75 62 72 6f 75 74 69 6e 65 |rsion subroutine| 000025a0 20 69 73 20 61 74 20 6c 69 6e 65 20 31 35 35 30 | is at line 1550| 000025b0 2e 20 20 4e 6f 74 65 20 74 68 61 74 20 74 68 65 |. 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The sign fla| 00002630 67 20 69 73 0d 63 6c 65 61 72 65 64 20 61 6c 73 |g is.cleared als| 00002640 6f 2e 0d 0d 57 65 20 67 6f 20 74 68 72 6f 75 67 |o...We go throug| 00002650 68 20 74 68 65 20 73 74 72 69 6e 67 20 69 6e 20 |h the string in | 00002660 74 68 65 20 69 6e 70 75 74 20 62 75 66 66 65 72 |the input buffer| 00002670 20 61 20 63 68 61 72 61 63 74 65 72 20 61 74 0d | a character at.| 00002680 61 20 74 69 6d 65 20 75 73 69 6e 67 20 58 20 61 |a time using X a| 00002690 73 20 61 20 63 6f 75 6e 74 65 72 2f 69 6e 64 65 |s a counter/inde| 000026a0 78 2e 20 20 49 66 20 61 20 63 68 61 72 61 63 74 |x. If a charact| 000026b0 65 72 20 69 73 20 6e 6f 74 0d 62 65 74 77 65 65 |er is not.betwee| 000026c0 6e 20 30 20 61 6e 64 20 39 20 77 65 20 62 72 61 |n 0 and 9 we bra| 000026d0 6e 63 68 20 74 6f 20 27 6e 6f 74 5f 6e 75 6d 62 |nch to 'not_numb| 000026e0 65 72 27 2e 20 20 54 68 65 72 65 2c 20 69 66 20 |er'. 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I c| 000028e0 6f 75 6c 64 20 74 75 72 6e 20 74 68 65 20 6e 75 |ould turn the nu| 000028f0 6d 62 65 72 20 6e 65 67 61 74 69 76 65 20 61 66 |mber negative af| 00002900 74 65 72 20 69 74 0d 77 61 73 20 65 6e 74 65 72 |ter it.was enter| 00002910 65 64 20 6f 72 20 49 20 63 6f 75 6c 64 20 61 63 |ed or I could ac| 00002920 74 75 61 6c 6c 79 20 62 75 69 6c 64 20 75 70 20 |tually build up | 00002930 61 20 6e 65 67 61 74 69 76 65 20 6e 75 6d 62 65 |a negative numbe| 00002940 72 0d 64 75 72 69 6e 67 20 74 68 65 20 6e 75 6d |r.during the num| 00002950 62 65 72 20 65 6e 74 72 79 20 72 6f 75 74 69 6e |ber entry routin| 00002960 65 2e 20 20 49 6e 20 74 68 65 20 65 6e 64 20 49 |e. In the end I| 00002970 20 64 65 63 69 64 65 64 20 74 6f 0d 63 6f 6e 76 | decided to.conv| 00002980 65 72 74 20 61 66 74 65 72 20 65 6e 74 72 79 2c |ert after entry,| 00002990 20 73 69 6e 63 65 20 74 68 69 73 20 6d 61 6b 65 | since this make| 000029a0 73 20 74 68 65 20 63 6f 64 65 20 73 69 6d 70 6c |s the code simpl| 000029b0 65 72 20 61 6e 64 0d 68 61 73 20 6f 6e 6c 79 20 |er and.has only | 000029c0 74 68 65 20 73 6c 69 67 68 74 20 73 69 64 65 20 |the slight side | 000029d0 65 66 66 65 63 74 20 6f 66 20 69 6e 76 61 6c 69 |effect of invali| 000029e0 64 61 74 69 6e 67 20 74 68 65 20 6c 61 72 67 65 |dating the large| 000029f0 73 74 0d 6e 65 67 61 74 69 76 65 20 6e 75 6d 62 |st.negative numb| 00002a00 65 72 20 28 26 38 30 30 30 30 30 30 30 29 20 61 |er (&80000000) a| 00002a10 73 20 49 20 6d 65 6e 74 69 6f 6e 65 64 20 62 65 |s I mentioned be| 00002a20 66 6f 72 65 2e 20 20 41 66 74 65 72 0d 74 68 65 |fore. 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Here | 00002dc0 69 73 20 61 6e 20 65 72 72 6f 72 20 74 72 61 70 |is an error trap| 00002dd0 20 74 68 61 74 20 63 61 74 63 68 65 73 20 61 6e | that catches an| 00002de0 79 0d 6e 75 6d 62 65 72 20 74 68 61 74 20 73 65 |y.number that se| 00002df0 74 73 20 74 68 65 20 74 6f 70 20 62 69 74 20 28 |ts the top bit (| 00002e00 62 65 63 6f 6d 65 73 20 6e 65 67 61 74 69 76 65 |becomes negative| 00002e10 29 20 64 75 72 69 6e 67 20 74 68 69 73 0d 61 64 |) during this.ad| 00002e20 64 69 74 69 6f 6e 2e 20 20 49 66 20 74 68 61 74 |dition. If that| 00002e30 20 68 61 70 70 65 6e 73 20 74 68 65 20 6e 75 6d | happens the num| 00002e40 62 65 72 20 68 61 73 20 6f 76 65 72 66 6c 6f 77 |ber has overflow| 00002e50 65 64 20 34 0d 62 79 74 65 73 2e 20 20 54 68 69 |ed 4.bytes. Thi| 00002e60 73 20 69 73 20 61 6c 73 6f 20 64 6f 6e 65 20 61 |s is also done a| 00002e70 74 20 6c 69 6e 65 20 33 30 35 30 2e 20 20 49 20 |t line 3050. 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Otherwise| 00003120 20 74 68 69 73 20 73 65 63 74 69 6f 6e 20 6f 66 | this section of| 00003130 20 63 6f 64 65 20 69 73 0d 61 20 73 74 72 61 69 | code is.a strai| 00003140 67 68 74 66 6f 72 77 61 72 64 20 6d 75 6c 74 69 |ghtforward multi| 00003150 70 6c 69 63 61 74 69 6f 6e 20 77 68 65 72 65 20 |plication where | 00003160 74 68 65 20 6d 75 6c 74 69 70 6c 69 65 72 20 69 |the multiplier i| 00003170 73 0d 6f 6e 6c 79 20 61 20 62 79 74 65 20 69 6e |s.only a byte in| 00003180 20 73 69 7a 65 20 61 6e 64 20 74 68 65 20 6d 75 | size and the mu| 00003190 6c 74 69 70 6c 69 63 61 6e 64 20 69 73 20 34 20 |ltiplicand is 4 | 000031a0 62 79 74 65 73 20 69 6e 20 73 69 7a 65 2e 0d 0d |bytes in size...| 000031b0 49 20 77 61 6e 74 65 64 20 74 6f 20 70 75 74 20 |I wanted to put | 000031c0 74 68 65 20 66 69 6e 61 6c 20 72 65 73 75 6c 74 |the final result| 000031d0 20 62 61 63 6b 20 69 6e 20 64 69 67 69 74 5f 77 | back in digit_w| 000031e0 73 20 73 6f 20 6c 69 6e 65 73 0d 33 32 37 30 20 |s so lines.3270 | 000031f0 74 6f 20 33 33 34 30 20 63 6f 70 79 20 69 74 20 |to 3340 copy it | 00003200 61 63 72 6f 73 73 20 74 68 65 72 65 2e 20 20 54 |across there. 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30-01-88/T\OSB12.m0
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.