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? Adam's Assembler Tutorial 1.0 ???
? ? ?
? PART V ? ?
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Revision : 1.5
Date : 15-03-1996
Contact : [[Email Removed]]
http://www.faroc.com.au/~blackcat
Note : Adam's Assembler Tutorial is COPYRIGHT, and all rights are
reserved by the author. You may freely redistribute only the
ORIGINAL archive, and the tutorials should not be edited in any
form.
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Well, another week or so seems to have gone by... Another week I should have
been using to accomplish something useful. Anyway, it seems that the
tutorials have gained a bit more popularity, which is good.
I've also received some demo code from someone who seems to have found the
tutorials of some use. Please, if you attempt something either with the help
of the tutorials or on your own, please send it to me. I like to see what
people have made of my work, or just how creative you all are. If you write
something that I think could be useful for others to learn from, or is just
pretty cool, I'll stick it up on my web site.
Note that I included a starfield demonstration in this week's tutorial just
for the hell of it. You can run STARS.EXE, or look at STARS.PAS for the full
source. It's only a simple demo, but it can be used to achieve some very
nice effects.
Now, this week we're firstly going to list a summary of all the instructions
that you should have learnt by now, and a few new ones as well. Then we'll
take a look at how the VGA is arranged, and cover a simple line routine.
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? ?
? THE INSTRUCTION SET SUMMARY ?
? ?
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? ADC <DEST>, <SOURCE> - Name: Add with Carry
Type: 8086+
Description: This instruction adds <SOURCE>
to <DEST> and adds the value stored in the
carry flag, which will be a one or a zero
to <DEST> also.
Basically, DEST = DEST + SOURCE + CF
EG: ADC AX, BX
? ADD <DEST>, <SOURCE> - Name: Add
Type: 8086+
Description: This instruction adds <SOURCE>
and <DEST>, storing the result in <DEST>.
EG: ADD AX, BX
? AND <DEST>, <SOURCE> - Name: Boolean AND
Type: 8086+
Description: This instruction performs a
bit by bit comparison of <DEST> and
<SOURCE>, storing the result in <DEST>.
EG: AND 0, 0 = 0
AND 0, 1 = 0
AND 1, 0 = 0
AND 1, 1 = 1
? BT <DEST>, <BIT NUMBER> - Name: Bit Test
Type: 80386+
Description: This instruction tests
<BIT NUMBER> of <DEST> which can either
be a 16 or 32-bit register or memory
location. If <DEST> is a 16-bit number
then <BIT NUMBER> can range from 0 - 15,
else if <DEST> is a 32-bit number, then
<BIT NUMBER> may have a value from 0 to 31.
The value held in <BIT NUMBER> of <DEST> is
then copied into the carry flag.
EG: BT AX, 3
JC WasEqualToOne
? CALL <DEST> - Name: Procedure Call
Type: 8086+
Description: This instruction simply calls
a subroutine. In more technical terms, it
pushes the address of the next instruction,
IP, onto the stack, and then sets the
instruction pointer, IP, to the value
specified by <DEST>.
EG: CALL MyProc
? CBW - Name: Convert Byte to Word
Type: 8086+
Description: This instruction extends the
byte in AL to AX.
EG: MOV AL, 01h
CBW
ADD BX, AX ; Do something with AX
? CLC - Name: Clear Carry Flag
Type: 8086+
Description: This instruction clears the
carry flag in the flags register to 0.
EG: CLC
? CLD - Name: Clear Direction Flag
Type: 8086+
Description: This instruction clears the
direction flag in the flags register to
0. When the direction flag is 0, any
string instructions increment the index
registers SI and DI.
EG: CLD
? CLI - Name: Clear Interrupt Flag
Type: 8086+
Description: This instruction clears the
interrupt flag in the flags register to
0, thus disabling hardware interrupts.
EG: CLI
? CMC - Name: Complement the Carry Flag
Type: 8086+
Description: This instruction checks the
value currently held in the carry flag.
If it is 0 - it becomes a 1 and if it is
1 - it becomes a 0.
EG: BT AX, 1 ; Test bit 1 of AX
JC WasOne
JMP Done
WasOne:
CMC ; Return CF to 0
Done:
? CMP <VALUE1>, <VALUE2> - Name: Compare Integer
Type: 8086+
Description: This instruction compares
<VALUE1> and <VALUE2> and reflects the
comparison in the flags.
EG: CMP AX, BX
See also the Jcc instructions.
? CWD - Name: Convert Word to Doubleword
Type: 8086+
Description: This instruction extends the
word in AX to the DX:AX pair.
EG: CWD
? DEC <VALUE> - Name: Decrement
Type: 8086+
Description: This instruction subtracts
one from the value held in <VALUE> and
stores the result in <VALUE>.
EG: DEC AX
? DIV <VALUE> - Name: Unsigned Division
Type: 8086+
Description: This instruction divides
<VALUE> by either AX for a byte, DX:AX for
a word or EDX:EAX for a doubleword.
For a byte, the quotient is returned in
AL and the remainder in AH, for a word the
quotient is returned in AX and the
remainder in DX and for a DWORD, the
quotient is returned in EAX and the
remainder in EDX.
EG: MOV AX, 12
MOV BH, 5
DIV BH
MOV Quotient, AL
MOV Remainder, AH
? IN <ACCUMULATOR>, <PORT> - Name: Input from I/O port
Type: 8086+
Description: This instruction reads a
value from one of the 65536 hardware ports
into the specified accumulator.
AX and AL are commonly used for input
ports, and DX is commonly used to
identify the port.
EG: IN AX, 72h
MOV DX, 3C7h
IN AL, DX
? INC <VALUE> - Name: Increment
Type: 8086+
Description: This instruction adds one to
the number held in <VALUE>, and stores
the result in <VALUE>.
EG: MOV AX, 13h ; AX = 13h
INC AX ; AX = 14h
? INT <INTERRUPT> - Name: Generate an Interrupt
Type: 8086+
Description: This instruction saves the
current flags and instruction pointer on
the stack, and then calls <INTERRUPT>
based on the value in AH.
EG: MOV AH, 00h ; Set video mode
MOV AL, 13h ; Video mode 13h
INT 10h ; Generate interrupt
? Jcc - Name: Jump if Condition
Type: 8086+
I'm not going to repeat myself for all 32 of them, just look in Tutorial
Three for the entire list of them. Bear in mind that it would be a good
idea to call CMP, OR, DEC or something similar before you use one of these
instructions. :)
EG: DEC AX
JZ AX_Has_Reached_Zero
? JMP <DEST> - Name: Jump
Type: 8086+
Description: This instruction simply
loads a new value, <DEST>, into the
instruction pointer, thus transferring
control to another part of the code.
EG: JMP MyLabel
? LAHF - Name: Load AH with Flags
Type: 8086+
Description: This instruction copies the
low bytes of the flags register into AH.
The contents of AH will look something
like the following after the instruction
has been executed:
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? Flag ? SF ? ZF ? -- ? AF ? -- ? PF ? -- ? CF ?
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? Bit ? 07 ? 06 ? 05 ? 04 ? 03 ? 02 ? 01 ? 00 ?
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You may now test the bits individually, or perform an
instruction similar to the follow to get an individual flag:
EG: LAHF
SHR AH, 6
AND AH, 1 ; AH now contains the ZF flag.
? LEA <DEST>, <SOURCE> - Name: Load Effective Address
Type: 8086+
Description: This instruction loads the
memory address that <SOURCE> resides in,
into <DEST>.
EG: I use LEA SI, Str in a procedure
of mine which puts a string on the
screen very fast.
? LOOP <LABEL> - Name: Decrement CX and Branch
Type: 8086+
Description: This instruction is a form
of the For...Do loop that exists in most
high-level languages. Basically it loops
back to a label, or memory offset, until
CX = 0.
EG: MOV CX, 12
DoSomeStuff:
;...
;...
;... This will be repeated 12 times
LOOP DoSomeStuff
? Lseg <DEST>, <SOURCE> - Name: Load Segment Register
Type: 8086+
Description: This instruction exists in
several forms. All accept the same
syntax, in which <SOURCE> specifies a
48-bit pointer, consisting of a 32-bit
offset and a 16-bit selector. The 32-bit
offset is loaded into <DEST>, and the
selector is loaded into the segment
register specified by seg.
The following forms exist:
LDS
LES
LFS * 32-bit
LGS * 32-bit
LSS
EG: LES SI, A_Pointer
? MOV <DEST>, <SOURCE> - Name: Move Data
Type: 8086+
Description: This instruction copies
<SOURCE> into <DEST>.
EG: MOV AX, 3Eh
MOV SI, 12h
? MUL <SOURCE> - Name: Unsigned Multiplication
Type: 8086+
Description: This instruction multiplies
<SOURCE> by the accumulator, which depends
on the size of <SOURCE>.
If <SOURCE> is a byte then:
* AL is the multiplicand;
* AX is the product.
If <SOURCE> is a word then:
* AX is the multiplicand;
* DX:AX is the product.
If <SOURCE> is a doubleword then:
* EAX is the multiplicand;
* EDX:EAX is the product.
Note: The flags are left in an un-touched
state except for OF and CF, which are
cleared to 0 if the high byte, word or
dword of the product is 0.
EG: MOV AL, 3
MUL 10
MOV Result, AX
? NEG <VALUE> - Name: Negate
Type: 8086+
Description: This instruction subtracts
<VALUE> from 0, resulting in a two's
complement negation of <VALUE>.
EG: MOV AX, 03h
NEG AX ; AX = -3
? NOT <VALUE> - Name: Boolean Complement
Type: 8086+
Description: This instruction inverts the
state of each bit in the operand.
EG: NOT CX
? OR <DEST>, <SOURCE> - Name: Boolean OR
Type: 8086+
Description: This instruction performs a
boolean OR operation between each bit of
<DEST> and <SOURCE>, storing the result
in <DEST>.
EG: OR 0, 0 = 0
OR 0, 1 = 1
OR 1, 0 = 1
OR 1, 1 = 1
? OUT <PORT>, <ACCUMULATOR> - Name: Output to Port
Type: 8086+
Description: This instruction outputs the
value in the accumulator to <PORT>. Using
the DX register to pass the port to OUT,
you may access up to 65,536 ports.
EG: MOV DX, 378h
OUT DX, AX
? POP <REGISTER> - Name: Pop Register
Type: 8086+
Description: This instruction pops the
current value off the stack, and places
it into <REGISTER>.
EG: POP AX
? POPA - Name: Pop All General Registers
Type: 80186+
Description: This instruction pops all
the 16-bit general purpose registers off
the stack, except for SP.
It is the same as:
POP AX
POP BX
POP CX
...
EG: POPA
? POPF - Name: Pop Stack into Flags
Type: 8086+
Description: This instruction pops the
low byte of the flags off the stack.
EG: POPF
? PUSH <REGISTER> - Name: Push Register
Type: 8086+
Description: This instruction pushes
<REGISTER> onto the stack.
EG: PUSH AX
? PUSHA - Name: Push All General Registers
Type: 80186+
Description: This instruction pushes all
16-bit general purpose registers onto the
stack.
It is the same as:
PUSH AX
PUSH BX
PUSH CX
...
EG: PUSHA
? PUSHF - Name: Push Flags
Type: 8086+
Description: This instruction pushes the
low byte of the flags of the stack.
EG: PUSHF
? REP - Name: Repeat String Prefix
Type: 8086+
Description: This instruction will repeat
the following instructing for the number
of times specified in the CX register.
EG: MOV CX, 6
REP STOSB ; Store 6 bytes
? RET - Name: Near Return from Subroutine
Type: 8086+
Description: This instruction returns IP
to the value it had held before the
last CALL instruction. RET, or RETF for a
far jump, must be called when using
stand alone assembler.
EG: RET
? ROL <DEST>, <VALUE> - Name: Rotate Left
Type: 8086+
Description: This instruction rotates
<DEST> <VALUE> times. A rotation is
achieved by shifting <DEST> once, then
transferring the bit shifted off the high
end to the low-order position of <DEST>.
EG: ROL AX, 3
? ROR <DEST>, <VALUE> - Name: Rotate Right
Type: 8086+
Description: This instruction rotates
<DEST> <VALUE> times. A rotation is
achieved by shifting <DEST> once, and
transferring the bit shifted off the low
end to the high-order position of <DEST>.
EG: ROR BX, 5
? SAHF - Name: Store AH in Flags
Type: 8086+
Description: This instruction loads the
contents of the AH register into bits
7, 6, 4, 2 and 0 of the flags register.
EG: SAHF
? SBB <DEST>, <SOURCE> - Name: Subtract with Borrow
Type: 8086+
Description: This instruction subtracts
<SOURCE> from <DEST>, and decrements
<DEST> by one if the carry flag is set,
storing the result in <DEST>.
Basically, <DEST> = <DEST> - <SOURCE> - CF
EG: SBB AX, BX
? SHL <DEST>, <VALUE> - Name: Shift Left
Type: 8086+
Description: This instruction shifts <DEST>
left by <VALUE>. I'm not going to go into
the theory behind shifts again. If you
are unsure as to what this instruction
does, please refer to Tutorial Four.
EG: SHL AX, 5
? SHR <DEST>, <VALUE> - Name: Shift Right
Type: 8086+
Description: This instruction shifts <DEST>
right by <VALUE>. Please refer to
Tutorial Four for the theory behind shifts.
EG: SHR DX, 1
? STC - Name: Set Carry Flag
Type: 8086+
Description: This instruction assigns the
value of the carry flag to one.
EG: STC
? STD - Name: Set Direction Flag
Type: 8086+
Description: This instruction sets the
value of the carry flag to one. This
instructs all string operations to
decrement the index registers.
EG: STD
REP STOSB ; DI is being decremented
? STI - Name: Set Interrupt Flag
Type: 8086+
Description: This instruction sets the
value of the interrupt flag to one, thus
allowing hardware interrupts to occur.
EG: CLI ; Stop interrupts
... ; Perform crucial function
STI ; Enable interrupts
? STOS - Name: Store String
Type: 8086+
Description: This instruction exists in the
following forms:
STOSB - Store a byte - AL
STOSW - Store a word - AX
STOSD - Store a doubleword - EAX
The instructions write the current contents
of the accumulator to the memory location
pointed to by ES:DI. It then increments
or decrements DI according to the operand
used, and the value in the direction flag.
EG: MOV AX, 0A000h
MOV ES, AX
MOV AL, 03h
MOV DI, 0
STOSB ; Store 03 at ES:DI,
; which just happens
; to be at the top of the screen in
; mode 13h
? SUB <DEST>, <SOURCE> - Name: Subtract
Type: 8086+
Description: This instruction subtracts
<SOURCE> from <DEST>, storing the result
in <DEST>.
EG: SUB ECX, 12
? TEST <DEST>, <SOURCE> - Name: Test Bits
Type: 8086+
Description: This instruction performs a
bit-by-bit AND operation on <SOURCE> and
<DEST>. The result is reflected in the
flags, and they are set as the would be
after an AND operation.
EG: TEST AL, 0Fh ; Check to see if any
; bits set in the low
; nibble of AL
? XCHG <VALUE1>, <VALUE2> - Name: Exchange
Type: 8086+
Description: This instruction exchanges the
values in <VLAUE1> and <VALUE2>.
EG: XCHG AX, BX
? XOR <DEST>, <SOURCE> - Name: Exclusive Boolean OR
Type: 8086+
Description: This instruction performs a
bit-by-bit exclusive OR operation on
<SOURCE> and <DEST>. The operation is
defined as follows:
XOR 0, 0 = 0
XOR 0, 1 = 1
XOR 1, 0 = 1
XOR 1, 1 = 0
EG: XOR AX, BX
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Phew! What a lot there are, and we only covered the basic ones! You are
not expected to understand each and every one of them though. You probably
saw words like 'Two's Complement', and thought - "What the hell does that
mean?".
Do not worry about them for now. We'll continue at our usual pace, and
introduce the new instructions above one by one, explaining them as we go. If
you already understand them now, then this is an added bonus. You will also
notice that there were a lot of 8086 instructions above. There are actually
very few instances where it is necessary to use a 386 or 486 instruction,
let alone Pentium instructions.
Anyway, before we press on with the VGA, I'll just list the speed at which
each of the above instructions execute at, so you can use this to gauge how
fast your Assembler routines are.
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Instruction 386 Clock Ticks 486 Clock Ticks
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ADC 2 1
ADD 2 1
AND 2 1
BT 3 3
CALL 7+m 3
CBW 3 3
CLC 2 2
CLD 2 2
CLI 5 3
CMC 2 2
CMP 2 1
CWD 2 3
DEC 2 1
DIV - -
- Byte 9-14 13-18
- Word 9-22 13-26
- DWord 9-38 13-42
IN 12/13 14
INC 2 1
INT depends depends
Jcc - -
- Branch 7+m 3
- No Branch 3 1
JMP 7+m 3
LAHF 2 3
LEA 2 1
LOOP 11 6
Lseg 7 6
MOV 2 1
MUL - -
- Byte 9-14 13-18
- Word 9-22 13-26
- DWord 9-38 13-42
NEG 2 1
NOT 2 1
OR 2 1
OUT 10/11 16
POP 4 1
POPA 24 9
POPF 5 9
PUSH 2 1
PUSHA 18 11
PUSHF 4 4
REP depends depends
RET 10+m 5
ROL 3 3
ROR 3 3
SAHF 3 2
SBB 2 1
SHL 3 3
SHR 3 3
STC 2 2
STD 2 2
STI 3 5
STOS 4 5
SUB 2 1
TEST 2 1
XCHG 3 3
XOR 2 1
Note: m = Number of components in next instruction executed.
Ugh, I never want to see another clock-tick again! Now, on with the fun stuff
- the VGA!
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You've probably noticed by now that your video card has more than 256K of RAM.
(If you haven't, then these tutorials are probably not for you.) Even if
you have only 256K of RAM, like my old 386, you'll still be able to get
into mode 13h - 320x200x256. However, this raises some questions.
Multiply 320 by 200 and you'll notice that you only need 64,000 bytes of
memory to store a single screen. (The VGA actually gives us 64K, which is
65,536 bytes for the unaware.) What happened to the remaining 192K or so?
Well, the VGA is actually arranged in bitplanes, like this:
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?????????2???????? ?
?????????1???????? ? ?
?????????0???????? ? ? ?
? ? ? ? ?
? ? ? ? ?
? 64,000 ? ? ???
? ? ???
? ???
??????????????????
Each plane being 64,000 bytes long. Here's how it works:
A pixel at 0, 0 is mapped in plane 0 at offset 0;
A pixel at 1, 0 is mapped in plane 1 at offset 0;
A pixel at 2, 0 is mapped in plane 2 at offset 0;
A pixel at 3, 0 is mapped in plane 3 at offset 0;
A pixel at 4, 0 is mapped in plane 0 at offset 1 ... and so on ...
Because of the pixels being chained across all four planes, it is impossible
to use multiple pages in mode 13h without having to resort to using a
virtual screen, or something similar.
The automatic mapping of the pixels is handled completely by the video card,
so you can blindly work away without even knowing about the four bitplanes if
you wish.
We'll go onto how we can get around this, by entering a special display mode,
known as Mode X, later, but for now, let's just see what we can do in plain
old mode 13h.
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????????????????????????????????????????????????????????????
? ?
? DRAWING LINES ?
? ?
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We've gone a little over the size that I'd planned to go to for this tutorial,
and I had intended to cover Bresenham's Line Algorithm, but that'll have to
wait till next week. However, I will cover how to draw a simple horizontal
line in Assembler.
An Assembler Horizontal Line Routine:
---------------------------------------
First we'll need to point ES to the VGA.
This should do the trick:
MOV AX, 0A000h
MOV ES, AX
Now, we'll need to read the X1, X2 and Y values into registers, so something
like this should work:
MOV AX, X1 ; AX now equals the X1 value
MOV BX, Y ; BX now equals the Y value
MOV CX, X2 ; CX now equals the X2 value
It will be necessary to work out how long the line is, so we'll use CX to
store this, seeing as: i) CX already holds the X2 value, and ii) we'll be
using a REP instruction, which will use CX as a counter.
SUB CX, AX ; CX = X2 - X1
Now we'll need to work out what DI will be for the very first pixel we'll be
plotting, so we'll use what we did in the PutPixel routine:
MOV DI, AX ; DI = X1
MOV DX, BX ; DX = Y
SHL BX, 8 ; Shift Y left 8
SHL DX, 6 ; Shift Y left 6
ADD DX, BX ; DX = Y SHL 8 + Y SHL 6
ADD DI, DX ; DI = Y x 320 + X
We have the offset of the first pixel now, so all we have to do is put the
color we want to draw in, in AL, and use STOSB to plot the rest of the line.
MOV AL, Color ; Move the color to plot with into AL
REP STOSB ; Plot CX pixels
Note that we used STOSB because it will increment DI for us, thus saving
a lot of MOV's and INC's. Now, depending on what language you'll use to
implement this in, you'll get something like:
void Draw_Horizontal_Line(int x1, int x2, int y, char color);
{
asm {
mov ax, 0xa000
mov es, ax ; Point ES to the VGA
mov ax, x1 ; AX = X1
mov bx, y ; BX = Y
mov cx, x2 ; CX = X2
sub cx, ax ; CX = Difference of X2 and X1
mov di, ax ; DI = X1
mov dx, bx ; DX = Y
shl bx, 8 ; Y SHL 8
shl dx, 6 ; Y SHL 6
add dx, bx ; DX = Y SHL 8 + Y SHL 6
add di, dx ; DI = Offset of first pixel
mov al, color ; Put the color to plot in AL
rep stosb ; Draw the line
}
}
????????????????????????????????????????????????????????????????????????????
We'll now we've covered how to draw a simple horizontal line. The above
routine isn't blindingly fast, but it isn't all that bad either. Just
changing the calculation of DI part to look like the fast PutPixel I gave out
in Tutorial Two would probably double the speed of this routine.
My own horizontal line routine is probably about 4 to 5 times as fast as this
one, so in the future, I'll show you how to optimize this one fully. Next
week we'll also cover how to get and set the palette, and how we can draw
circles. I'm sorry it didn't make it into this tutorial, but this one sort of
grew a bit...
THINGS TO DO:
---------------
1) Write a vertical line routine based on the one above. Clue: You'll
need to increment DI by 320 at some stage.
2) Go over the list of Assembler instructions, and learn as many as you
can.
3) Have a look at the Starfield I wrote, and try to fix the bugs in it.
See what you can do with it.
????????????????????????????????????????????????????????????????????????????
Sorry again that I didn't include the things I said I would last week, but
as I said, the tutorial just grew, and I'm a bit behind with some other
projects I'm supposed to be working on.
Next week's tutorial _will_ include:
? Line algorithms and examples;
? A circle algorithm;
? The palette;
? Something else that you ought to know...
If you wish to see a topic discussed in a future tutorial, then mail me, and
I'll see what I can do.
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Don't miss out!!! Download next week's tutorial from my homepage at:
? http://www.faroc.com.au/~blackcat
See you next week!
- Adam.
