small & short Assembler tutorial...

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small & short Assembler tutorial...

Beitragvon CharlieChaplin » Di 19. Mai 2009, 23:01

here is a short & small assembler tutorial, written by Chris Crawford 1985. It is divided into 8 lessons, so here is part one... the next part will be uploaded in a few days... -Andreas Koch.



** LESSON 1 **

Machine code is nothing more than a
bunch of numbers that mean something
to the CPU. It's hard to work with
pure numbers, so we use a little
code that makes it easier for us to
understand the codes that the
computer uses. This programmer-
friendlier code is called assembly
language, It is a direct, one-to-one
translation of machine code. Here is
an example of the two side by side:

Machine Code : Assembly Language
133 $9C : STA COUNT

The code on the right may not look
very readable, but you must agree,
it's far more readable than the code
on the left. And they both mean
exactly the same thing.
Unfortunately, the computer cannot
read the assembly code, only the
machine code. Therefore, we need a
translator program that will
translate the easier-to-understand
code on the right into the
impossible-to-understand code on the
left. This translator program is
called an assembler.

A program that goes in the reverse
direction, translating machine code
to assembly, is called a
disassembler. It may seem like a
bother to go through all the hassle
of using an assembler, but it is
actually much easier. Assembly
language is not only more readable
than machine code, but it is also
assembly-time relocatable; this
means you can move it around in RAM
freely before you start the assembly
process. A good assembler also
offers a number of extra features
that make it easier to keep track of
your program or modify it quickly.

There are three steps involved in
writing an assembly language
program: editing, assembling, and
debugging. Editing is the process of
typing in your assembly language
statements. Assembling is the
invocation of the assembler.
Debugging is the process of running
your program and analyzing why it
doesn't work. Thus, the entire
process of writing an assembly-
language process can be described by
a fictitious BASIC program:

FOR I=1 to 1,000,000,000...

The first item in the 6502 that I
will describe is the accumulator.
This is a single one-byte register
in the 6502. It is the central
workbench of the microprocessor;
almost everything happens in the
accumulator. Your first three
instructions on the 6502 are:

LDA address (Load the Accumulator
with the contents of address)

This instruction loads the
accumulator with the contents of the
memory location specified by the
value of address. The address can be
specified by either an outright
value, such as $0600, or a symbolic
reference, such as FISH, where the
value of FISH has been previously
declared by, say, an ORG statement
or an equate statement.

LDA #value (Load the Accumulator
with value)

This is much like the earlier
statment; it loads the accumulator
with a number, only the number
loaded is specified immediately
rather than stored in a memory
location. Thus, the command LDA # 9
will put a 9 into the accumulator

STA address (Store the Accumulator
into address)

This command will store the
contents of the accumulator into the
RAM location whose address is
specified in the command. It is just
like the first command, except that
the direction of data motion is
reversed. The LDA command is like a
read, which the STA is like a write.

You are now equipped to move data
around inside the computer. These
commands will allow you to read data
from one area of memory and store it
into another. LDA and STA are the
two most common instructions used in
any 6502 program. Exercise: Write a
program that will read the contents
of address $FE00 and store the
result into address $680. Your
biggest problem here will be just
getting your assembler to work.
Therefore, I will give the answer

ORG $600

That's the program. Try to get it
running with your assembler.

End of lesson 1.

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Beitragvon cas » Mi 20. Mai 2009, 08:57

Hallo Andreas,

danke fuer den Kurs.

Ich bereite den Kurs im Wiki auf --> ... age+Course

Bitte poste die weiteren Teile des Kurses.


Zuletzt geändert von cas am Mi 20. Mai 2009, 13:06, insgesamt 1-mal geändert.
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Beitragvon eda70 » Mi 20. Mai 2009, 11:49

Wäre so was nicht auch etwas fürs Abbuc-Magazin (in dt. natürlich) ???
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Beitragvon CharlieChaplin » Mi 20. Mai 2009, 22:20

ich poste natürlich auch die weiteren Teile - aber nicht alle auf einmal, sondern einen nach dem anderen, damit die Leute auch Zeit zum Lesen haben. Als Nicht-Programmierer kann ich das Tutorial aber leider in keinster Weise kommentieren oder moderieren. Also der zweite Teil folgt morgen, ok ?!?

Gibt ja insgesamt nur acht Teile (und wer sich über das Layout mit ca. 37 Zeichen pro Zeile wundert, habe ich so von einer Atari Diskette übernommen und lediglich von ATASCII in ASCII umgewandelt). Sollte das Tutorial Fehler oder Ungereimtheiten beinhalten, dann bitte *freundlich* darauf hinweisen, ich kann das mangels Programmierkenntnissen eh nicht entdecken oder korrigieren...

Sollte jemand dieses Tutorial für das Abbuc Magazin übersetzen wollen, dann nur zu ! Es ist aber wirklich nur eine kleine Einführung in Assembler, mehr nicht. Wer mehr als nur eine Einführung will, muss Bücher wälzen oder sich auf div. Internetseiten schlau machen (Carsten hat ja bereits jede Menge Links an anderer Stelle gepostet)...

Gruß, Andreas Koch.
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Beitragvon CharlieChaplin » Do 21. Mai 2009, 14:05

lets continue with part two of Chris Crawford`s assembler tutorial...
-Andreas Koch.



** 6502 ARITHMETIC **

In this lecture I will take up the
problem of arithmetic on the 6502. I
choose this topic only because it is
fairly simple to do on the 6502.
There are a couple of nerve-jangling
problems associated with 6502
arithmetic, but I will breeze over
those in a very cavalier fashion.

Before we can do arithmetic,
though, you must know a little bit
about number systems. There are
three that you must know: decimal,
binary, and hexadecimal.
Decimal is the standard numbers you
have used since grade school. You
count 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,
and then you reach 0 again, so you
put down a 1 in the tens place and
resume counting from 0.

Binary works the same way, except
that there are only two digits, not
ten. The two digits are 0 and 1. You
count 0, then 1, then you reach 0
again, so you put down a 1 in the
twos place and resume counting from
0. Thus, counting from 0 to ten in
binary like this:

Decimal Binary
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001
10 1010

In binary, instead of having ones,
tens, and hundreds places, we have
ones, twos, fours and eights places.
It takes a lot more digits to
express a number in binary, but then
again, we have only the two
numberals 0 and 1 to work with, so
what does one expect?

The hexadecimal number system is a
base-16 system. In this system, you
count from 0 to 16 like so:
The 10 in hexadecimal really means
16 in decimal. So 10 is 16, right?
Black is white, truth is
likes....stay with assembly language
long enough and you'll believe

Actually, it's easy to avoid
confusion. We use little prefixes to
tell you and the computer whether a
number is expressed in decimal,
binary, or hexadecimal. No prefix
means decimal. A $ prefix means
hexadecimal; a % means binary. Thus
%10 means 2 while $10 means 16, but
10 means just plain old 10.
Hexadecimal is not hard to learn at
all; if you go into any store you
will see that they use hexadecimal
on all their signs.


Addition with the 6502 is very
simple; it uses the ADC instruction.
This instruction means "Add with
Carry"; I'll get to the Carry part
in just a moment. For now, let me
explain the instruction. The ADC
instruction has an operand, normally
a location in memory. When the
instruction is executed, it takes
the contents of that memory location
and adds that value to the value in
the accumulator.

It leaves the sum of the two
numbers in the accumulator. This of
course destroys the old value in the
accumulator. You can use the
immediate mode of addressing with
the ADC instruction, in which case
it adds the value itself. Thus, "ADC
# 9" will add a 9 to the contents of
the accumulator, while "ADC FISH"
will add the contents of address
FISH to the accumulator.

Subtraction is just like addition.
The instruction to use is SBC, which
means "Subtract Borrowing Carry".
Again, I'll tell you about the Carry
part in a moment. This instruction
subtracts the operand from the
contents of the accumulator, leaving
the reslt in the accumulator. It
also can be done in either immediate
mode (e.g. SBC#5) or absolute mode
(e.g., SBC GOAT).


If that were all there were to
arithmetic with the 6502,
programmers would be paid a lot
less. The first killer problem is
that the 6502 uses 8-bit words; that
is, the numbers that the 6502 stores
and works with are only 8 bits wide.
This means that the biggest number
the 6502 can comprehend is 255.

Uh-oh! What happens if you want to
have a checkbook balancing program
and you have more than 255? What
happens if you get more than 255
points in your "Decapitate the
Orphans" game? In fact, what happens
if you just ignore the limit and
add, say, 10 to 250?

Well, believe it or not, the 6502
will give you an answer of 4. Why?
The number system that the 6502 uses
is like a wheel, with 0 at the top,
counting clockwise 1,2,3,...all the
way up to 255, which lies right next
to the 0. If you go up from 255 you
just wrap around past the 0 and
start all over. Similarly, if you
subtract 2 from 0, you'll get 254.

The solution to all this is
provided by the Carry bit,
discussion of which I've been
putting off until now. The Carry bit
is a flag that the 6502 uses to
remember when it has done arithmetic
that carried it over the boundary
between 0 and 255. By using it
properly, you can solve your
arithmetic problems.

The first trick to using the Carry
bit is to use multi-byte words. This
means that, instead of using a
single byte to store a number, you
use several. For example, if you use
two bytes to remember a number, you
can store a number as large as
65,535. Three bytes lets you to to
16,777,215. Four bytes lets you go
to 4,294,967,295. Big enough for

To use multi-byte arithmetic, you
set up a series of additions or
subtractions. Suppose, for example,
that you want to add two two-byte
words. The program fragment to do
this would look like this:


This little fragment of code
assumes that the first two-byte
value is called (LOGOAT, HIGOAT),
the second two-byte value is called
(LOFISH, HIFISH), and the two-byte
result is (LOANSR, HIANSR). The new
instruction, CLC, stands for "Clear
Carry" and it means that the Carry
bit should be set to 0. It should
always be used with all additions
except chained additions like this

The code does the following: first
it adds the two low values. If the
addition resulted in a wraparound
(result greater than 255), then the
Carry bit was set; otherwise, it was
cleared. Then it performed the
second addition, adding in the value
of the Carry bit (That's why we call
it "Add with Carry"). Thus, if a
wraparound occurred, an additional
one was added into the high sum.
This system insures that multi-byte
addition works properly.

For subtraction, you use the SEC
instruction ("Set Carry").
Otherwise, you handle subtraction
the same way that you handle
addition. In both addition and
subraction, though, the low bytes
must be handled first, then the
higher bytes in the proper order
(lower to higher).


There are two variations on
standard 6502 arithmetic. Both are
so rarely used that I will not treat
them here. The first is decimal
arithmetic using the Decimal flag.
This allows you to set up an
automatic decimal adjust mode. This
is useful in certain types of
arithmetic, primarily BCD

If you don't know what this is,
don't bother with the Decimal flag.
Your program should always begin
with the instruction CLD, which
means "Clear Decimal Flag". I will
tell you this just once: failure to
clear the decimal flag is the source
of the most frustrating and
impossible-to-trace bug in the 6502.
Every single program should start
with the instruction CLD.

The second arcane bit of 6502
arithmetic is signed arithemetic. It
uses the V flag ("oVerflow"). Signed
arithmetic is always confusing and
seldom useful. In 7 years of working
with the 6502, I have never had need
of it. Don't bother.


There are quite a few limitations
on 6502 arithmetic. There is no
facility for multiplication and
division; you have to write
subroutines to do that. You must
design your programs to make do with
8-bit words; failing in that, you
must use multi-byte arithemetic,
with its consequent price in speed
and RAM. All in all, arithmetic is a
real pain on the 6502. This is the
major reason why most 6502 programs
do so little arithmetic.

End of lesson 2.

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Beitragvon cas » Do 21. Mai 2009, 16:39

Teil 2 ist auch im Wiki eingepflegt! Danke Andreas
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Beitragvon cas » Do 21. Mai 2009, 16:55

Teil 0 des Kurses

Assembly language is the great barrier that divides the professional programmer from the amateur. It is the most powerful language available for a microcomputer.
There are four reasons for learning to program in assembly language. First, the speed of execution of assembly language is very high -- about ten time higher than BASIC on the average, perhaps a thousand times faster on certain operations. Even ACTION, the fastest high-level language, is only about half as fast as assembly language.
Second, assembly language tends to be more compact than many languages. Again, ACTION! provides a good comparison. Code produced by ACTION! is about twice as large as equivalent assembly language.
The third reason to program in assembly language is that assembly gives you access to features of the machine that simply are not available in high-level languages. Interrupts are the most notable examples.
Finally, the most important reason for learning to program in assembly language is that it will help you to understand the machine better. And that is a very good place to begin, for you cannot learn assembly language unless
you know a little bit about computers.
I am now going to describe how computers work, in very rough terms. Computers operate on a hierarchy of concepts that spans a great range, rather like the hierarchy that starts with protons and electrons, moves through atoms, molecules, cells, people to civilizations.
A civilization is composed of protons and electrons, but to understand how it is so composed one must know a great deal about the intermediate steps. So too is a computer composed of transistors. There are four intermediate steps between the transistor and the computer.
A transistor is an electrically operated switch. We can assemble transistors into gates that will turn circuits on or off depending on the states of other circuits. There are a variety of gates reflecting the various Boolean operations: AND, OR, NOT, NAND, NOR and EOR.
Gates can be assembled into latches, decoders, and adders. A latch is the simplest memory element: it remembers one bit of information. A decoder translates a number encoded in binary form on a few wires into a selection of one of many wires. An adder will add two one-bit values, with a carry, and generate a carry of its own. We can next broaden each of these devices into an eight-bit device by simply slinging the devices side by side. Eight one-bit latches slung side-by-side give one byte of RAM. Eight adders make an eight-bit adder.
We can thus create a RAM module by building many butes of RAM. We access this RAM module with three buses: a data bus, an address bus, and a control bus. The data bus carries information between the central processing unit and the RAM module.
The address bus is sixteen bits wide; a decoder in the RAM module takes the numeric value on the address bus and decodes it to select the single byte of RAM that is indicated by the address. The control bus establishes the direction of the data flow on the data bus and the timing of data transfer.
The central processing unit (CPU) represents the highest intellectual level of the computer. It is composed of four parts: the Arithmetic and Logic Unit (ALU), the registers, the address bus controller, and the instruction decoder. The ALU is composed of adders and gate arrays that crunch numbers. The particular device to use is selected with a decoder.
The registers are simply on-board RAM. The address bus controller is a device that puts the desired RAM address onto the address bus. The real heart of the CPU is the instruction decoder, a very complex decoder that takes the program instructions out of RAM and translates them into action. It does this by feeding the instructions (which are numbers) into decoder circuits that activate the desired gateways in the CPU.
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Beitragvon CharlieChaplin » Sa 23. Mai 2009, 12:52

and here is part 3... -Andreas Koch.




A great deal of programming
involves the use of Boolean logic.
This is a standardized system for
handling logical manipulations. It's
sort of like algebra for logic. You
must understand Boolean logic if you
are to write assembly language
programs, so let's get started.

Where algebra deals with numbers,
Boolean logic deals with
propositions. A proposition is just
a statement such as "Fred eats
worms". It can take only two
possible values -- True or False. In
our programs we seldom bother with
broad and glorious propositions such
as "Love is the universal language
of truth" or "War is the extension
of policy by other means". Instead,
we normally deal with propositions
such as "The joystick trigger has
been pressed", or "There is a
diskette in the disk drive".

When we use Boolean logic with a
computer, we may think in terms of
true and false, but the computer is
actually working with 1's and 0's.
We use the following convention: a 1
corresponds to a Boolean value of
"true", while a 0 correspons to a
Boolean "false". Using this system
we can represent propositions inside
the computer. However, programming
requires more than the mere
representation of data; we must also
be able to manipulate that data.
This brings us to the Boolean
operators. There are four common
Boolean operations necessary for
most programming practices.

This is the simplest of Boolean
operators. It takes a single Boolean
value as an input and produces as
its output the logical converse of
the input. Thus, a true input yields
output, while a false input
generates a true input.

This Boolean operator takes two
Boolean values as its input and
generates a single Boolean value as
its output. The value of the output
depends on the values of the inputs
according to the following rule: If
one input is true OR the other value
is true, then the output is true.
Otherwise, the output is false.

This Boolean operator is just like
the or-operator, except that it uses
a different rule. Its rule is: If
one input is true AND the other
input is true, then the output is
true; otherwise the output is false.

This Boolean operator is just like
the or-operator, except that its
rule is: If one input is true, OR
the other input is true, BUT not
both are true, then the output is
true; otherwise, the output is

When we use the 6502 for Boolean
operations, you must remember that
the operations are eight bits wide.
Instead of working with one bit at a
time, we use all eight bits of a
word in parallel. The bits in a byte
are independent and do not affect
each other in any way -- at least as
far as Boolean operations are
concerned. The 6502 has three
instructions for performing Boolean
operations. These are AND, EOR, and
ORA. The first performs an
and-operation. For example, consider
the following code:


This will first Load the
accumulator with the value of FISH.
It will then And the contents of the
accumulator with the contents of
GOAT. The result of the
and-operation will be left in the

The AND instruction can use an
immediate operand if you desire,
just as the ADC-instruction can.
The EOR instruction provides the
exclusive-or operator. It works just
like the AND-instruction.

The ORA instruction provides the
OR-operator in just the same way.
If you wish to obtain the NOT
operation, just use EOR #$FF; this
will invert each bit in the
accumulator. Because NOT is so
easily reproduced with EOR, there is
no special NOT instruction in the

If you have any sense at all, you
are probably asking, "What good is
all this Boolean nonsense? What
would I use it for?" Four
applications are available:

Many times our programs encounter
rather complex logical situations.
The program must be able to load a
file; if the FMS is in place and
there is a diskette in the disk
drive, and the diskette has the file
we are looking for, or the file
specification calls for a cassette
load, then we will load the program.
Many programming problems involve
such Boolean operations, Keeping
them straight is certainly a

Sometimes we need to isolate
particular bits in a byte. For
example, in Eastern Front (1941) I
used the character value to store
the unit type. The color of the unit
was encoded in the upper two bits of
the byte, the type in the lower six
bits. If I wanted to get only the
unit type, I had to mask out the
upper two bits. This I did with the
following code:

AND #$3F

The AND-instruction eliminated the
upper two bits, leaving me with just
the unit type. Bit-masking like this
is useful in many situations. We use
it frequently when we pack bits into
a byte to save memory. It is also
handy with input handling. If you
want to read the joystick port, you
frequently mask out the bits in turn
to see which is active. By the way,
you mask out bits set to 1 with the
AND-instruction. You mask out bits
set to 0 with the ORA instruction.
The logic is reversed.

We also use the AND and ORA
instructions to set or clear
individual bits within a byte. This
is most often useful for handling
arrays of flag bits. This little
fragment of code will fold bytes


This is a magical piece of code.
See if you can figure out what it
does. Experiment with two values of
MASK: $OF and $FO.

The 6502 also has instructions that
allow you to shift the bits around
inside a byte. The first of these
are the shift instructions. One,
ASL, shifts a byte to the left; the
other, LSR, shifts a byte to the
right. Thus, the byte %01101011,
when shifted left, becomes
%11010110. Each bit is shifted one
position to the left. The leftmost
bit is rudely pushed right out of
the byte and falls away
("Aaaaaaaaarrrrrggggg!"). A zero is
shifted into the rightmost bit. The
LSR instruction does the same thing
in the opposite direction. Note that
ASL also doubles the value of the
byte, while LSR halves it. Two ASL's
multiply by four; three multiply by
eight. This makes it easy to do
simple multiplication, but be
careful with round-off error here.

What happens if you try to multiply
by 256? What do you get if you halve
3? A variation on the shift
instructions are the rotate
instructions. There are two: rotate
left (ROL) and rotate right (ROR).
These function just like the shift
instructions, except that the bit
that gets shoved into the bottom is
not necessarily a zero; it is the
contents of the Carry bit. The bit
that gets pushed off the edge of the
byte goes into the Carry bit, so it
is not lost. Thus, if you rotate
either way nine times, you'll be
right back where you started.

Rotate instructions are a handy way
to get a particular bit into the
carry bit where you can work on it.
Conversely, once you get your
desired bit into the carry bit the
way you want it, you can put it back
into a byte with some rotate

The last instructions I will cover
are the increment and decrement
instructions. These allow you to add
one (increment) or subtract one
(decrement) from a memory location.
These are not considered to be
arithmetic operations so they do not
affect the Carry flag, nor are they
affected by it. You cannot increment
or decrement the accumulator, only
RAM locations.

End of lesson 3.

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Beitragvon cas » Sa 23. Mai 2009, 13:50

AUch wieder aufbereitet im WIki unter ... age+Course
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Beitragvon CharlieChaplin » So 24. Mai 2009, 16:26

And now,
lesson 4... greetings, Andreas Koch.




One of the most important ideas in
computing is the concept of
conditional execution. This is the
ability of the program to execute
different routines depending on
conditions at the time of execution.
The significance of this capability
is best realized by considering how
programs would operate in its
absence. A program without
conditional execution would not be
able to change its program flow in
response to conditions.

In other words, it would always
execute exactly the same code in
exactly the same order. Every run of
the program would follow exactly the
same sequence and perform exactly
the same operations. Not very
interesting, right? To get a grip on
conditional execution, we need to
look at it in its simplest
expression. The simplest type of
conditional execution is binary in
nature. We have a chunk of code; the
6502 will either execute it or it
will not execute it. The decision is
made on the basis of a boolean
value; a true value will tell us to
execute the chunk, while a false
value will tell the 6502 not to
execute the chunk.

The basic mechanism for doing this
is through an instruction that
performs a transfer of control. This
involves nothing more than altering
the program counter. You may recall
that the program counter is a
register in the 6502 that points to
the address of the currently
executed instruction. When that
instruction has been executed, the
program counter is increased by the
length of the instruction (1,2, or 3
bytes, depending on the instruction).
It now points to the next

This little system allows the 6502
to step through a program in
sequence. But there are also
instructions that will alter the
value of the program counter,
alowing the 6502 to jump to another
area of memory and another part of
the program. The simplest of these
is the JMP instruction. It takes the
form JMP LABEL. This loads the value
of the LABEL into the program
counter. Its effect is to make the
6502 jump to the address of LABEL
and continue execution from there.

It is directly analagous to a GOTO
instruction in Basic. For
conditional execution we need
something more. We need the 6502 to
have capability to make a binary
decision based on a binary value.
The solution used by the 6502
involves flags. These are single-bit
Boolean values stored together in a
single byte of the 6502 called the
processor status register(SR). The
status register is eight bits wide
but stores only seven flags. These
seven flags are labelled N,V, B, D,
I, Z, and C. You have already
encountered the C(Carry) flag and
the D (Decimal) flag. In this
chapter, we are concerned only with
the N, V, Z, and C flags. The magic
instruction that makes possible
conditional execution can take many
forms. Its general form is Bfv
LABEL. The B stands for "branch".

The "f" stands for a flag, and the
"v" stands for the value of the
flag, either true or false. However,
in this case, we do not use the
terminology "true or false". Instead
we use the terms "set" or "clear".
"Set" means the same thing as "1" or
"true", while "clear" means "0" or
"false". The label is the address to
which the 6502 should branch if the
condition is satisfied. If the
condition is not satisfied, then the
6502 will simply skip this branch
instruction and go to the following

For example, suppose that we have
the following instruction sequence:

LDA #0
LDA #5

This will first load the
accumulator with a zero. Then the
6502 encounters the BCS ("Branch on
Carry Set") instruction. It looks at
the Carry flag. If this flag is set
then the 6502 will indeed branch to
the label KARELIA. (For all you
geography buffs, Karelia used to be
in Finland). In other words, if the
Carry flag is set, the 6502 will
skip over the LDA #5 instruction.

Thus, a zero will be stored into
FISH. However, if the Carry flag is
clear, then the 6502 will not take
the branch. It will instead continue
executing the next instruction,
which will load a 5 into the
accumulator. Then it will come to
the label KARELIA and store that 5
into FISH. Thus, the value of the
Carry flag determines whether a zero
or a five is stored into FISH.

The converse of BCS is BCC ("Branch
on Carry Clear"). This will cause
the 6502 to take the branch if the
Carry flag is clear. There is also a
pair of similar instructions for the
V-flag. These are BVS and BVC. They
will cause the 6502 to branch on the
value of the V-flag. Now the
situation gets unncessarily
confusing. The istructions for the
Z-flag should be BZS and BZC --
"Branch on Z Set" and "Branch on Z
Clear". Unfortunately, the dumb
designer of the 6502 thought he
would get cute at this point, so
instead he called these instructions
BEQ and BNE, for "Branch on Equal"
and "Branch on Not Equal". He never
mentioned what he thought is
supposed to be equal to what.

We're stuck with it, so make the
best of it. Just remember what these
instructions really mean BZS and
BZC. If you think in terms of the
Z-flag, it will work out just fine.
If you try to think in terms of
equal or not equal, your attention
will be diverted from the real truth
of the matter and you may make
mistakes. So keep your eye on the
ball and think in terms of Z!

The next pair of branch
instructions use the N-flag. These
are even more insidious than the
previous two. They are called BMI
and BPL, meaning "Branch on Minus"
and "Branch on Plus".
At first glance, these appear to be
reasonable substitutions for BNS and
BNC. After all, if you load the
accumulator with a signed number,
and the number is negative, then the
N-flag will be set, while if the
number is positive, the N-flag will
be clear. Thus, it would seem that
BMI is truly equivalent to BNS and
BPL is truly equivalent to BNC. This
is the source of many a bug in
beginner's programs. Consider the
following fragment of code:


This code is supposed to branch to
POSANSR if FISH is greater than
GOAT. And indeed, if FISH is greater
than GOAT, then when you subtract
GOAT from FISH, you will get a
positive result, right? Not
necessarily! Suppose, for example,
that the value in FISH is $C1 and
the value in GOAT is 1. When the
6502 subtracts GOAT from FISH, it
will get a result of $C0. Note that
the highest bit of $C0 is set to 1.
This is the value that will go into
the N-flag. In other words, even
though FISH is greater than GOAT,
the 6502 will not take the branch,
and this code will fail. The moral
of his tale is, don't take those
instructions literally. They are
misleadingly named. When you see
BPL, don't think "Branch on Plus",
think "Branch on N Clear".

Otherwise, you'll screw up someday.
By the way, the correct branch to
use in the above problem is BCS. Now
for a catch with the branch
instructions. A JMP instruction is a
simple absolute jump -- you specify
the target address and it goes
there. The designers of the 6502
realized that the vast majority of
branch instructions only go a short
distance. They therefore decided to
implement the branch instruction as
a relative branch. The machine code
doesn't specify the target of the
branch, it only specifies an offset.
In other words, instead of saying,
"jump there", it says, "jump so many
bytes forward or backward". The
allowable range is 126 bytes forward
or backward. Thus, you can't branch
anywhere you want, only to nearby
locations. If you must branch
further, reverse the logic of the
branch and use the branch to skip
over a JMP statement.

End of lesson 4.

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Beitragvon cas » So 24. Mai 2009, 19:54


diese Lektion gibt es auch wieder im Wiki: ... HBRANCHING

Chris schreibt:
Unfortunately, the dumb
designer of the 6502 thought he
would get cute at this point, so
instead he called these instructions
BEQ and BNE, for "Branch on Equal"
and "Branch on Not Equal". He never
mentioned what he thought is
supposed to be equal to what.

Die Entwickler des 6502 waren nicht dumm. Hier ist es Chris der einige Hintergründe nicht kannte oder nicht verstanden hatte. Vergleiche zwischen Zahlen werden in CPUs oft als Substraktion ausgeführt. Ist das Ergebniss der Substraktion 0, so sind beide Werte des vergleiches "equal", also gleich. Beispiel:

Code: Alles auswählen
A = 10
B = 10
C = A - B

weiter im Text schreibt Chris:

are even more insidious than the
previous two. They are called BMI
and BPL, meaning "Branch on Minus"
and "Branch on Plus".

Die Namen der Memnonics für BMI und BPL kommen von der Tatsache, das beim 6502 negative 8bit Werte über das BIT 7 dargestellt werden. Damit kann man den Wertebereich von +128 bis -127 abdecken. Wichtig ist dies z. B. bei der Berechnung von relativen Sprüngen per Branch Befehl. Alle Werte mit gesetzten BIT 7 sind wenn als Vorzeichenbehafteter Wert interpretiert negativ (minus), alle Werte in denen das BIT 7 nicht gesetzt ist, sind positiv (plus).

Dieser Abschnitt des Assemblerkurses ist nicht optimal, da der Autor her die 6502 Assembler Memnonics und die Arbeitsweise des 6502 nicht richtig verstanden hatte.


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Beitragvon Dietrich » So 24. Mai 2009, 22:42

Stimmt der Code ist ziemlich komisch. Statt


sollte es besser heißen:


BPL und BMI benutze ich selber fast nie in Zusammenhang mit CMP und SBC, sondern meist nur bei einfachen Schleifen (DEY: BPL LOOP) und zum Testen von Bit 7 eines Bytes (BIT MEM: BMI ...)
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Beitragvon CharlieChaplin » Di 26. Mai 2009, 19:33

here is part / lesson 5... -Andreas Koch.




We are now going to expand the
model of the 6502 that you have been
using. Until now, the 6502 I have
described had nothing more than a
status register, program counter,
and accumulator. Now I am going to
reveal the existence of two new
registers in the 6502: the X- and

These two registers are eight-bit
registers just like the accumulator.
You can load numbers into them and
store them out just as you can with
the accumulator. You cannot do
arithmetic or Boolean operations
with them as you can with the
accumulator. But you can do a number
of very special things that greatly
increase the power of the 6502.

Let's start with the simple move
instructions. The first are LDX and
LDY, which load the X- and Y-
registers the same way that LDA
loads the accumulator. Then there
are STX and STY, which store the X-
and Y-registers the same way that
STA stores the accumulator. There
are also four commands for
transferring bytes between
registers; these are TAX (transfer A
to X), TAY (transfer A to Y), TXA
(transfer X to A) , and TYA
(transfer Y to A).

Then there are four special
instructions that you will use very
often. These are INX and INY, which
increment (add one to) the X- and
Y-registers, and DEX and DEY, which
decrement (subtract one from) the
X-and Y-registers.
Finally, we have the CPX and CPY
commands, which compare X or Y with
the operand of the instruction.
These two instructions operate in
exactly the same way that the CMP
instruction works, except that they
use the X- and Y-registers instead
of the accumulator.

What are these two registers used
for? Well, they are sometimes used
as temporary registers. If you are
in the middle of a lengthy
computation, and you need to save a
value currently in the accumulator
to make room for something else, the
X- and Y-registers are a handy place
to stuff values away for temporary
storage. Programmers do this all the

However, temporary storage is not
the real purpose and value arise
from their utility as index
registers. Index registers go hand
in hand with loops; the best way to
show you how they are used is to
dump the whole schmeer at once and
then explain it.
So consider the following problem:
your program has to deal with the
possibility of user errors. Suppose
you require the user to type in a
file name for your program to read.
What happens if this file is not on
the disk? You have to put an error
message on the screen that says,
"FILE NOT ON DISK!" How do you print
the message? Here's a sample bit of
code that will do it:

SBC #$20

Let's take apart this code and
explain it step by step. First thing
we do is load the X-register with
the number of characters (minutes
one) in the message. The expression
(ENDMSG-ERRMSG-1) will calculate
that length at assembly time. This
turns out to be 17 characters long.
If we were pedestrian about it we
could have just written LDX #16, but
this way, if we decide to change the
message we don't have to remember to
go back and change the LDX command.
Neat, huh?

OK, so now we have a 16 in the
X-register. Now the 6502 comes to
the next command -- LDA, ERRMSG, X.
This command tells it to load the
accumulator with the byte at
(address ERRMSG, indexed by X). What
this means is as follows: the 6502
will take the address ERRMSG and add
the value of the x-register to that
address. It will then go to the
address so calculated and load the
accumulator with the contents of
that address. Since X contains a 16,
the 6502 will go to the 16th byte
after the first byte in the table
ERRMSG. If you count characters, you
will see that the 16th byte is the
exclamation point. So the 6502 will
load the ASCII code for an
exclamation point into the

The next two instructions (SEC, SBC
#$20) are necessary to correct for
the Atari's nonstandard handling of
ASCII codes. They make sure that the
exclamation will be printed on the
screen as an exclamation point.
The next instruction (STA SCREEN,X)
stores the reslt indexed by X. The
6502 will add the contents of X
(still 16) to the address SCREEN. It
will then store the contents of the
accumulator into that address. If
that address is part of screen RAM,
then you will see an exclamation
point appear on the screen.

The next instruction that the 6502
encounters is the DEX instruction.
This instruction subtracts one from
the X-register, making it a 15.
Next, the 6502 comes to the
instruction BPL, LOOP 1. This will
branch if the N-flag is clear. The
vaue of the N-flag is affected by a
DEX instruction. The value of bit D7
of the result is transferred to the
N-flag. Bit Dd7 of 15 is a zero,
hence the N-flag is clear, hence the
6502 will indeed take the branch.
Note that it branches back up to

Now it will repeat the process,
only this time X contains a 15, not
a 16. It will therefore grab the
15th character, an ASCII 'K', and
store that to the screen position
just before the exclamation point.
Then it will subtract one form X to
get a 14, and will continue the

This process will continue, with
the 6502 grabbing bytes in reverse
order form the table and storing
them onto the screen, until after
the 6502 does the seroth byte. When
X contains a zero, and the 6502
executes a DEX, it obtains the
result $FF. This sets the N-flag.
When the 6502 encounters the BPL
command, it will NOT take the
branch; instead, it will skip the
branch and go on to the JMP
statement. The loop is terminated.
In this one fragment of code you
have seen two major ideas: indexed
addressing and looping. They are so
closely related that it is hard to
talk about one without talking about
the other.

You can use indexed addressing with
either the X-register or the
Y-register. You most commonly use
indexed addressing with the LDA and
STA commands, but you can also use
it with many of the other 6502
commands: ADC, SBC, CMP, AND, ORA,
EOR, LSR, ROR, ADL, and ROL can all
be used with indexed addressing.
Indexed addressing allows you to
work with tables or arrays of data.
There is one ugly catch: all of
your arrays must be less than 257
bytes long, because the index
registers are only eight bits wide.

Most of the time this is not a
serious problem. However, if you
must address a larger table or
array, you can use indirect
addressing. To do this, you
calculate the address that you
desire to access, store that address
in two contiguous bytes on page zero
(low, then high) -- we call these
two bytes a pointer -- and then
refer to the pointer like so:

LDA (POINTER), Y This instruction
will take the address out of
pointer, add the value of Y to it,
and load the accumulator with the
contents of the address so
calculated. If POINTER contains
$4567 and Y contains a 2, then the
6502 will load the acumulator with
the contents of address $4569. You
are still restricted by the size of
Y, but you can always go back and
change the POINTER if you need to
span larger arrays. In this case,
you frequently just leave Y equal to
zero and do all of your indexing
directly with changes to POINTER.

The last topic I will take up is
termination techniques. Every loop
must somehow be terminated if you
are to avoid the problem of the
Sorcerer's Apprentice. You will note
that the programming example I gave
used a rather odd approach. I
started at the end of the array and
worked backwards. Why not start at
the beginning and work forwards?
It's slightly more efficient going
backwards than forwards. When you go
forwards, you have to terminate the
loop with:

CPX #17

Whereas when you go backwards, you
need only use:


Going backwards you save one
instruction. However, if this
confuses you, feel free to count
forward; that works, too, only it's
a little less efficient.

There is also a problem on choosing
whether to BNE or BPL. BPL restricts
you to a range of only 127 bytes,
but BNE allows you a range of 129
bytes by starting the branch two
bytes after the branch test byte(a
reverse branch starts at the same
place but is a minimum of three
bytes to jump over the branch code
itself). Just remember to start the
index from ERRMSG-1 and SCREEN-1
instead of ERRMSG and SCREEN.

There are lots of other sneaky ways
to terminate loops, but they fall
into advanced topics.

End of lesson 5.

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Beitragvon cas » Mi 27. Mai 2009, 08:56

Danke Andreas.

Auch dieser Abschnitt ist wieder im Wiki unter ... S26LOOPING
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Beitragvon CharlieChaplin » Do 28. Mai 2009, 19:39

here is part / lesson 6... -Andreas Koch.




We now take up the first topic in
this series that is not absolutely
essential to writing programs:
subroutines. The loops and indexed
addressing discussed in the previous
lecture are truly essential: it is
hardly possible to write a useful
program that has no loops.

Subroutines are a matter of
convenience, not necessity. It is
quite possible to write an entirely
adequate program without using a
single subroutine. However, you will
find that the convenience of
subroutines with large programs is
so great that you would never want
to write such a program without

The primary purpose of a subroutine
is to perform some function that is
frequently needed at many points in
the program. Instead of having to
repetitively insert the same code
over and over again, we simply write
it once, place it in a subroutine,
and call that subroutine many times
from the main program. The use of
subroutines dramatically reduces the
size of a program.

Subroutines are implemented on the
6502 in a fashion very similar to
that used by BASIC. You may recall
the two BASIC commands for
subroutines: "GOSUB lineno" and
"RETURN". The two corresponding 6502
commands are "JSR label" and RTS".
The label in "JSR label" is the
label of the beginning of the
subroutines. Thus, writing and using
subroutines in 6502 is trivially
simple. First, you write the
subroutine. You give it a name (say,
"MYSUBR") and stick that label in
front of the first instruction. You
put an RTS command after the last
normal command of the subroutine. To
call the subroutine, you just put
JSR MYSUBR. That's all it takes!

However, in order to understand how
it works is not so easy. Here's the
problem we must solve when the 6502
jumps to a subroutine, the JSR
instruction tells it the destination
address to which the 6502 must jump.
But when the 6502 hits the RTS
instruction, how does it know the
address to which it must return? The
RTS doesn't say, "Return to THIS
address"; it says only "Return".

Moreover, how could the 6502 know
where to return? If the subroutine
can be called from, say, five
different points in the program, how
would the 6502 know which of those
points to which it must return? What
if we gave the 6502 a special
register for remembering return
addresses? That is, whenever the
6502 encounters a JSR instruction,
it stores the current address into
its return address register. Then
when it encounters an RTS
instruction, it simply takes the
address out of the return address
register. There is only one problem
with this: what if we use nested
subroutines (one subroutine calls
another)? The second subroutine call
will erase the return address for
the first subroutine call. Trouble!

The solution to all this is called
a stack. A stack is a chunk of RAM
allocated for certain special
operations such as subroutines. The
6502 stack is stored on page one --
that is, addresses $0100 to $01FF.

The stack operates like 128 return
address registers arranged in
sequence (remember: two bytes per
address). The 6502 keeps a stack
pointer register to keep track of
which byte in the stack is currently
being used. I will now trace through
the operation of the stack in a
subroutine. We start with the stack
pointer set equal to $FF. That means
that the stack is empty; the stack
pointer is at the very top of the
stack. The 6502 encounters a JSR
instruction. It takes the current
value of the program counter and
breaks it into two bytes. It pushes
the first byte onto the stack. This
means that it stores the first byte
at $01FF, then decrements the stack
pointer. Now the stack pointer is
$FE. The 6502 then pushes the second
byte of the return address onto the
stack, storing that byte at $01FE
and decrementing the stack pointer
to $FD. Then the 6502 jumps to the
subroutine. When it encounters the
RTS instruction, it pulls the two
address bytes off of the stack
(increments stack pointers and loads
byte at address $0100,SP). Those two
bytes go directly into the program
counter, returning the 6502 to the
original entry point.

The advantage of this approach is
that it allows very deep nesting of
subroutines. If one subroutine calls
another, the 6502 simply stores more
values onto the stack. The addresses
won't be confused because you always
exit subroutines in exactly the
reverse of the order that you
entered them. You can use the stack
yourself, if you wish. You have six
instructions that allow you to play
with the stack: PHA, PLA, PHP, PLP,
TSX, and TXS. The PHA instruction
pushes the value of the accumulator
onto the stack and decrements the
stack pointer. The PLA instruction
increments the stack pointer and
pulls the current stack value into
the accumulator. These two
instructions allow you to store and
retrieve values onto the stack. They
must be exactly balanced, though, or
you will generate that most feared
of bugs, the stack crashes.

Consider: you are in a subroutine.
You push a value onto the stack, but
forget to pull it off. When the 6502
attempts to return to its original
location, it pulls two address bytes
off the stack -- but they're the
wrong two bytes. One of them is the
value you pushed but didn't pull.
Result: the 6502 return to the wrong
address. Your program goes haywire
and the computer crashes. This is
caled a stack crash. This type of
crash tends to the particularly
difficult to recover from.

Prevention is the best medicine
here. The rule for preventing stack
crashes is simple and absolute: each
and every push onto the stack must
be balanced by one pull from the
stack. Violate this rule and you
will certainly experience a stack

The next pair of stack manipulation
instructions are PHP and PLP. These
push and pull the process status
register pmtp the stack. They are
useful for two purposes. First, you
may wish to save the values of the
various flags before performing some
operation, then restore them so that
you can branch on a previously
created condition. Second, it is
sometimes handy to PHP, than PLA to
get the processor status register
into the accumulatorada where you
can more directly manipulate it.
Again, each push must be balanced by
one pull. The third stack
manipulation pair of commands do not
modify the stack. They are TSX and
TXS. These transfer the stack
pointer to and from the x-register.

Once in the x-register, you can
change the value of the stack value
and then TXS to jump over sections
of the stack. This can be a very
handy way to pass parameters to
subroutines, but it is also very
tricky. If you make a mistake, you
will generate a stack crash. So be
careful with this one. I have always
avoided these commands like the
plague. They are very dangerous and
never essential.

End of lesson 6.

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Beitragvon cas » Fr 29. Mai 2009, 07:10

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Beitragvon CharlieChaplin » Sa 30. Mai 2009, 19:09

And now,
part 7... -Andreas Koch.




We now approach one of the most
difficult topics in the world of
assembly: interrupts. This is such a
messy topic that very few high-level
languages make any provision for
interrupts. Moreover, interrupts are
one of the best ways around to crash
your program hard. Programmers using
interrupts must be very careful.

The standard way to handle this
problem is with a technique called
polling. Your program runs out every
now and then to check whether the
high-priority situation has arisen.
If it has, then the program responds
to it. If not, it returns to its
original work.

The problem with polling arises
from the choice of polling interval.
If you choose a long(infrequent)
polling interval, then you may not
respond to a demand quickly enough.
If you choose a short (frequent)
polling interval, then you will
respond quickly to the demand, but
you will never have any time for
your regular computations.

You may think this type if
situation is infrequent, but I can
list quite a few situations where
this is fairly common. Most I/O
operations involve short bursts of
computation at infrequent intervals,
but they must be attended to on a
tight schedule. For example, talking
to a cassette deck involves very
little real work from the CPU, but
they must be done according to a
precise schedule. Even a disk drive
is very slow by the standards of a
6502. Or how about keyboard
response? When the human operator
presses a button, he wants to see a
response NOW, not two or three
seconds from now. Yet he could press
on that button at any time. So
should your program sit on its hands
waiting for a keypress, or should it
ignore the human operator?

The solution to all of these
problems is the interrupt. An
interrupt is rather like a
subroutine that can be called by a
hardware action. There is a wire
going into the 6502 called
IRQ(Interrupt Request). That wire is
normally quiet. But when something
important happens, like a keypress,
the computers' hardware puts a
signal on that wire to interrupt the

Here's what happens next: The 6502
is busy running a program, but when
it gets the interrupt signal it
first checks the 1-bit (internal) in
the processor status register. If
the 1-bit is set, it decides to
ignore the interrupt, but if it is
clear, it proceeds to the next step.
It saves the process status register
and the current value of the program
counter onto the stack.

Then it loads the program counter
with the address stored a special
place in ROM -- it's either $FFFC or
$FFFE, I can never get it straight.
It thus jumps to the address
specified in ROM. It expects to find
an interrupt service routine there,
which presumably will deal with the
keypress in the appropriate manner.

This routine will probably start by
pushing the A, X, and Y registers
onto the stack to preserve them.
When done, the routine will the pull
them off the stack and execute an
RTI instruction, which causes the
6502 to pull the processor status
register off the stack, and then
pull the program counter off and
resume operating.

The important thing about the
rather complex sequence is that is
allows the 6502 to drop whatever its
doing, service the interrupt, and
the return to its earlier
functioning without skipping a beat.
The overriding goal of all this is
to be absolutely certain that, when
the 6502 returns from the interrupt,
it returns in EXACTLY the same state
that it was in when the interrupt
hit. Otherwise, all sorts of
horrible, untraceable bugs would
result(and the computer would never

Imagine -- you are in the middle of
some huge computation when a
interrupt strikes. It subtly changes
some very tiny parameter, just
enough to insure that when the
computation resumes, it will be
slightly incorrect. When you try to
find the bug, you discover that
sometimes the code works perfectly
and sometimes it fouls up, and you
can't figure out why it should do
that. Very bad business! Moral:
interrupts must follow a very tight
discipline if they are to be of any

Now let's get into some of the
technical gore involving interrupts.
First, there are two interrupts on
the 6502. They are called IRQ
(Interrupt Request) and NMI(Non-
Maskable Interrupt). The idea is
that the IRQ can be masked out by
setting the 1-bit with the SEI
instruction. Then you use the CLI
instruction to clear the 1-bit.

Thus, IRQ is used for interrupts
that have second priority. NMI is
reserved for first-priority
interrupts, it is not maskable.
However, the designers of the Atari
computers routed IRQ and NMI through
the POKEY and ANTIC chips
respectively. And they put mask
registers into these chips. Thus,
the NMI can be masked out after all
-- but only on Atari computers.
Other 6502-based computers don't
allow that.

The NMI and IRQ interrupts have
separate interrupt vectors in ROM,
so they can be treated differently.
These vectors route the interrupts
to the OS, but the OS is smart
enough to route interrupt flow
thorugh some RAM locations. This
means that you can intercept these
two interrupts by altering the
contents of the RAM-vectors. (I
won't list them here, there are a
number of them for different
situations). You must be careful,
though, when altering such a RAM
vector. What happens if an interrupt
strikes after you have changed one
byte of the address and before you
have changed the other byte?

The 6502 will fly off into
never-never land and you have
crashed. Sure, it's unlikely, but
good programmers don't count on luck
to make their programs work. You
have to guarantee that the interrupt
won't occur before you mess with the
vector. Use SETVBV from the OS.

The two primary applications of
interrupts with the Atari computers
are for VBIs (Vertical Blank
Interrupts) and DLIs (Display List
Interupts). These are very involved
topics covered quite thoroughly in
the book 'De Re Atari'. VBIs are
most often used for animation
control, input handling, and other
time-critical operations. For
example, the entire player I/O of my
game Eastern Front (1941) is handled
by VBIs. The scrolling, giving of
orders, identifying units, and so
forth is all done by VBIs. The
mainline routine meanwhile figures
the artificial intelligence. DLIs
are used to enhance the graphics on
the screen. You can get more colors,
more use out of players, more
scrolling, and more character sets
with proper use of DLIs. Again,
consult 'De Re Atari' for a full
treatment of this complex subject.

Interrupts are extremely difficult
to debug because they tend to crash
the system when they fail. You must
exercise the strictest discipline in
writing interrupt code. Timing
problems, seldom of concern in
mainline programming, can become
critical with interrupts. What
happens, for example, if your
interrupt service routine takes so
much execution time that more
interrupts arrive than you can
service? Bad things, I assure you.

You must always ask yourself, what
happens if an interrupt strikes
here? Or there? You must assume that
an interrupt will strike at the
worst possible time, and write your
code to deal with that possibility.

The most important discipline to
follow in writing interrupt service
routines is this: keep your
interrupt database separate from
your mainline database. If the ISR
can freely write to variables used
by the mainline, you will certainly
have problems when the mainline
attempts to work with variables
whose values change in unpredictable
ways. You must set up ironclad rules
about when the ISR can mess with
variables used by the mainline, what
it can do to them, and how it
notifies the mainline routines that
it has indeed altered them.

Approach interrupts with extreme
caution. They are very powerful, but
every programmer can tell you horror
stories about debugging interrupt

End of lesson 7.

Night&Day Poster
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Beitragvon cas » So 31. Mai 2009, 10:18

Auch diesmal in Wiki ... INTERRUPTS

Schoene Pfingsten
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Beitragvon CharlieChaplin » Mo 1. Jun 2009, 21:36

here is the last part / lesson... -Andreas Koch.




We have covered all of the
traditional material associated with
6502 assembly language PROGRAMMING.
However, there remain a number of
topics that should be addressed
before we finish. They are not
closely associated with each other,
so I will take them in random order.

The first topic is perhaps the most
difficult one for a beginning
assembly-language programmer: Where
do I begin? How do I put together an
entire assembly language project?

The problem here is seldom a
technical one. Most beginners are
stopped by their own lack of goals
rather than any lack of technical
expertise. One does not just write
an assembly language program because
one knows assembly language -- that
is putting the cart before the
horse. One starts with goals and
then considers means.

A story from my early days with
micros will illustrate this point. I
did not have anybody to teach me
assembly language. I decided in 1976
that I wanted to do wargames on
computers. Accordingly I bought a
KIM-1, an early 6502-based
single-board computer. I received it
in January 1977. I studied the
manuals and taught myself 6502
machine language. I had my first
wargame up and running in six weeks.
That means that I not only taught
myself 6502 in six weeks, but I
wrote and debugged a program at the
same time.

Now, the point of this story is NOT
"Wow, isn't Chris Crawford the
smartest programmer who ever lived!"
The point of this story is that
goal-oriented learning is far more
effective than goal-less learning.
Had I sat in on some technical
course on 6502, I would have taken
months and months to learn the
material. Because I had a clear
goal, I learned very quickly.

My advice to you, the beginning
assembly programmer, is this: You
have acquainted yourself with the
rudiments of 6502 programming. If
you have some project you would like
to pursue, some goal you would like
to achieve, then do it. If not,
don't waste your time trying to use
a tool for its own sake.

Assuming you pass this first test,
there remains the broad problem of
organizng your assembly language
program. I suggest that you break
your program up into six modules,
each forming a separate source code
file. These six modules would be:

EQUATES file: this file defines all
of the equates used by the program:
the data areas, the page zero and
page six usage, and perhaps some of
the large graphics and screen

DATA file: this file contains all
of the static tables used by the
program. This would include all the
text messages that would be printed
onto the screen, bitmaps of graphics
images, graphics character set
definitions, and so forth.

INITIALIZATION code: this file
contains the routines that
initialize the program when it first
fires up. They set up the screen,
clear out all the special graphics
and sound registers, zero out all
the arrays that need to be cleared,
and do all the other legwork
associated with clearing the decks
for a program.

INTERRUPT code: this module
contains the code associated with
any interrupts used by your program.
This would most commonly involve
vertical blank interrupts and
display list interrupts. In as much
as your interrupts should be
well-separated from your other code,
you might as well keep the code in a
separate file.

MAINLINE code. This includes the
main program loop that controls the
primary behavior of the program. If
you have problems imagining this,
think of it as nothing more than a
series of subroutine calls arranged
in a loop, with each subroutine
handling one chunk of the overall

SUBROUTINE code: After a while you
build up a collection of subroutines
for handling standard processes in
the program. Keep them here.

The second topic I would like to
talk about is the place of the 6502
in the larger world of
microprocessors. The 6502 is
undoubtedly the most successful of
microprocessors to date, having been
installed in more systems than any
other microprocessor. It is also a
very old microprocessor, having
first appeared in 1976. That makes
it quite old.

A very simple way to approach the
world of microprocessors is to group
them into two sets -- the Sixes and
the Eights. The Eights represent the
earliest group of microprocessors,
they trace their lineage all the way
back to the 4004, the first
microprocessor. The 4004 was
followed by the 8008, the first
eight-bit microprocessor. The 8008
was superseded by the 8080, which
was in turn followed by the Z-80.

The Z-80 was the most advanced
eight-bit processor in the Eights
line. The next step was to go to 16
bits with the 8088 and 8086. These
were followed by the more powerful
80186, 80286, and 80386.

The fundamental philosophy of all
the Eights can be expressed in two
words: features and compatibility.
The designers of the Eights were
always adding new features to the
microprocessors with each successive
generation. The goal seemed to be to
pack as many bells and whistles in
as would fit. The second goal,
compatibility, meant compatibility
with the previous microprocessor in
the series. This insured that
software developed for previous
versions would still run on the
newer versions.

The result of this design
philosophy was a series of powerful
microprocessors that were quite
complex in layout and rather
difficult to learn. The features
were piled up on each other in a
bewildering array. Once you learn
the system, it seems natural enough.
But it is something of a mess.

The Sixes include the 6800, the
6502, the 6809, and the 68000. The
two key words guiding the design of
the Sixes are cleanliness and speed.
The idea was to make the instruction
sets clean, powerful, and fast. The
hope was that the processors would
be so easy to learn that
compatibility would not be a
problem. The design approach was to
use just a few simple instructions,
but give them variations that
greatly extend their power. Thus,
the 6502 has a LDA instruction that
can be used with a great many
addressing modes.

The 68000 is the 32-bit entry into
the Sixes line. It carries the idea
of cleanliness even further than the
6502. The 68000 uses a single
instruction with different modes to
replace the 6502 instructions LDA,
TYA, PHA and PLA. That's quite a

The 68000 also boasts sixteen
registers, each 32 bits wide. That's
a total of 512 bits of register
space, the 6502 has 32 bits of
equivalent register space. Those
sixteen registers eliminate many of
the data-shuffling problems so
common with the 6502.

The 68000 has a linear address
space 24 bits wide -- that's sixteen
megabytes! Thus, a 68000 can
directly address 16 megabytes of RAM
and ROM. The 6502, by contrast, can
only address 64K directly -- it must
use paging sytems that slow it down
to address more memory.

Finally, the 68000 has a number of
advanced capabilities that make
possible a number of special
capabilities. I will describe just
one stack frames. The 68000 makes it
easy to set up a local, temporary
stack when you enter a subroutine.
Thus, subroutines can have their own
local variables stored on the stack,
accessed via a special stack pointer
register. The 68000 will manage all
the housekeeping necessary to keep
such a system straight.

Did I mention that 68000 has
hardware multiply/divide?

End of lesson 8.

***** THE END *****

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Beitragvon Speak » Di 2. Jun 2009, 12:57

Hallo @cas @charlie,
ich weiss nicht ob ich nur für mich spreche, aber wäre eine deutsche Übersetzung nicht auch sinnvoll?
Wenn es denn machbar wäre ...
Gruss Speak
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