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27_11_87/T\OSB06
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 » telesoftware3.adl |
| Filename: | 27_11_87/T\OSB06 |
| Read OK: | ✔ |
| File size: | 3095 bytes |
| Load address: | 0000 |
| Exec address: | 0000 |
File contents
OSBITS - An Exploration of the BBC Micro at Machine Level
By Programmer
...........................................................
Part 6: Rotating and Shifting
Continuing along the machine code trail this week I'll do
two things. Firstly some more byte manipulation with bits
being rotated and shifted about, and second some thoughts on
where to keep numbers temporarily during your programs.
Right then, let's have some bit manipulation. This is
something you can't do in BASIC although it is needed for
multiplication and division in machine code.
To start with here is a byte (the dots represent the ends of
the byte):
. b7 b6 b5 b4 b3 b2 b1 b0 .
each bit being the number b0 to b7, representing 2^n and
enabling us to make numbers up to 255 out of the byte.
There are four rotating/shifting operations we can carry out
on this byte. First of all we can shift it left one place.
We now have the following result:
b7 . b6 b5 b4 b3 b2 b1 b0 0 .
A new byte holding the old bits b6 down to b0 and with a
zero added in at the end. The value of b7 has now set or
cleared the Carry flag. This operation is called an
Arithmetic Shift Left and the mnemonic for it is ASL. You
will probably also have seen that this operation also
effectively multiplies the number held in the byte by 2.
The equivalent operation but going right is a Logical Shift
Right with mnemonic LSR. This time the value of bit 0 is
put in the carry flag and bit 7 replaced by a zero whilst
all the bits are shifted down one. This effectively halves
the number in the byte.
OK so you may now be wondering in which two other directions
can we shift things. Well obviously we only have left and
right but the other two operations rotate the carry flag
into the byte instead of shifting in a zero.
So we have ROL and ROR for a rotation left or right
respectively. With ROL bits 0 to 6 are moved up one, the
carry flag is put into bit 0 and the original bit 7 is put
in the carry flag. With ROR bits 7 to 1 are moved down one,
the Carry flag is put into bit 7, and the original bit 0 is
put in the carry flag.
Obviously it is important that the state of the carry flag
is known before you ROL or ROR.
These four operations can be carried out on memory, e.g.
ROR temp
ASL &70
or on the Accumulator e.g.
LSR A
Besides having a place in multiplication and division these
operations enable you to examine a byte bit by bit. If you
rotate a byte in the accumulator or in a memory location bit
by bit you can then test the Carry flag to see what each bit
is. The program with this week's module, B/osb06, includes
a section that examines a byte in this way and prints out
its value in binary.
Most of the operations we have discussed so far, and
similarly for the others we will come to, can operate on
memory locations as well as the accumulator. Now I have
said already that you use a label to indicate a memory
location and that such a label behaves as a BASIC variable
name. Remember though that the variable holds the location
not what is in it.
Let's think about holding variables for a moment. In
assembly language programming, as with many computer
languages (but not BBC BASIC ) you have to allocate a slot
for your variables before you actually store something. I
use the term variable a little loosely to also include
constants that occupy a memory location throughout the whole
program.
We've already come across this allocation of space and have
used EQU or FNEQU methods for allocating it as the program
is assembled. It's worth noting that you can use a slightly
dirty but nonetheless legal alternative like this:
.variable BRK
which sets the byte at address 'variable' to zero (the
mnemonic BRK assembles to zero). As the actual value you
want to store there changes you just overwrite the byte at
'variable'.
As with other methods of allocating space this must not lie
in the path of your program, or the bytes stored will be
interpreted as instructions with unpredictable results.
If you want your code to run at any memory location
(relocatable code, something I'll discuss in detail in a
later module) you cannot put your variables within the
program space so you could deliberately put them at fixed
locations elsewhere in memory, such as within the zero page
space allocated for the user between &70 and &8F. In this
case the sensible thing to do is define BASIC variables at
the start of your program, outside the assembly bit, just as
I've done with OS routines.
When you put variables into memory you can read the location
as many times as you like and the byte stored will remain
the same. I often use a memory location I call 'accustore'
to store the byte currently being processed. In this way,
no matter what I use the accumulator for I can retrieve the
main value from this location as many times as I like. This
method is used in B/osb06 to hold a byte you type in with
one key stroke while the program prints it to the screen in
binary, hexadecimal and then decimal form.
There is a third and more specialised storage method. This
uses the processor stack. Now the 6502 reserves for itself
a page of memory, from &100 to &1FF, which it uses as a
last-in first-out store. It is rather like a stack of
plates, hence its name. The stack has an important function
in that it automatically holds the return address while
software is executing a subroutine. In fact the RTS command
takes the top two memory locations off the stack and puts
them in the program counter. I mention all this as a
warning that the storage trick I am about to divulge must
never be used where a subroutine call or return might
interfere. Again you would get those notorious
unpredictable results.
Say you had just entered a subroutine and knew that the
routine was going to mess up the values of the accumulator
and X and Y registers. Now you could save them to memory if
you wished and in many instances that would gain you a
little speed, but a neat thing to do is to use the stack.
You would simply push the relevant register values onto the
stack while you did your computation in the subroutine and
then pull (or pop) them back again afterwards. It is
important to make sure you do it the right way round. This
means pulling back off the stack in the reverse order to the
way you pushed onto it, like this:
On the way in - PHA
TXA
PHA
TYA
PHA
On the way back - PLA
TAY
PLA
TAX
PLA
PHA means PusH A onto the stack and PLA means PulL A from
the stack. Once you have pulled a value from the stack it
is effectively gone and the next pull will produce a
different value, unlike normal memory storage. The
instructions TAX and TAY transfer A to X and Y while TXA and
TYA transfer X and Y to A. There are also instructions to
push and pull the processor flags on and off the stack (PHP
and PLP) which are important if you are concerned about a
subroutine affecting the flags.
Those pushes and pulls must always balance out otherwise
your program will have, you guessed, unpredictable results.
You could end up with your registers mixed up or with some
funny numbers getting into the program counter.
If you ever have subroutines that are recursive (call
themselves) then you have to use the stack for storage since
using a fixed location will be disastrous with each call
writing in a new value before the next outermost call can
reuse the value it put in there in the first place. There
will also be times when you can not get at memory easily, if
you have written a sideways ROM for instance, and in these
cases the stack will be very useful. However it is only 256
bytes long and too much use will make it overflow which will
almost certainly crash the machine.
In the case of both stack and memory storage if you want to
store anything larger than a byte you have to do so a byte
at a time. In memory your label would represent the lowest
byte in memory which, in 6502 convention, stores the lowest
significant byte. On the stack you can do as you wish,
providing it balances.
In B/osb06 there are a couple of places where I have used
the stack for brief storage, lines 550 + 610 and 1070 +
1190. In the first instance I should really just have used
'accustore' since this is what it is for. The latter though
is a reasonable instance of stack use as a temporary storage
for an interim result while the code checks for zeros.
Let's look at B/osb06 in more detail. What it does is take
a byte that you type in by pressing a key and print its
value out in three ways. Binary, hexadecimal and decimal.
The JSR osrdch at line 190 reads your keypress and puts it
in the accumulator. (I'll be dealing with some more OS
routines next week). It is then put into the 'accustore'
for the duration of the routine.
Now since I wanted this code to be self-contained and not
return to BASIC after each key press I had to put in some
way of exiting. Any key value would do but ESCAPE seemed
most appropriate, although you have to carry out an escape
in an orderly manner. So at line 220 the code checks for an
escape and if it detects one, by A being set to 27, OSBYTE
126 is called which carries out housekeeping to acknowledge
an ESCAPE after which we return to BASIC. Because the
ESCAPE has been acknowledged you return to BASIC without an
error condition and so there is no printing of ESCAPE. You
could try missing out the trap for 27, or omit the
acknowledgement and compare the results.
[It is possible to trigger an ESCAPE from another ASCII code
than 27, although 27 is the standard. OSBYTE 220 will do
this. You might do this to enable a device to generate
ESCAPE's down the RS423.]
If the key pressed is not ESCAPE then the three sections of
the program are executed. Notice that three locations for
variables have been set at lines 290-310 using the BRK short
cut I mentioned earlier.
The first section prints out binary and it does this by
rotating the byte eight times, each time looking at the
carry flag and printing a 1 or 0 accordingly. The loop that
does the work is between 400 and 460 and uses the X register
as a counter. Since it is the Carry flag that is being used
we can make use of the ADC instruction to add, with carry, 0
to the ASCII value of 0.
Hexadecimal printing is more difficult. First we separate
the top nybble of the byte (the top 4 bits) by shifting the
byte right four times. This nybble must then hold a value
between 0 and 15, or 0 and F in hex. A little trick with
binary coded decimal is of help here.
In binary coded decimal the four bits of a nybble can only
ever hold between 0 and 9, the other numbers just don't
exist and if you add 1 to 9 you get 10 and not A as you
would normally. To tell the processor you are working in
BCD you set the decimal flag with an SED (line 710).
If the accumulator contains your nybble then adding 90 to it
in BCD mode gives you a number in the range 90 to 99 (Carry
clear) if you had 0 to 9 in there and 0 to 5 (Carry set) if
you had 10 to 15 in there. If we then add 40 with carry we
will get 30 to 39 or 41 to 46 (because of the Carry).
Returning to normal binary (with CLD) these become &33 to
&39 or &41 to &46 which gives us the ASCII values of 0-9 or
A-F. Thus you print the nybble value.
Regrabbing the original byte we then look at the lower
nybble by ANDing with &F which (more details in a week or
so) wipes the top four bits. The nybble printing routine is
then repeated.
Decimal printing is more difficult still, and I don't know a
short-cut. One extra complication I have introduced is that
in this case we don't print leading zeros, except for the
final figure. In this case I have a subroutine called
'digit' which takes the value in the accumulator and prints
out the number of times the number in the Y register will go
into it. This is found out by repeated subtraction.
In all cases here we are dealing only with byte sized
numbers. In fact the first two routines will adapt easily
to any size of number but that is not so true of the decimal
one.
Next week some more OS calls and I'll show you how to use a
look-up table and indexed addressing to provide proportional
spacing.
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 36 3a 20 |........Part 6: | 00000090 52 6f 74 61 74 69 6e 67 20 61 6e 64 20 53 68 69 |Rotating and Shi| 000000a0 66 74 69 6e 67 0d 0d 0d 43 6f 6e 74 69 6e 75 69 |fting...Continui| 000000b0 6e 67 20 61 6c 6f 6e 67 20 74 68 65 20 6d 61 63 |ng along the mac| 000000c0 68 69 6e 65 20 63 6f 64 65 20 74 72 61 69 6c 20 |hine code trail | 000000d0 74 68 69 73 20 77 65 65 6b 20 49 27 6c 6c 20 64 |this week I'll d| 000000e0 6f 0d 74 77 6f 20 74 68 69 6e 67 73 2e 20 20 46 |o.two things. F| 000000f0 69 72 73 74 6c 79 20 73 6f 6d 65 20 6d 6f 72 65 |irstly some more| 00000100 20 62 79 74 65 20 6d 61 6e 69 70 75 6c 61 74 69 | byte manipulati| 00000110 6f 6e 20 77 69 74 68 20 62 69 74 73 0d 62 65 69 |on with bits.bei| 00000120 6e 67 20 72 6f 74 61 74 65 64 20 61 6e 64 20 73 |ng rotated and s| 00000130 68 69 66 74 65 64 20 61 62 6f 75 74 2c 20 61 6e |hifted about, an| 00000140 64 20 73 65 63 6f 6e 64 20 73 6f 6d 65 20 74 68 |d second some th| 00000150 6f 75 67 68 74 73 20 6f 6e 0d 77 68 65 72 65 20 |oughts on.where | 00000160 74 6f 20 6b 65 65 70 20 6e 75 6d 62 65 72 73 20 |to keep numbers | 00000170 74 65 6d 70 6f 72 61 72 69 6c 79 20 64 75 72 69 |temporarily duri| 00000180 6e 67 20 79 6f 75 72 20 70 72 6f 67 72 61 6d 73 |ng your programs| 00000190 2e 0d 0d 52 69 67 68 74 20 74 68 65 6e 2c 20 6c |...Right then, l| 000001a0 65 74 27 73 20 68 61 76 65 20 73 6f 6d 65 20 62 |et's have some b| 000001b0 69 74 20 6d 61 6e 69 70 75 6c 61 74 69 6f 6e 2e |it manipulation.| 000001c0 20 20 54 68 69 73 20 69 73 0d 73 6f 6d 65 74 68 | This is.someth| 000001d0 69 6e 67 20 79 6f 75 20 63 61 6e 27 74 20 64 6f |ing you can't do| 000001e0 20 69 6e 20 42 41 53 49 43 20 61 6c 74 68 6f 75 | in BASIC althou| 000001f0 67 68 20 69 74 20 69 73 20 6e 65 65 64 65 64 20 |gh it is needed | 00000200 66 6f 72 0d 6d 75 6c 74 69 70 6c 69 63 61 74 69 |for.multiplicati| 00000210 6f 6e 20 61 6e 64 20 64 69 76 69 73 69 6f 6e 20 |on and division | 00000220 69 6e 20 6d 61 63 68 69 6e 65 20 63 6f 64 65 2e |in machine code.| 00000230 0d 0d 54 6f 20 73 74 61 72 74 20 77 69 74 68 20 |..To start with | 00000240 68 65 72 65 20 69 73 20 61 20 62 79 74 65 20 28 |here is a byte (| 00000250 74 68 65 20 64 6f 74 73 20 72 65 70 72 65 73 65 |the dots represe| 00000260 6e 74 20 74 68 65 20 65 6e 64 73 20 6f 66 0d 74 |nt the ends of.t| 00000270 68 65 20 62 79 74 65 29 3a 0d 0d 20 20 20 20 20 |he byte):.. | 00000280 20 20 20 20 20 20 20 2e 20 20 62 37 20 62 36 20 | . b7 b6 | 00000290 62 35 20 62 34 20 62 33 20 62 32 20 62 31 20 62 |b5 b4 b3 b2 b1 b| 000002a0 30 20 20 2e 0d 0d 65 61 63 68 20 62 69 74 20 62 |0 ...each bit b| 000002b0 65 69 6e 67 20 74 68 65 20 6e 75 6d 62 65 72 20 |eing the number | 000002c0 62 30 20 74 6f 20 62 37 2c 20 72 65 70 72 65 73 |b0 to b7, repres| 000002d0 65 6e 74 69 6e 67 20 32 5e 6e 20 61 6e 64 0d 65 |enting 2^n and.e| 000002e0 6e 61 62 6c 69 6e 67 20 75 73 20 74 6f 20 6d 61 |nabling us to ma| 000002f0 6b 65 20 6e 75 6d 62 65 72 73 20 75 70 20 74 6f |ke numbers up to| 00000300 20 32 35 35 20 6f 75 74 20 6f 66 20 74 68 65 20 | 255 out of the | 00000310 62 79 74 65 2e 0d 0d 54 68 65 72 65 20 61 72 65 |byte...There are| 00000320 20 66 6f 75 72 20 72 6f 74 61 74 69 6e 67 2f 73 | four rotating/s| 00000330 68 69 66 74 69 6e 67 20 6f 70 65 72 61 74 69 6f |hifting operatio| 00000340 6e 73 20 77 65 20 63 61 6e 20 63 61 72 72 79 20 |ns we can carry | 00000350 6f 75 74 0d 6f 6e 20 74 68 69 73 20 62 79 74 65 |out.on this byte| 00000360 2e 20 20 46 69 72 73 74 20 6f 66 20 61 6c 6c 20 |. First of all | 00000370 77 65 20 63 61 6e 20 73 68 69 66 74 20 69 74 20 |we can shift it | 00000380 6c 65 66 74 20 6f 6e 65 20 70 6c 61 63 65 2e 0d |left one place..| 00000390 57 65 20 6e 6f 77 20 68 61 76 65 20 74 68 65 20 |We now have the | 000003a0 66 6f 6c 6c 6f 77 69 6e 67 20 72 65 73 75 6c 74 |following result| 000003b0 3a 0d 0d 20 20 20 20 20 20 20 20 20 20 20 20 62 |:.. b| 000003c0 37 20 2e 20 62 36 20 62 35 20 62 34 20 62 33 20 |7 . b6 b5 b4 b3 | 000003d0 62 32 20 62 31 20 62 30 20 30 20 20 2e 0d 0d 41 |b2 b1 b0 0 ...A| 000003e0 20 6e 65 77 20 62 79 74 65 20 68 6f 6c 64 69 6e | new byte holdin| 000003f0 67 20 74 68 65 20 6f 6c 64 20 62 69 74 73 20 62 |g the old bits b| 00000400 36 20 64 6f 77 6e 20 74 6f 20 62 30 20 61 6e 64 |6 down to b0 and| 00000410 20 77 69 74 68 20 61 0d 7a 65 72 6f 20 61 64 64 | with a.zero add| 00000420 65 64 20 69 6e 20 61 74 20 74 68 65 20 65 6e 64 |ed in at the end| 00000430 2e 20 20 54 68 65 20 76 61 6c 75 65 20 6f 66 20 |. The value of | 00000440 62 37 20 68 61 73 20 6e 6f 77 20 73 65 74 20 6f |b7 has now set o| 00000450 72 0d 63 6c 65 61 72 65 64 20 74 68 65 20 43 61 |r.cleared the Ca| 00000460 72 72 79 20 66 6c 61 67 2e 20 20 54 68 69 73 20 |rry flag. This | 00000470 6f 70 65 72 61 74 69 6f 6e 20 69 73 20 63 61 6c |operation is cal| 00000480 6c 65 64 20 61 6e 0d 41 72 69 74 68 6d 65 74 69 |led an.Arithmeti| 00000490 63 20 53 68 69 66 74 20 4c 65 66 74 20 61 6e 64 |c Shift Left and| 000004a0 20 74 68 65 20 6d 6e 65 6d 6f 6e 69 63 20 66 6f | the mnemonic fo| 000004b0 72 20 69 74 20 69 73 20 41 53 4c 2e 20 20 59 6f |r it is ASL. Yo| 000004c0 75 0d 77 69 6c 6c 20 70 72 6f 62 61 62 6c 79 20 |u.will probably | 000004d0 61 6c 73 6f 20 68 61 76 65 20 73 65 65 6e 20 74 |also have seen t| 000004e0 68 61 74 20 74 68 69 73 20 6f 70 65 72 61 74 69 |hat this operati| 000004f0 6f 6e 20 61 6c 73 6f 0d 65 66 66 65 63 74 69 76 |on also.effectiv| 00000500 65 6c 79 20 6d 75 6c 74 69 70 6c 69 65 73 20 74 |ely multiplies t| 00000510 68 65 20 6e 75 6d 62 65 72 20 68 65 6c 64 20 69 |he number held i| 00000520 6e 20 74 68 65 20 62 79 74 65 20 62 79 20 32 2e |n the byte by 2.| 00000530 0d 0d 54 68 65 20 65 71 75 69 76 61 6c 65 6e 74 |..The equivalent| 00000540 20 6f 70 65 72 61 74 69 6f 6e 20 62 75 74 20 67 | operation but g| 00000550 6f 69 6e 67 20 72 69 67 68 74 20 69 73 20 61 20 |oing right is a | 00000560 4c 6f 67 69 63 61 6c 20 53 68 69 66 74 0d 52 69 |Logical Shift.Ri| 00000570 67 68 74 20 77 69 74 68 20 6d 6e 65 6d 6f 6e 69 |ght with mnemoni| 00000580 63 20 4c 53 52 2e 20 20 54 68 69 73 20 74 69 6d |c LSR. This tim| 00000590 65 20 74 68 65 20 76 61 6c 75 65 20 6f 66 20 62 |e the value of b| 000005a0 69 74 20 30 20 69 73 0d 70 75 74 20 69 6e 20 74 |it 0 is.put in t| 000005b0 68 65 20 63 61 72 72 79 20 66 6c 61 67 20 61 6e |he carry flag an| 000005c0 64 20 62 69 74 20 37 20 72 65 70 6c 61 63 65 64 |d bit 7 replaced| 000005d0 20 62 79 20 61 20 7a 65 72 6f 20 77 68 69 6c 73 | by a zero whils| 000005e0 74 0d 61 6c 6c 20 74 68 65 20 62 69 74 73 20 61 |t.all the bits a| 000005f0 72 65 20 73 68 69 66 74 65 64 20 64 6f 77 6e 20 |re shifted down | 00000600 6f 6e 65 2e 20 20 54 68 69 73 20 65 66 66 65 63 |one. This effec| 00000610 74 69 76 65 6c 79 20 68 61 6c 76 65 73 0d 74 68 |tively halves.th| 00000620 65 20 6e 75 6d 62 65 72 20 69 6e 20 74 68 65 20 |e number in the | 00000630 62 79 74 65 2e 0d 0d 4f 4b 20 73 6f 20 79 6f 75 |byte...OK so you| 00000640 20 6d 61 79 20 6e 6f 77 20 62 65 20 77 6f 6e 64 | may now be wond| 00000650 65 72 69 6e 67 20 69 6e 20 77 68 69 63 68 20 74 |ering in which t| 00000660 77 6f 20 6f 74 68 65 72 20 64 69 72 65 63 74 69 |wo other directi| 00000670 6f 6e 73 0d 63 61 6e 20 77 65 20 73 68 69 66 74 |ons.can we shift| 00000680 20 74 68 69 6e 67 73 2e 20 20 57 65 6c 6c 20 6f | things. Well o| 00000690 62 76 69 6f 75 73 6c 79 20 77 65 20 6f 6e 6c 79 |bviously we only| 000006a0 20 68 61 76 65 20 6c 65 66 74 20 61 6e 64 0d 72 | have left and.r| 000006b0 69 67 68 74 20 62 75 74 20 74 68 65 20 6f 74 68 |ight but the oth| 000006c0 65 72 20 74 77 6f 20 6f 70 65 72 61 74 69 6f 6e |er two operation| 000006d0 73 20 72 6f 74 61 74 65 20 74 68 65 20 63 61 72 |s rotate the car| 000006e0 72 79 20 66 6c 61 67 0d 69 6e 74 6f 20 74 68 65 |ry flag.into the| 000006f0 20 62 79 74 65 20 69 6e 73 74 65 61 64 20 6f 66 | byte instead of| 00000700 20 73 68 69 66 74 69 6e 67 20 69 6e 20 61 20 7a | shifting in a z| 00000710 65 72 6f 2e 0d 0d 53 6f 20 77 65 20 68 61 76 65 |ero...So we have| 00000720 20 52 4f 4c 20 61 6e 64 20 52 4f 52 20 66 6f 72 | ROL and ROR for| 00000730 20 61 20 72 6f 74 61 74 69 6f 6e 20 6c 65 66 74 | a rotation left| 00000740 20 6f 72 20 72 69 67 68 74 0d 72 65 73 70 65 63 | or right.respec| 00000750 74 69 76 65 6c 79 2e 20 20 57 69 74 68 20 52 4f |tively. With RO| 00000760 4c 20 20 62 69 74 73 20 30 20 74 6f 20 36 20 61 |L bits 0 to 6 a| 00000770 72 65 20 6d 6f 76 65 64 20 75 70 20 6f 6e 65 2c |re moved up one,| 00000780 20 74 68 65 0d 63 61 72 72 79 20 66 6c 61 67 20 | the.carry flag | 00000790 69 73 20 70 75 74 20 69 6e 74 6f 20 62 69 74 20 |is put into bit | 000007a0 30 20 61 6e 64 20 74 68 65 20 6f 72 69 67 69 6e |0 and the origin| 000007b0 61 6c 20 62 69 74 20 37 20 69 73 20 70 75 74 0d |al bit 7 is put.| 000007c0 69 6e 20 74 68 65 20 63 61 72 72 79 20 66 6c 61 |in the carry fla| 000007d0 67 2e 20 20 57 69 74 68 20 52 4f 52 20 62 69 74 |g. With ROR bit| 000007e0 73 20 37 20 74 6f 20 31 20 61 72 65 20 6d 6f 76 |s 7 to 1 are mov| 000007f0 65 64 20 64 6f 77 6e 20 6f 6e 65 2c 0d 74 68 65 |ed down one,.the| 00000800 20 43 61 72 72 79 20 66 6c 61 67 20 69 73 20 70 | Carry flag is p| 00000810 75 74 20 69 6e 74 6f 20 62 69 74 20 37 2c 20 61 |ut into bit 7, a| 00000820 6e 64 20 74 68 65 20 6f 72 69 67 69 6e 61 6c 20 |nd the original | 00000830 62 69 74 20 30 20 69 73 0d 70 75 74 20 69 6e 20 |bit 0 is.put in | 00000840 74 68 65 20 63 61 72 72 79 20 66 6c 61 67 2e 0d |the carry flag..| 00000850 0d 4f 62 76 69 6f 75 73 6c 79 20 69 74 20 69 73 |.Obviously it is| 00000860 20 69 6d 70 6f 72 74 61 6e 74 20 74 68 61 74 20 | important that | 00000870 74 68 65 20 73 74 61 74 65 20 6f 66 20 74 68 65 |the state of the| 00000880 20 63 61 72 72 79 20 66 6c 61 67 0d 69 73 20 6b | carry flag.is k| 00000890 6e 6f 77 6e 20 62 65 66 6f 72 65 20 79 6f 75 20 |nown before you | 000008a0 52 4f 4c 20 6f 72 20 52 4f 52 2e 0d 0d 54 68 65 |ROL or ROR...The| 000008b0 73 65 20 66 6f 75 72 20 6f 70 65 72 61 74 69 6f |se four operatio| 000008c0 6e 73 20 63 61 6e 20 62 65 20 63 61 72 72 69 65 |ns can be carrie| 000008d0 64 20 6f 75 74 20 6f 6e 20 6d 65 6d 6f 72 79 2c |d out on memory,| 000008e0 20 65 2e 67 2e 0d 0d 20 20 20 20 20 20 20 20 20 | e.g... | 000008f0 20 20 20 20 20 20 20 20 20 20 52 4f 52 20 74 65 | ROR te| 00000900 6d 70 0d 20 20 20 20 20 20 20 20 20 20 20 20 20 |mp. | 00000910 20 20 20 20 20 20 41 53 4c 20 26 37 30 0d 0d 6f | ASL &70..o| 00000920 72 20 6f 6e 20 74 68 65 20 41 63 63 75 6d 75 6c |r on the Accumul| 00000930 61 74 6f 72 20 65 2e 67 2e 0d 0d 20 20 20 20 20 |ator e.g... | 00000940 20 20 20 20 20 20 20 20 20 20 20 20 20 20 4c 53 | LS| 00000950 52 20 41 0d 0d 42 65 73 69 64 65 73 20 68 61 76 |R A..Besides hav| 00000960 69 6e 67 20 61 20 70 6c 61 63 65 20 69 6e 20 6d |ing a place in m| 00000970 75 6c 74 69 70 6c 69 63 61 74 69 6f 6e 20 61 6e |ultiplication an| 00000980 64 20 64 69 76 69 73 69 6f 6e 20 74 68 65 73 65 |d division these| 00000990 0d 6f 70 65 72 61 74 69 6f 6e 73 20 65 6e 61 62 |.operations enab| 000009a0 6c 65 20 79 6f 75 20 74 6f 20 65 78 61 6d 69 6e |le you to examin| 000009b0 65 20 61 20 62 79 74 65 20 62 69 74 20 62 79 20 |e a byte bit by | 000009c0 62 69 74 2e 20 20 49 66 20 79 6f 75 0d 72 6f 74 |bit. 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S| 00002890 69 6e 63 65 20 69 74 20 69 73 20 74 68 65 20 43 |ince it is the C| 000028a0 61 72 72 79 20 66 6c 61 67 20 74 68 61 74 20 69 |arry flag that i| 000028b0 73 20 62 65 69 6e 67 20 75 73 65 64 0d 77 65 20 |s being used.we | 000028c0 63 61 6e 20 6d 61 6b 65 20 75 73 65 20 6f 66 20 |can make use of | 000028d0 74 68 65 20 41 44 43 20 69 6e 73 74 72 75 63 74 |the ADC instruct| 000028e0 69 6f 6e 20 74 6f 20 61 64 64 2c 20 77 69 74 68 |ion to add, with| 000028f0 20 63 61 72 72 79 2c 20 30 0d 74 6f 20 74 68 65 | carry, 0.to the| 00002900 20 41 53 43 49 49 20 76 61 6c 75 65 20 6f 66 20 | ASCII value of | 00002910 30 2e 0d 0d 48 65 78 61 64 65 63 69 6d 61 6c 20 |0...Hexadecimal | 00002920 70 72 69 6e 74 69 6e 67 20 69 73 20 6d 6f 72 65 |printing is more| 00002930 20 64 69 66 66 69 63 75 6c 74 2e 20 20 46 69 72 | difficult. 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A li| 000029f0 74 74 6c 65 20 74 72 69 63 6b 20 77 69 74 68 0d |ttle trick with.| 00002a00 62 69 6e 61 72 79 20 63 6f 64 65 64 20 64 65 63 |binary coded dec| 00002a10 69 6d 61 6c 20 69 73 20 6f 66 20 68 65 6c 70 20 |imal is of help | 00002a20 68 65 72 65 2e 0d 0d 49 6e 20 62 69 6e 61 72 79 |here...In binary| 00002a30 20 63 6f 64 65 64 20 64 65 63 69 6d 61 6c 20 74 | coded decimal t| 00002a40 68 65 20 66 6f 75 72 20 62 69 74 73 20 6f 66 20 |he four bits of | 00002a50 61 20 6e 79 62 62 6c 65 20 63 61 6e 20 6f 6e 6c |a nybble can onl| 00002a60 79 0d 65 76 65 72 20 68 6f 6c 64 20 62 65 74 77 |y.ever hold betw| 00002a70 65 65 6e 20 30 20 61 6e 64 20 39 2c 20 74 68 65 |een 0 and 9, the| 00002a80 20 6f 74 68 65 72 20 6e 75 6d 62 65 72 73 20 6a | other numbers j| 00002a90 75 73 74 20 64 6f 6e 27 74 0d 65 78 69 73 74 20 |ust don't.exist | 00002aa0 61 6e 64 20 69 66 20 79 6f 75 20 61 64 64 20 31 |and if you add 1| 00002ab0 20 74 6f 20 39 20 79 6f 75 20 67 65 74 20 31 30 | to 9 you get 10| 00002ac0 20 61 6e 64 20 6e 6f 74 20 41 20 61 73 20 79 6f | and not A as yo| 00002ad0 75 0d 77 6f 75 6c 64 20 6e 6f 72 6d 61 6c 6c 79 |u.would normally| 00002ae0 2e 20 20 54 6f 20 74 65 6c 6c 20 74 68 65 20 70 |. To tell the p| 00002af0 72 6f 63 65 73 73 6f 72 20 79 6f 75 20 61 72 65 |rocessor you are| 00002b00 20 77 6f 72 6b 69 6e 67 20 69 6e 0d 42 43 44 20 | working in.BCD | 00002b10 79 6f 75 20 73 65 74 20 74 68 65 20 64 65 63 69 |you set the deci| 00002b20 6d 61 6c 20 66 6c 61 67 20 77 69 74 68 20 61 6e |mal flag with an| 00002b30 20 53 45 44 20 28 6c 69 6e 65 20 37 31 30 29 2e | SED (line 710).| 00002b40 0d 0d 49 66 20 74 68 65 20 61 63 63 75 6d 75 6c |..If the accumul| 00002b50 61 74 6f 72 20 63 6f 6e 74 61 69 6e 73 20 79 6f |ator contains yo| 00002b60 75 72 20 6e 79 62 62 6c 65 20 74 68 65 6e 20 61 |ur nybble then a| 00002b70 64 64 69 6e 67 20 39 30 20 74 6f 20 69 74 0d 69 |dding 90 to it.i| 00002b80 6e 20 42 43 44 20 6d 6f 64 65 20 67 69 76 65 73 |n BCD mode gives| 00002b90 20 79 6f 75 20 61 20 6e 75 6d 62 65 72 20 69 6e | you a number in| 00002ba0 20 74 68 65 20 72 61 6e 67 65 20 39 30 20 74 6f | the range 90 to| 00002bb0 20 39 39 20 28 43 61 72 72 79 0d 63 6c 65 61 72 | 99 (Carry.clear| 00002bc0 29 20 69 66 20 79 6f 75 20 68 61 64 20 30 20 74 |) if you had 0 t| 00002bd0 6f 20 39 20 69 6e 20 74 68 65 72 65 20 61 6e 64 |o 9 in there and| 00002be0 20 30 20 74 6f 20 35 20 28 43 61 72 72 79 20 73 | 0 to 5 (Carry s| 00002bf0 65 74 29 20 69 66 0d 79 6f 75 20 68 61 64 20 31 |et) if.you had 1| 00002c00 30 20 74 6f 20 31 35 20 69 6e 20 74 68 65 72 65 |0 to 15 in there| 00002c10 2e 20 49 66 20 77 65 20 74 68 65 6e 20 61 64 64 |. If we then add| 00002c20 20 34 30 20 77 69 74 68 20 63 61 72 72 79 20 77 | 40 with carry w| 00002c30 65 0d 77 69 6c 6c 20 67 65 74 20 33 30 20 74 6f |e.will get 30 to| 00002c40 20 33 39 20 6f 72 20 34 31 20 74 6f 20 34 36 20 | 39 or 41 to 46 | 00002c50 28 62 65 63 61 75 73 65 20 6f 66 20 74 68 65 20 |(because of the | 00002c60 43 61 72 72 79 29 2e 20 0d 52 65 74 75 72 6e 69 |Carry). .Returni| 00002c70 6e 67 20 74 6f 20 6e 6f 72 6d 61 6c 20 62 69 6e |ng to normal bin| 00002c80 61 72 79 20 28 77 69 74 68 20 43 4c 44 29 20 74 |ary (with CLD) t| 00002c90 68 65 73 65 20 62 65 63 6f 6d 65 20 26 33 33 20 |hese become &33 | 00002ca0 74 6f 0d 26 33 39 20 6f 72 20 26 34 31 20 74 6f |to.&39 or &41 to| 00002cb0 20 26 34 36 20 77 68 69 63 68 20 67 69 76 65 73 | &46 which gives| 00002cc0 20 75 73 20 74 68 65 20 41 53 43 49 49 20 76 61 | us the ASCII va| 00002cd0 6c 75 65 73 20 6f 66 20 30 2d 39 20 6f 72 0d 41 |lues of 0-9 or.A| 00002ce0 2d 46 2e 20 20 54 68 75 73 20 79 6f 75 20 70 72 |-F. Thus you pr| 00002cf0 69 6e 74 20 74 68 65 20 6e 79 62 62 6c 65 20 76 |int the nybble v| 00002d00 61 6c 75 65 2e 0d 0d 52 65 67 72 61 62 62 69 6e |alue...Regrabbin| 00002d10 67 20 74 68 65 20 6f 72 69 67 69 6e 61 6c 20 62 |g the original b| 00002d20 79 74 65 20 77 65 20 74 68 65 6e 20 6c 6f 6f 6b |yte we then look| 00002d30 20 61 74 20 74 68 65 20 6c 6f 77 65 72 0d 6e 79 | at the lower.ny| 00002d40 62 62 6c 65 20 62 79 20 41 4e 44 69 6e 67 20 77 |bble by ANDing w| 00002d50 69 74 68 20 26 46 20 77 68 69 63 68 20 28 6d 6f |ith &F which (mo| 00002d60 72 65 20 64 65 74 61 69 6c 73 20 69 6e 20 61 20 |re details in a | 00002d70 77 65 65 6b 20 6f 72 0d 73 6f 29 20 77 69 70 65 |week or.so) wipe| 00002d80 73 20 74 68 65 20 74 6f 70 20 66 6f 75 72 20 62 |s the top four b| 00002d90 69 74 73 2e 20 20 54 68 65 20 6e 79 62 62 6c 65 |its. 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27_11_87/T\OSB06.m0
27_11_87/T\OSB06.m1
27_11_87/T\OSB06.m2
27_11_87/T\OSB06.m4
27_11_87/T\OSB06.m5
.