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--------------------------
CHAPTER no. 3:
* FIRST PROGRAMS *
--------------------------
*** INTRODUCTION WITH AN EXAMPLE ***
------------------------------------
- Consider the following 68000 ASSEMBly program (#1):
-------------------
; program #1 in 68000 Assembler
TEXT ;start Text zone
A EQU 18 ;A=18
B EQU 2 ;B=2
move.w A,destination ;move word A to destination
move.b B,other ;move bit B to other
move.w other,d0 ;move what's in other to
;register d0
add.l A,d0 ;add A to d0
lea A,a0 ;put A's address in
;register a0
move.w (a0)+,save_A ;increment a0 and save a0
add.l save_A,d1 ;add save_A to d1
add.l d0,d1 ;add d0 to d1
move.l d1,result ;put d1 as result
move.w DAT,d0 ;put dat in d0
add.l DAT,result ;add dat to result
clr.l result ;clear result
DATA ;start Data zone
DAT DC.W 6 ;data 6 in DAT
BSS ;start Bss zone
save_A DS.L 1 ;reserve a L-W in save_A
result DS.L 1 ;reserve a L-W in result
destination DS.L 1 ;same in destination
other DS.B 1 ;reserve one BYTE in other
END ;end of the listing
--------------------
1) EXPLANATIONS:
-------------
- Don't look for any particular meaning in this listing, it is only meant
to introduce the various concepts of ASS programming...
- As you can see, an ASS. program is structured.
This structure is divided into 3 columns.
.1st column: LABELS or TAG names:
-----------
Their purpose is to set a memory address, so one
can call an ADDRESS by naming the LABEL when an instruction
requires it. (Like a line number in BASIC)
The actual address assigned to the Label is given after linking,
these addresses serve as reference points in
the program, and do not aim at definable addresses.
(Unless the program is relocated to a specific location in
memory, but this is not of real interest)
.2nd column: INSTRUCTIONS and their OPERANDS:
-----------
The instructions tell the computer the steps to follow, these
instructions can be followed by operands if their syntax
demands it (as for MOVE x,y)
In ASS 68000, there are 56 basic instructions.
.3rd column: COMMENTS:
-----------
All text located after the Operands is no longer recognized as
an instruction, and can therefore be used to describe the listing
by including useful information...
With some editors (PROFIMAT), a comma must be put
in front of comments, otherwise it causes an error during
assembly.
Blank (or empty) lines or those starting with *
(for METACOMCO ) or ';' (for PROFIMAT) are also treated
as COMMENTS.
NB: .For a LABEL to be recognized as such, it must be written on
--- the first column, the other columns must be separated
by at least one space (' ').
.No more than one instruction per line.
2) DETAILED COMMENTARY OF THE LISTING:
--------------------------------
-line 1 : '; Program #1 in 68000 Assembler '
---------
.As you can see, it's a 'comment', here
it refers to the name of the program...
-line 2 : ' '
---------
.A blank line... i.e. nothing at all (yes indeed!)
-line 3 : ' TEXT'
---------
.There is an instruction (column 2).
.In fact, it's an ASSEMBLY DIRECTIVE.
.DIRECTIVES are functions specific to the ASSEMBLER used,
hence their syntax can vary with the editor used.
For most directives, the syntax is identical. (I will name
the exceptions, but the manual of your assembler should
contain the names and descriptions of directives it uses.)
.DIRECTIVES are placed in the 2nd column of the listing, just like
instructions or MACRO-INSTRUCTIONS.
.The 'TEXT' directive forces the initialization of the P.C.,
the Program Counter (to 0 or to its value at its last
initialization if there are multiple 'text' sections)
----------------
- But what is the 'P.C.'? :
------------
- It's the Program Counter
It is a 32-bit REGISTER which contains the address (even number) of the WORD
in which the CODE (in BINARY) of the next instruction to be executed is found.
In practice, only the 24 least significant bits are used in this
particular register.
Therefore, the P.C. is incremented after each instruction by an even number
of bytes (depending on the size of the instruction).
Jump (jmp...) or branch (bsr..) instructions aim to modify the P.C.
and thus cause a jump to the address pointed to by the P.C.
so: PC | code of the instruction in BIN
-----
???????????????
PC ---------->
???????????????
PC ---------->
???????????????
PC ---------->
etc...
representation of the PC:
---------------------
32 23 0
********[][][][][][][][][][][][][][][][][][][][][][][][]
------------------
.The 'TEXT' directive is therefore meant to initialize the P.C.:
It precedes the instructions that form the listing, hence
the name 'text'.
-line 4: ' A EQU 18 '
-------- (I stop counting blank lines)
.We find a LABEL: A, a DIRECTIVE: EQU, and its OPERAND: 18
.the directive EQU aims to assign a value to the label
it is associated with (an integer number).
In our listing: 18 is associated with 'A' address.
-line 5: ' B EQU 2 '
--------
.2 is associated with 'B' address.
-line 6: ' move.w A,destination '
--------
.There is an INSTRUCTION (move) and two operands (the source
and the destination).
.The 'move' instruction transfers data from a source operand
to a destination operand.
.The 'move' instruction is followed by the suffix '.w': this
indicates that the instruction operates on a WORD (or Word)
.There are 3 suffixes that can be added to certain instructions
(we'll detail which ones)
- .L :the instruction deals with a L-W (Long)
- .W :it involves a WORD (Word)
- .B :it deals with a BYTE
NB: If an instruction accepts one of these suffixes and it's not included:
--- for example, if you write 'MOVE #1,d0', the .W suffix is implied by
default, meaning if you write 'MOVE #3,d2', it equates to writing
'MOVE.W #3,d2')
----------------------
What are ADDRESSING MODES?:
-------------------------
This is the most fundamental point of ASS programming.
- ASS allows easy movement of data in memory (e.g.,
with the 'move' instruction) or to 'point' to instructions or data
identified in memory.
One of the richnesses of ASS compared to other languages is that ASS
utilizes several ADDRESSING MODES: 14 in 68000.
That is, in ASS, it is possible to move (directly or indirectly)
data in memory or act on data located in memory, in 14 different ways!
- The elements that come into play in the different addressing modes
are: The REGISTERS, the PC (and also the SR, a very special
register, which we will study in detail)
the REGISTERS:
--------------
. We distinguish the DATA REGISTERS:
---------------------
There are 8 and they are designated by their addresses: d0,d1,d2,d3,d4,d5,d6,d7
They are 32 bits in size and USED TO STORE NUMERICAL DATA. ------------------------------
so, if we write: MOVE.L #1,d0
ADD.L #1,d0
We place 1 in the 32-bit d0 register (all 32 bits of the register are
affected by the 'move.l' instruction because of the suffix '.L'), then
we add (add.l) 1 to this register (the 32 bits of the register are once
again affected), so d0 will contain '2' and will be represented in
memory as:
00000000000000000000000000000010 ( %10=2 )
And the ADDRESS REGISTERS:
---------------------
There are 9, 8 are available to the programmer: they are designated
by their addresses: a0,a1,a2,a3,a4,a5,a6,a7 and ARE USED TO STORE
ADDRESSES. -----------------------
--------
thus, if we write: MOVE.L NAME,a0
The a0 register is loaded with the value of the address 'NAME' (the 32
bits of a0 register are affected because of the suffix '.L')
WARNING: Only WORDS or L-Ws can be transferred into an
---------- ADDRESS REGISTER (no BYTE, this is very important!)
NB the a7 register is special, it is used as SYSTEM STACK pointer
-- (or SP from 'Stack Pointer') which is a special area of
memory used by certain jump instructions that store there
the return address to the instruction calling the subroutine (in real-
ity it's the PC that is saved then reloaded, so put back to its initial
value at the end of the subroutine which causes a return to the instr-
uction following the jump instruction, we will study this in depth
later)
there is the PC, which is also A REGISTER:
-----
.We have seen that it is composed of 32 bits of which 24 are used and that
it points to the even-numbered address of the next instruction to execute.
And the SR (the Status Register):
--
It is a 16-bit register that is divided into 2 distinct bytes:
- A user byte (LSB)
- A supervisor byte (MSB)
Here is its structure:
supervisor | user
--------------------------+------------------------
SR: [T][ ][S][ ][ ][i2][i1][i0][ ][ ][ ][X][N][Z][V][C]
---
Bits n° 15 8 7 0
- The supervisor byte: is writable only in SUPERVISOR
MODE by setting the 'S' bit. (There is a Gemdos function
that does this when called)
It is only in supervisor mode that one can access the SYSTEM
STACK and certain privileged instructions.
The 'T' bit allows or the microprocessor to function in TRACE
mode (step-by-step program execution after each instruction, we
will talk about it in the chapter on DEBUGGERS)
The bits i2, i1, i0 constitute the interrupt mask.
(I will come back to this in detail...)
- The user byte: can be used in both MODES (user and supervisor)
This BYTE is also called the CONDITION CODES REGISTER or CCR from
'Condition Codes Register' ---
- It is modified by most 68000 instructions.
* The 'N' bit (n°3) is 1 if the result of an arithmetic
operation is Negative, otherwise it is set to 0.
* The 'Z' bit (n°2) is set to 1 if the result of an operation
is zero (Zero), otherwise it is set to 0.
* The 'V' bit (n°1) is set to 1 if the result of an operation
cannot be represented in the size of the defined operand (overflow)
otherwise it is set to 0.
* The 'C' bit (n°0) is set to 1 if an operation causes a
carry beyond the most significant bit of the result operand
(such as division), otherwise it is set to 0.
* The 'X' bit (n°4) is the eXtension bit, its use is
limited to certain instructions that we will study.
.Now that you have familiarized yourself with the different 68000 Registers,
let's define the different ADDRESSING MODES.
Addressing modes allow you to modify the values of the PC, SP, SR, and system stack.
I will use the MOVE instruction (allows to move the source operand to the
destination operand) and ADD (adds the source operand to its destination operand)
to illustrate the different types of addressing modes.
*** THE ADDRESSING MODES OF THE 68000 ***
------------------------------
1) IMMEDIATE addressing (shown as #...)
--------------------
A) NUMERIC: (The source operand is data)
----------
It's written:
-----------
+-------------------------------------+
| Instruction #data,destination |
+-------------------------------------+
And reads as:
----------
Place the source data in (to) the destination operand
Examples:
--------- MOVE #12,d1
( i.e. MOVE.W #12,d1)
We place the number 12, coded on a WORD in the least significant WORD of
the d1 register:
0000000000001100 ( WORD=%12 )
|
\|/
................0000000000000000( d1 register, only the least significant WORD is affected because
we wrote:'move.W' )
and we get:
--------------
................0000000000001100 ( %12 In the least significant WORD of d1 )
Bits n° 31 15 0
Exp 2: ADD.L #3,d1
------
We add 3, coded on a L-W to the contents of d1 and all 32 bits of d1
participate in the operation:
00000000000000000000000000000011 ( L-W=%3 )
|
\|/
00000000000000000000000000000000 ( d1 register )
Bits n° 31 0
and we get:
--------------
00000000000000000000000000000011 ( d1 register=%3)
Bits n° 31 0
B) SYMBOLIC: (The source operand is a LABEL)
----------
It's written:
----------- +--------------------------------------+
| Instruction #Label,destination |
+--------------------------------------+
And reads as:
----------
Place the address of the Label in (to) the destination operand.
Example:
--------
MOVE.L #tag,a0
We place the L-W (even number) containing the address of 'tag' in the
address register a0, all 32 bits of the register are affected.
if we take the address of 'tag' =00000000000000001101101011010000
it would give us:
00000000000000001101101011010000 ( address of 'tag')
|
\|/
00000000000000000000000000000000 ( a0 register )
Bits n° 31 0
and we would get:
---------------
00000000000000001101101011010000 ( address of 'tag'
in a0 register )
Bits n° 31 0
2) SIMPLE INDIRECT addressing: (shown as (an) )
----------------------------
It's written:
-----------
+-------------------------------------+
| Instruction (an),destination |
+-------------------------------------+
OR
--
+-------------------------------------+
| Instruction source,(an) |
+-------------------------------------+
And reads as:
----------
Move the data pointed by the address register an into
(to) the destination operand.
OR
--
Move the source data to the address pointed to by an
Example:
--------
MOVE.B (a2),d2
We place the BYTE located at the address pointed to by the address register a2
in the least significant BYTE of the data register d2.
If a2 points to an address that contains the byte 01101001
We get:
........................01101001 (d2 register)
Bits n° 31 7 0
NB: note that the .B operation size is allowed for this mode
of addressing, whereas it is forbidden to move an address
into a register.
3) INDIRECT ADDRESSING WITH POST-INCREMENTATION: (schematized (an)+)
---------------------------------------------
It is written as:
-----------
+-----------------------------------+
| Instruction (an)+,destination |
+-----------------------------------+
OR
--
+-----------------------------------+
| Instruction source,(an)+ |
+-----------------------------------+
And is read as:
----------
Take the data pointed by the address register 'an', then
increment (increase) the value of 'an' according to the SUFFIX of
the instruction (by 1 for .B, 2 for .W, 4 for .L) and move
the pointed data to the destination operand.
OR
--
Move the source operand to the address pointed by the regis-
ter an, then increment the value of 'an' according to the SUFFIX
of the instruction (by 1 for .B, 2 for .W, 4 for .L)
That is to say: If you write 'MOVE.B #%10101011,(A2)+'
-------------
Place the BIT '10101011' at the address pointed by register a2,
then the address register a2 is INCREMENTED by 1 unit (.B).
If a2 points for example to the address $FFA0:
we have: 10101011
|
\|/
|--------|--------| $FF9F
$FFA0 |--------|--------| $FFA1
memory: $FFA2 |--------|--------| $FFA3
-------- |--------|--------|
register a2: 00000000000000001111111110100000 (=$ffa0)
------------
and we obtain:
|--------|--------| $FF9F
$FFA0 |10101011|--------| $FFA1
memory: $FFA2 |--------|--------| $FFA3
-------- |--------|--------|
register a2: 00000000000000001111111110100001 (=$ffa1)
------------
.The byte 10101011 is placed at $FFA0
.The address of a2 is Incremented by 8 Bits (one Byte) because the
suffix of the instruction is ' .B ', as memory is
addressable by the byte, a2 is increased by one unit.
And if we now wrote: MOVE.B #%11101010,(a2)
What would happen?
Answer: MOVE.B #%11101010,(a2) is to place the byte 11101010 at
-------- the address pointed by the register a2:
As a2 now equals $FFA1, we would obtain:
|--------|--------| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |--------|--------| $FFA3
-------- |--------|--------|
register a2: 00000000000000001111111110100001 (=$ffa1)
------------
And if I now write: MOVE #%00000001,(a2) ?
There would be an ERROR!, a2 points to $FFA1, it's an odd address,
therefore improper for placing a WORD (MOVE without suffix = MOVE.W)
And if I wrote: MOVE.B (a2)+,(a2) ?
Take the byte that a2 points to, increase a2 by one unit (.B) i.e.,
a2 points to $FFA2 and place this byte at the address pointed by
a2, i.e., at $FFA2 since a2 has just been incremented.
it gives:
|--------|--------| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |--------|--------| $FFA3
-------- |--------|--------|
register a2: 00000000000000001111111110100001 (=$ffa1)
------------
after MOVE.B (a2)+,(a2)
|--------|--------| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |11101010|--------| $FFA3
-------- |--------|--------|
register a2: 00000000000000001111111110100001 (=$ffa1)
------------
If now I change the value of a2, and a2 equals $FFA0 and if
I write: MOVE (a2)+,(a2)+
Take the WORD at $FFA0, increment a2, a2 therefore equals $FFA2 (as
MOVE = MOVE.W and a WORD=2 Bytes, as memory is addressable at
the byte level: a2=a2+2) and place it at the address pointed by
register a2 (i.e., $FFA2) then increment the address register a2 by 2
(Bytes) so a2 will finally equal $FFA4.
|--------|--------| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |11101010|--------| $FFA3
-------- $FFA4 |--------|--------|
register a2: 00000000000000001111111110100000 (=$ffa0)
------------
MOVE (a2)+,(a2)+
|--------|--------| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |10101011|11101010| $FFA3
-------- $FFA4 |--------|--------|
register a2: 00000000000000001111111110100100 (=$ffa4)
------------
4) INDIRECT ADDRESSING WITH PRE-DECREMENTATION: (schematized -(an) )
--------------------------------------------
It is written as:
----------- +------------------------------------+
| Instruction -(an),destination |
+------------------------------------+
OR
--
+------------------------------------+
| Instruction source,-(an)|
+------------------------------------+
And is read as:
----------
Decrement (decrease) the value of the address register an according to
the suffix of the instruction (by 1, 2, or 4), take the data pointed
by this new value of 'an' and place it in (to) the destination operand.
OR
--
Take the source operand, decrement the value of the register 'an' according to
the suffix of the instruction (by 1, 2, or 4) then place it in
to the address pointed by this new value of 'an'.
NB: Important, note that for this addressing mode the decrement of
--- 'an' occurs before (sign '-' before '(an)'), this is where the
name PRE-decrementation comes from.
For the addressing mode (an)+, the '+' sign is placed after '(an)',
explaining the name POST-incrementation.
This addressing mode is very similar to the post-incrementation
(an)+ mode, except that here the address register 'an' decreases in
value according to the suffix of the instruction.
Example: Consider the following memory portion:
--------
If the a2 register points to $FFA4:
|--------|--------| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |11101010|00000001| $FFA3
-------- $FFA4 |--------|--------|
register a2: 00000000000000001111111110100100 (=$ffa4)
------------
and if I write: MOVE.L -(a2),-(a2)
Decrement a2 by 4 (bytes) because we have move.L, a2 now points to
$FFA0, take the L-M located at $FFA0, decrement a2 again
by 4, a2 now points to $FF9C and move the L-M to $FF9C
resulting in:
|--------|--------| $FF9B
$FF9C |10101011|11101010| $FF9D
$FF9E |11101010|00000001| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |11101010|00000001| $FFA3
-------- $FFA4 |--------|--------|
register a2: 00000000000000001111111110011100 (=$ff9c)
------------
If now I make a2 point to $FFA0 and write:
MOVE.B #23,-(a2)
Take the byte 23 (=%00010111), decrement a2 by 1 (.B), a2
will then be $FF9F and place the byte 00010111 at this new ad-
dress.
So:
|--------|--------| $FF9F
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |11101010|00000001| $FFA3
-------- $FFA4 |--------|--------|
register a2: 00000000000000001111111110100000 (=$ffa0)
------------
MOVE.B #23,-(a2)
results in:
|--------|00010111| $FF9F (23 at $FF9F)
$FFA0 |10101011|11101010| $FFA1
memory: $FFA2 |11101010|00000001| $FFA3
-------- $FFA4 |--------|--------|
register a2: 0000000000000000111111111011111 (=$ffa0)
------------
5) INDIRECT ADDRESSING WITH DISPLACEMENT: (schematized d(an) )
--------------------------------------
It is written as:
----------- +------------------------------------+
| Instruction d(an),destination |
+------------------------------------+
And is read as:
----------
Add the value (signed) of the displacement 'd' to the contents of the
address register 'an', the data pointed by this new value of 'an' is
placed in (to) the destination operand.
So d(an)= d+(an)
'd' is a signed integer (+ if MSB is zero, - if it's active)
contained in a WORD: ( -32768 <= d < 32768 ).
Example: If the address register a3 points to $FFA0
--------
|--------|--------| $FF9F
$FFA0 |--------|--------| $FFA1
memory: $FFA2 |--------|--------| $FFA3
-------- $FFA4 |--------|--------|
register a3: 00000000000000001111111110100000 (=$ffa0)
------------
and I write: MOVE.W #458,2(a3)
Place 458=%0000000111001010 at the address $FFA0+2=$FFA2
resulting in:
|--------|--------| $FF9F
$FFA0 |--------|--------| $FFA1
memory: $FFA2 |00000001|11001010| $FFA3
-------- $FFA4 |--------|--------|
register a3: 00000000000000001111111110100000 (=$ffa0)
------------
The register a3 remains unchanged!
6) INDIRECT ADDRESSING WITH INDEX AND DISPLACEMENT: (schematized d(an,rn) )
-----------------------------------------------
It is written as:
----------- +---------------------------------------+
| Instruction d(an,rn),destination |
+---------------------------------------+
And is read as:
----------
Add the signed value of the displacement 'd' contained in a WORD and
the value of register 'rn' (address or data) to the register 'an', then
move the pointed data to (into) the destination operand.
so d(an,rn)= d+(an)+(an) or =d+(an)+(dn)
If the register rn takes the suffix .w, its low-order word participates in
the operation, if it takes the suffix .l, it participates in full. (default size .w)
Example: If the address register a2 points to $FFA0 and the data register
-------- d5 equals 122.
If I write MOVE.B #2,5(a2,d5)
Place the byte 2=%00000010 at 5+$FFA0+122=$1001F
00000010
|
\|/
|--------|--------| $FF9F
$FFA0 |--------|--------| $FFA1
memory: $FFA2 |--------|--------| $FFA3
-------- $FFA4 |--------|--------|
register a2: 00000000000000001111111110100000 (=$ffa0)
------------
register d5: 00000000000000000000000001111010 (=122)
------------
resulting in:
|--------|--------| $1001D
$1001E |--------|00000010| $1001F
memory: $10020 |--------|--------| $10021
-------- $10022 |--------|--------| $10023
register a2: 00000000000000001111111110100000 (=$ffa0)
------------
register d5: 00000000000000000000000001111010 (=122)
------------
Both registers a2 and d5 remain unchanged!
NB: ATTENTION, some Assemblers (Metacomco) do not accept the syntax
--- (an,rn) if d=0, you must write 0(an,rn) when d=0!
-----------------
PIECHOCKI Laurent
8, impasse Bellevue
57980 TENTELING continued in COURS.DOC
---------
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