[Back to TUTOR SWAG index]  [Back to Main SWAG index]  [Original]

A introduction to x86 assembly
Written by Gavin Estey		          email: gavin@senator.demon.co.uk        
This document is version: 

0.9 Beta 2/19/95

This will hopefully be the final version of this unless somebody finds some 
mistakes in it. I have spent over 620 minutes creating this document (not 
including writing or testing the code or the original text document) so I hope 
you will find it useful.

This version is still in testing which means that I cannot be sure that all the code 
will compile and work correctly. I have done my best to make sure that all the 
information contained in this document is correct. If any mistakes are found 
please could you notify me be email.

If someone who has a computer that doesnt have a VGA card/monitor could you 
please compile the code on P29 that checks what graphics the computer can handle 
and tell me if it returned DX as 0. 

This document was written by Gavin Estey. No part of this can be reproduced and sold 
in any commercial product without my written permission (email me for more 
information). I am not responsible for any damage caused in anyway from this 

If you enjoy this document then a donation would be gratefully received. No payment 
is necessary but none will turned down. If any mistakes are noticed then I would like to 
be notified. If you would like any other help or have suggestions for a later version of 
this document.	     

Firstly you will need a suitable assembler to compile your programs. All examples have 
been tested with two compilers: A86 and Turbo Assemblerâ  ver 4. A86 is a very 
good shareware assembler capable of producing code up to 80286. This can be found 
on your local simtel mirror (try ftp.demon.co.uk or ftp.cdrom.com) under 
simtel/msdos/asmutil called A86V322.ZIP. 

Firstly I am going to talk about SEGMENTS and OFFSETS. This (for me anyway) is 
probably the hardest part of assembly to understand. 

Segments and offsets:

The original designers of the 8088 decided that nobody will every need to use more 
that one megabyte of memory. So they built the chip so it couldn't access above that. 
The problem is to access a whole megabyte 20 bits are needed (one bit being either a 
one or a zero). Registers only have 16 bits and they didn't want to use two because 
that would be 32 bits and they thought that this would be too much for anyone. They 
decided to do the addressing with two registers but not 32 bits. Is this confusing? 
Blame the designers of the 8088.

SEGMENT = OFFSET / 16 (the lower 4 bits are lost)

SEGMENT * 16	|0010010000010000 - ---|  range (0 to 65535) * 16
OFFSET         		|-- - -0100100000100010|  range (0 to 65535)
20 bit address		|00101000100100100010|  range 0 to 1048575 (1 MEG)
		             =====  DS =====
		                      ====== SI ===== 

(note DS and SI overlap). This is how DS:SI is used to make a 20 bit address. The 
segment is in DS and the offset is in SI.
Segment registers are: CS, DS, ES, SS. On the 386+ there are also FS & GS

Offset registers  are: BX, DI, SI, BP, SP, IP.  In 386+ protected mode, ANY general 
register (not a segment register) can be used as an  Offset register.  (Except IP, which 
you can't access.)


AX, BX, CX and DX are general purpose registers. On a 386+ they are 32 bits; EAX, 
EBX, ECX and EDX. The all can be split up into high and low parts, each being 8 bits.

EAX 	32 bits
AX 	16 bits   
AH	8 bits
AL	8 bits

This means that you can use AH and AL to store different numbers and treat them as 
separate registers for some tasks.

BX (BH/BL): same as AX         
CX (CH/CL): Same as AX (used for loops)
DX (DH/DL): Same as AX (used for multiplication/division)

DI and SI are index registers and can be used as offset registers.

SP: Stack Pointer. Does just that. Points to the current position in the stack. Don't 
alter unless you REALLY know what you are doing or want to crash your computer.

The Stack:

This is and area of memory that is like a stack of plates. The last one you put on is the 
first one that you take off (LOFO). If another piece of data is put on the stack it grows 
Diagram 1: This shows how the stack is organised 


Push puts a value onto the stack and pop takes it back off. Here's some code. (you 
can't compile this - yet.)

	push cx         ;put cx on the stack
	push ax         ;put ax on the stack
	.               ;this is going to pop them back in 
	.			 ;the wrong so they are reversed
	pop cx          ;put value from stack into cx
	pop ax          ;put value from stack into ax

This means that cx is equal to what ax is and ax is equal to what cx was.

Some instructions:

MOV: this moves a value from one place to another.

MOV destination, source

for example:
	MOV ax,10       ;moves an immediate value into ax 
	MOV bx,cx       ;moves value from cx into bx

INT: calls a DOS or BIOS function, mainly to do something you would rather not 
write a function for e.g. change video mode, open a file etc.

INT interrupt number

Note: an h after a number denotes that that number is in hex.

	INT 21h         ;Calls DOS standard interrupt 21h
	INT 10h         ;Calls the Video BIOS interrupt

All interrupts need a value to specify what subroutine to use. This is usually in AH. To 
print a message on the screen all you need to do is this:

	MOV ah,9      ;subroutine number 9
	INT 21h       ;call the interrupt

But first you have to specify what to print. This function needs DS:DX to be a far 
pointer to where the string is. The string has to be terminated with a dollar sign ($). 

This example shows how it works:
	MyMessage db    "This is a message!$"
	mov dx,OFFSET MyMessage
	mov ax,SEG MyMessage
	mov ds,ax
	mov ah,9
	int 21h

DB declares a piece of data. This is the same as in BASIC:
	LET MyMessage$ = "This is a message!"

DB declares a byte, DW declares a word and DD declares a Dword.

This is your first assembly program. Cut this in and then assemble it. If you are using 
A86, type: 
	A86 name of file
If you are using Turbo Assembler then type:
	tasm name of file
	tlink name of file minus extention
With tlink you can make a .COM file by using a /t switch.

	.MODEL SMALL ;Ignore the first two lines
			   ;I'll explain what this means later.
	.STACK	   ;allocate a stack
	.CODE        ;a directive to start code segment
	START:       ;generally a good name to use as an 	
			   ;entry point

	JMP BEGIN    ;jump to BEGIN: (like goto)

	MyMessage db "This is a message!$" 
	BEGIN:       ;code starts here

	mov dx,OFFSET MyMessage	;put the offset of message 
						;in DX
	mov ax,SEG MyMessage	;put the segment that the 
						;message is in in AX
	mov ds,ax				;copy AX into DS
	mov ah,9				;move 9 into ah to call
	int 21h				;function 9 of interrupt 
						;21h - display a string to 

	MOV ax,4c00h ;move the value of 4c00 hex into AX
	INT 21h      ;call interrupt 21h, subroutine 4C 
;which returns control to dos, otherwise it would of 
;crashed. The 00 on the end means what errorlevel it 
;would return to DOS. This can be checked in a batch file
	END START    ;end here

Some instructions that you need to know:

This is just a list of some basic assembly instructions that are very important and are 
used often.

ADD	Add the contents of one number to another


	ADD operand1,operand2

This adds operand2 to operand1. The answer is stored in operand1. Immediate data 
cannot be used as operand1 but can be used as operand2.

SUB	Subtract one number from another

	SUB operand1,operand2

This subtracts operand2 from operand1. Immediate data cannot be used as operand1 
but can be used as operand2.

MUL	Multiplies two unsigned integers (always positive)
IMUL	Multiplies two signed integers (either positive or negitive)

	MUL register or variable
	IMUL register or variable

This multiples the register given by the number in AL or AX depending on the size of 
the operand. The answer is given in AX. If the answer is bigger than 16 bits then the 
answer is in DX:AX (the high 16 bits in DX and the low 16 bits in AX). 

On a 386, 486 or Pentium the EAX register can be used and the answer is stored in 

DIV	Divides two unsigned integers(always positive)
IDIV 	Divides two signed integers (either positive or negitive)

	DIV register or variable
	IDIV register or variable

This works in the same way as MUL and IMUL by dividing the number in AX by the 
register or variable given. The answer is stored in two places. AL stores the answer 
and the remainder is in AH. If the operand is a 16 bit register than the number in 
DX:AX is divided by the operand and the answer is stored in AX and remainder in 

The way we entered the address of the message we wanted to print was a bit 
cumbersome. It took three lines and it isnt the easiest thing to remember

	mov dx,OFFSET MyMessage						
	mov ax,SEG MyMessage						
	mov ds,ax		

We can replace all this with just one line. This makes the code easier to read and it 
easier to remember. This only works if the data is only in the data is in one segment i.e. 
small memory model.

	lea dx,MyMessage
or	mov dx,OFFSET MyMessage
Using lea is slightly slower and results in code which is larger.

LEA means Load Effective Address.

LEA destination,source

Desination can be any 16 bit register and the source must be a memory operand (bit of 
data in memory). It puts the offset address of the source in the destination.

Keyboard input:

We are going to use interrupt 16h, function 00h to read the keyboard. This gets a key 
from the keyboard buffer.  If there isn't one, it waits until there is. It returns the SCAN 
code in AH and the ASCII translation in AL.

        MOV ah,00h		;function 00h of
   INT 16h		;interrupt 16h

All we need to worry about for now is the ascii value which is in al.

Printing a character:

The problem is that we have the key that has been pressed in ah. How do we display 
it? We can't use function 9h because for that we need to have already defined the string 
which has to end with a dollar sign. This is what we do instead:

;after calling function 00h of interrupt 16h
        MOV dl,al       ;move al (ascii code) into dl
        MOV ah,02h      ;function 02h of interrupt 21h
        INT 21h         ;call interrupt 21h

If you want to save the value of ah then push it before and pop it afterwards.

Control flow:
Firstly, the most basic command:
	JMP label 
This is the same as GOTO in basic.

        	JMP ALabel
     	;code to do something

What do we do if we want to compare somthing. We have just got a key from the user 
but we want to do something with it. Lets print something out if it is equal to 
somethine else. How do we do that? Its easy. We use the jump on condition 
commands. Here is a list of them:

Jump on condition commands:

jumps if the first number was above the second number

same as above, but will also jump if they are equal

jumps if the first number was below the second

same as above, but will also jump if they are equal

jumps if the first number was NOT above  (same as JBE)

jumps if the first number was NOT above or the same as (same as JB)

jumps if the first number was NOT below (same as JAE)

jumps if the first number was NOT below or the same as (same as JA)

jumps if the two numbers were equal

same as JZ, just a different name

jumps if the two numbers are NOT equal

same as above

[NOTE: There are quite a few more but these are the most useful. If you want the full 
list then get a good assembly book]

They are very easy to use.

        CMP register containing value, a value
    jump command destination    

An example of this is:
      cmp al,'Y'      ;compare the value in al with Y
      je ItsYES       ;if it is equal then jump to ItsYES

The following program is an example of how to use control and input and output.

.STACK 					;define a stack
Start:					;a good place to start.
      lea dx,StartUpMessage  	;display a message on th e
      mov ah,9                ;using function 09h
      int 21h                 ;of interrupt 21h
      lea dx,Instructions     ;display a message on the 
      mov ah,9                ;using function 09h
      int 21h                 ;of interrupt 21h
	 mov ah,00h              ;function 00h of
      int 16h                 ;interrupt 16h gets a 	
						;character from user     
      mov bl,al               ;save bl    
      mov dl,al               ;move ascii value of key 
				        	;pressed to dl
	 mov ah,02h              ;function 02h of 
      int 21h                 ;interrupt 21h displays a 
				          ;character to screen
      cmp bl,'Y'              ;is al=Y?
      je Thanks    		     ;if yes then goto Thanks
 	 cmp bl,'y'              ;is al=y?
      je Thanks          	;if yes then goto Thanks
      jmp TheEnd
      lea dx,ThanksMsg        ;display message
      mov ah,9                ;using function 9
      int 21h                 ;of interrupt 21h
      lea dx,GoodBye          ;print goodbye message
      mov ah,9                ;using function 9
      int 21h                 ;of interrupt 21h
      mov AX,4C00h            ;terminate program and 	
						;return to DOS using
      INT 21h                 ;interrupt 21h function 4CH

;0Dh,0Ah adds a enter at the beginning

StartUpMessage DB "A Simple Input Program$"     
Instructions  DB 0Dh,0Ah,"Just press a Y to continue...$                  
ThanksMsg      DB 0Dh,0Ah,"Thanks for pressing Y!$"    
GoodBye        DB 0Dh,0Ah,"Have a nice day!$"



Assembly, like C and Pascal can have procedures. These are very useful for series of 
commands that have to be repeated often. 

This is how a procedure is defined:

PROC AProcedure
    .          ;some code to do something       
    RET		;if this is not here then your computer 
			;will crash
ENDP AProcedure

You can specify how you want the procedure to be called by adding a FAR or a 
NEAR after the procedure name. Otherwise it defaults to the memory model you are 
using. For now you are better off not doing this until you become more experianced. I 
usually add a NEAR in as compilers cant decide between a near and a far very well. 
This means if the jump needs to be far the compiler will warn you and you can change 

This is a program which uses a procedure.

;a simple program with a procedure that prints a message 
;onto the screen. (Use /t switch with tlink). Should 
;display Hello There! on the screen.

ORG     100h

	   JMP Start 		  ;skip the data
HI      DB "Hello There!$" ;define a message
Start:				  ;a good place to start.
        Call Display_Hi    ;Call the procedure
        MOV AX,4C00h       ;terminate program and return 
					  ;to DOS using
        INT 21h            ;interrupt 21h function 4Ch

Display_Hi PROC            ;Defines start of procedure
        lea dx,HI          ;put offset of message into DX
        mov ah,9           ;function 9 of 
        int 21h            ;interrupt 21h
        RET                ;THIS HAS TO BE HERE
Display_Hi ENDP            ;Defines end of procedure

Main    ENDP
        END MAIN

Memory Models:

We have been using the .MODEL directive to specify what type of memory model we  
use, but what does this mean?

	.MODEL MemoryModel

Where MemoryModel can be SMALL, COMPACT, MEDIUM, LARGE, HUGE, 

This means that there is only one segment for both code and data. This type of 
program can be a .COM file.

This means that by default all code is place in one physical segment and likewise all 
data declared in the data segment is also placed in one physical segment. This means 
that all proedures and variables are addressed as NEAR by pointing at offsets only.

This means that by default all elements of code (procedures) are placed in one physical 
segment but each element of data can be placed in its own physical segment. This 
means that data elements are addressed by pointing at both at the segment and offset 
addresses. Code elements (procedures) are NEAR and varaibles are FAR.

This is the opposite to compact. Data elements are near and procedures are FAR.

This means that both procedures and variables are FAR. You have to point at both the 
segment and offset addresses.

This isnt used much as it is for 32 bit unsegmented memory space. For this you need a 
dos extender. This is what you would have to use if you were writing a program to 
interface with a C/C++ program that used a dos extender such as DOS4GW or 


(All code examples given are for macros in Turbo Assembler. For A86 either see the 
documentation or look in the A86 macro example later in this document).

Macros are very useful for doing something that is done often but for which a 
procedure cant be use. Macros are substituted when the program is compiled to the 
code which they contain.

This is the syntax for defining a macro:

Name_of_macro	macro		
;a sequence of instructions 

These two examples are for macros that take away the boring job of pushing and 
popping certain registers:

	SaveRegs macro
		pop ax
		pop bx
		pop cx
		pop dx
	RestoreRegs macro
		pop dx
		pop cx
		pop bx
		pop ax
Note that the registers are popped in the reverse order to they wer popped.

To use a macro in you program you just use the name of the macro as an ordinary 
macro instruction:

	;some other instructions

This example shows how you can use a macro to save typing in. This macro simply 
prints out a variable to the screen.

The only problems with macros is that if you overuse them it leads to you program 
getting bigger and bigger and that you have problems with multiple definition of labels 
and variables. The correct way to solve this problem is to use the LOCAL directive for 
declaring names inside macros.

	LOCAL name 
Where name is the name of a local variable or label.

If you have comments in a macro everytime you use that macro the comments will be 
added again into your source code. This means that it will become unesescarily long. 
The way to get round this is to define comments with a ;; instead of a ;. This example 
illustrates this.

	;a normal comment
	;;a comment in a macro to save space

Macros with parameters:
Another useful property of macros is that they can have parameters. The number of 
parameters is only restricted by the length of the line. 

Name_of_Macro macro par1,par2,par3
;commands go here

This is an example that adds the first and second parameters and puts the result in the 

	AddMacro macro num1,num2,result
		push ax		;save ax from being destroyed
		mov ax,num1	;put num1 into ax
		add ax,num2	;add num2 to it
		mov result,ax	;move answer into result
		pop ax		;restore ax

On the next page there is an example of some a useful macro to exit to dos with a 
specified . There are two versions of this program because both A86 and Turbo 
Assembler handle macros differently.

;this is a simple program which does nothing but use a 
;macro to exit to dos. This will only work with tasm and 
;should be compiled like this: tasm /m2 macro.asm 
;because tasm needs more than one pass to work out what 
;to do with macros. It does cause a warning but this can 
;be ignored.

.MODEL small 
.CODE        ;start the code segment                                  

Start:       ;a good a place as any to start this

;now lets go back to dos with another macro
	BackToDOS 0     ;the errorlevel will be 0

BackToDOS macro errorlevel  

;;this is a macro to exit to dos with a specified 
;;errorlevel given. No test is done to make sure that a 
;;procedure is actually passed or it is within range.
	mov ah,4Ch          ;;terminate program and return                              
					;;to DOS using
	mov al,errorlevel   ;;put errorlevel into al        
	int 21h             ;;interrupt 21h function 4Ch 

	endm				;;end macro
end					;end program

Procedures that pass parametres:
Procedures can be made even more useful if they can be made to pass parameters to 
each other. There are three ways of doing this and I will cover all three methods:
  In registers,
  In memory,
  In stack.

In registers:

Advantages:		Easy to do and fast.
Disadvantages: 	The is not many registers.

This is very easy to do, all you have to do is to is move the parameters into registers 
before calling the procedure. This example adds two numbers together and then 
divides by the third it then returns the answer in dx.

	push ax		;save value of ax
	push bx		;save value of bx
	push cx		;save value of cx
	mov ax,10		;first parameter is 10			
	mov bx,20		;second parameter is 20
	mov cx,3		;third parameter is 3
	Call ChangeNumbers ;call procedure
	pop cx		;restore cx
	pop bx		;restore bx
	pop ax		;restore dx 
ChangeNumbers PROC	;Defines start of procedure
	add ax,bx		;adds number in bx to ax
	div cx		;divides ax by cx
	mov dx,ax		;return answer in dx
ChangeNumbers ENDP	;defines end of procedure

How to use a debugger:

This is a good time to use a debugger to find out what your program is actually doing. 
I am going to demonstrate how to use Turbo Debugger to check if this program is 
working properly. First we need to compile this program to either a .EXE or .COM 
file. Then type:
	td name of file
Turbo Debugger then loads. You can see the instructions that make up your programs,  
for example the first few lines of this program is shown as:
	cs:0000 50		push   ax
	cs:0001 53		push   bx
	cs:0002 51		push   cx
(This might be slighly different than is shown on your screen but hopefully you will get 
the main idea)
This diagram shows what the Turbo Debuggerâ  screen looks like

The numbers that are moved into the registers are different that the ones that we typed 
in because they are represented in their hex form (base 16) as this is the easiest base to 
convert to and from binary and that it is easier to understand than binary also. 

At the left of this display there is a box showing the contents of the registers. At this 
time all the main registers are empty. Now press F7 this means that the first line of the 
program is run. As the first line pushed the ax register into the stack, you can see that 
the stack pointer (SP) has changed. Press F7 until the line which contains mov 
ax,000A is highlighted. Now press it again. Now if you look at the box which contains 
the contents of the registers you can see that AX contains A. Press it again and BX 
now contains 14, press it again and CX contains 3. Now if you press F7 again you can 
see that AX now contains 1E which is A+14. Press it again and now AX contains A 
again, A being 1E divided by 3 (30/3 = 10). Press F7 again and you will see that DX 
now also contains A. Press it three more times and you can see that CX,BX and AX 
are all set back to their origional values of zero.

Passing through memory:

Advantages:		Easy to do and can use more parameters
Disadvantages: 	Can be slower 

To pass parameters through memory all you need to do is copy them to a variable 
which is stored in memory. You can use a varable in the same way that you can use a 
register but commands with registers are a lot faster. This table shows the timing for 
the MOV command with registers and then variables and then the amount of clock 
cycles (the speed - smaller faster) it takes to do them. 

KEY: 	reg8 means an 8 bit register eg AL
	mem8 means an 8 bit variable declared with DB
	reg16 means an 16 bit register eg AX
	mem16 means an 16 bit variable declared with DW
	imm8 means an immediate byte eg MOV al,8
	imm16 means an immediate word eg MOV ax,8


MOV reg/mem8,reg8

MOV reg,mem16,reg16

MOV reg8,reg/mem8

MOV reg16,reg/mem16

MOV reg8,imm8

MOV reg16,imm16

MOV reg/mem8,imm8

MOV reg/mem16,imm16

These timings are taking from the Borlandâ Turbo Assemblerâ Quick Reference

This shows that on the 8086 using variables in memory can make the instuction four 
times as slow. This means that you should avoid using too many variables 
unnecessarily. On the 486 it doesnt matter as both instructions take the same amount 
of time.

The method actually used is nearly identical to passing parameters in registers. This 
example is just another version of the example given for passing in registers.

FirstParam  db 0  ;these are all variables to store
SecondParam db 0  ;the parameters for the program	
ThirdParam  db 0
Answer	  db 0
	mov FirstParam,10	;first parameter is 10		
	mov SecondParam,20	;second parameter is 20
	mov ThirdParam,3	;third parameter is 3
	Call ChangeNumbers
ChangeNumbers PROC	     ;Defines start of procedure
	push ax			;save ax
	push bx			;save bx
	mov ax,FirstParam	;copy FirstParam into ax
	mov bx,SecondParam	;copy SecondParam into bx
	add ax,bx			;adds number in bx to ax
	mov bx,ThirdParam	;copy ThirdParam into bx 
	div bx			;divides ax by bx
	mov Answer,ax		;return answer in Answer
	pop bx			;restore bx
	pop ax			;restore ax
ChangeNumbers ENDP		;defines end of procedure
This way may seem more complicated but it is not really suited for small numbers of 
this type of parameters. It is much more useful when dealing with strings or large 
numbers of big values.

Passing through Stack:

This is the most powerful and flexible method of passing parameters. This example 
shows the ChangeNumbers procedure that has been rewritten to pass its parameters 
through the stack. 
	mov ax,10		;first parameter is 10			
	mov bx,20		;second parameter is 20
	mov cx,3		;third parameter is 3
	push ax		;put first parameter on stack
	push bx		;put second parameter on stack
	push cx		;put third parameter on stack
	Call ChangeNumbers
ChangeNumbers PROC	;Defines start of procedure
	push bp
	mov bp,sp
	mov ax,[bp+8]	;get the parameters from bp
	mov bx,[bp+6]  ;remember that first it is last out
	mov cx,[bp+4]	;so number is larger
	add ax,bx		;adds number in bx to ax
	div cx		;divides ax by cx
	mov dx,ax		;return answer in dx
ChangeNumbers ENDP	;defines end of procedure

This diagram shows the contents of the stack for a program with two parameters:

To get a parameter from the stack all you need to do is work out where it is. The last 
parameter is at BP+2 and then the next and BP+4.

Files and how to used them:

Files can be opened, read and written to. DOS has some ways of doing this which save 
us the trouble of writing our own routines. Yes, more interrupts. Here is a list of 
helpful functions of interrupt 21h that we are going to need to use for our simple file 

Function 3Dh: open file
Opens an existing file for reading, writing or appending on the specified drive and 

        AH = 3Dh
        AL = access mode
             bits 0-2   000 = read only
                        001 = write only
                        010 = read/write
             bits 4-6   Sharing mode (DOS 3+)
                        000 = compatibility mode
                        001 = deny all (only current program can access file)
                        010 = deny write (other programs can only read it)
                        011 = deny read (other programs can only write to it)
                        100 = deny none 
        DS:DX = segment:offset of ASCIIZ pathname

        CF = 0 function is succesful
        AX = handle
        CF = 1 error has occured
        AX = error code
                01h missing file sharing software
                02h file not found
                03h path not found or file does not exist
                04h no handle available
                05h access denied
                0Ch access mode not permitted

What does ASCIIZ mean? An ASCIIZ string is a ASCII string with a zero on the end 
(instead of a dollar sign).

Important: Remember to save the file handle it is needed for later.

How to save the file handle:
It is important to save the file handle because this is needed to do anything with the 
file. Well how is this done? There are two methods, which is better? 

  Copy the file handle into another register and don't use that register.
  Copy it to a variable in memory.

The disadvantages with the first method is that you will have to remember not to use 
the register you saved it in and it wastes a register that can be used for something more 
useful. We are going to use the second. This is how it is done:

	FileHandle DW 0     ;use this for saving the file 

     MOV FileHandle,ax   ;save the file handle (same as 

Function 3Eh: close file
Closes a file that has been opened.

        AX = 3Eh
        BX = file handle
        CF = 0 function is sucsessful
        AX = destroyed
        CF = 1 function not sucsessful
        AX = error code - 06h file not opened or unautorized handle.

Important: Don't call this function with a zero handle because that will close the 
standard input (the keyboard) and you won't be able to enter anything.
Function 3Fh: read file/device
Reads bytes from a file or device to a buffer.

        AH = 3Fh
        BX = handle
        CX = number of bytes to be read
        DS:DX = segment:offset of a buffer

        CF = 0 function is successful
        AX = number of bytes read
        CF = 1 an error has occured
                05h access denied
                06h illegal handle or file not opened

If CF = 0 and AX = 0 then the file pointer was already at the end of the file and no 
more can be read. If CF = 0 and AX is smaller than CX then only part was read 
because the end of the file was reached or an error occured. 

This function can also be used to get input from the keyboard. Use a handle of 0, and it 
stops reading after the first carriage return, or once a specified number of characters 
have been read. This is a good and easy method to use to only let the user enter a 
certain amount of characters.

Note: If you are using A86 this will cause an error. Change @data to data to make it 

.MODEL small
	mov ax,@data     ;base jaddress of data
	mov ds,ax        ;segment
	lea dx,FileName  ;put address of fileneame in dx        
	mov al,2         ;access mode - read and write
	mov ah,3Dh       ;function 3Dh -open a file
	int 21h          ;call DOS service
	mov Handle,ax    ;save file handle for later
	jc ErrorOpening
	mov dx,offset Buffer  ;address of buffer in dx
	mov bx,Handle         ;handle in bx
	mov cx,100            ;amount of bytes to be read
	mov ah,3Fh            ;function 3Fh - read from file
	int 21h               ;call dos service
	jc ErrorReading           

	mov bx,Handle    ;put file handle in bx 
	mov ah,3Eh       ;function 3Eh - close a file
	int 21h          ;call dos service
	mov di,OFFSET Buffer+101  ;Where to put the "$"
	mov byte ptr [di],"$"     ;Put it at es:di
	mov ah,9                  ;write a string to the 
	mov dx,OFFSET Buffer	 ;screen using function 9h
	int 21h				 ;of interrupt 21h

	mov AX,4C00h     ;terminate program and return to                                                                       
				  ;DOS using
	INT 21h          ;interrupt 21h function 4CH

	mov dx,offset OpenError ;display an error                                                                            
	mov ah,09h       ;using function 09h
	int 21h          ;call dos service

     mov ax,4C01h ;end program with an errorlevel of 1   
	int 21h
	mov dx,offset ReadError ;display an error                                                                            
	mov ah,09h       ;using function 09h
	int 21h          ;call dos service
	mov ax,4C02h ;end program with an errorlevel of 2   
        int 21h

Handle        DW ? ;variable to store file handle   
FileName      DB "C:\test.txt",0 ;file to be opened

OpenError     DB "An error has occured(opening)!$"
ReadError     DB "An error has occured(reading)!$"

Buffer        DB 101 dup (?) ;buffer to store data from                                             
					    ;file one bigger for $

Function: 3Ch: Create file
Creates a new empty file on a specified drive with a specified pathname.

        AH = 3Ch
        CX = file attribute
                bit 0 = 1 read-only file
                bit 1 = 1 hidden file
                bit 2 = 1 system file
                bit 3 = 1 volume (ignored)
                bit 4 = 1 reserved (0) - directory
                bit 5 = 1 archive bit
                bits 6-15 reserved (0)
    DS:DX = segment:offset of ASCIIZ pathname

    CF = 0 function is successuful
    AX = handle
    CF = 1 an error has occured
                03h path not found
                04h no availible handle
                05h access denied

Important: If a file of the same name exists then it will be lost. Make sure that there 
is no file of the same name. This can be done with the function below.

Function 4Eh: find first matching file
Searches for the first file that matches the filename given.

    AH = 4Eh
    CX = file attribute mask (bits can be combined)
                bit 0 = 1 read only
                bit 1 = 1 hidden
                bit 2 = 1 system
                bit 3 = 1 volume label
                bit 4 = 1 directory
                bit 5 = 1 archive
                bit 6-15 reserved
    DS:DX = segment:offset of ASCIIZ pathname

    CF = 0 function is successful
    [DTA] Disk Transfer Area = FindFirst data block

Example of checking if file exists:
File    DB "C:\file.txt",0   ;name of file that we want

        LEA dx,File      ;address of filename
        MOV cx,3Fh       ;file mask 3Fh - any file
        MOV ah,4Eh      	;function 4Eh - find first file
        INT 21h          ;call dos service
        JC NoFile
        ;print message saying file exists
        ;continue with creating file

This example program creates a file and then writes to it.
	mov ax,@data     ;base jaddress of data
	mov ds,ax        ;segment

	mov dx,offset StartMessage    ;display the starting 
     mov ah,09h     ;using function 09h
     int 21h        ;call dos service
     mov dx,offset FileName ;put offset of filename in dx        
     xor cx,cx      ;clear cx - make ordinary file
     mov ah,3Ch     ;function 3Ch - create a file
     int 21h        ;call DOS service
     jc Error       ;jump if there is an error
     mov dx,offset FileName ;put offset of filename in dx
     mov al,2       ;access mode -read and write
     mov ah,3Dh     ;function 3Dh - open the file
     int 21h        ;call dos service
     jc Error       ;jump if there is an error
     mov Handle,ax  ;save value of handle        

     mov dx,offset WriteMe 	;address of information to 
						;write to the file
     mov bx,Handle  ;file handle for file
     mov cx,38      ;38 bytes to be written
     mov ah,40h     ;function 40h write to file
     int 21h        ;call dos service
     jc error       ;jump if there is an error
     cmp ax,cx      ;was all the data written? Does 	
     jne error      ;if it wasn't then there was an error

     mov bx,Handle  ;put file handle in bx 
     mov ah,3Eh     ;function 3Eh - close a file
     int 21h        ;call dos service
     mov dx,offset EndMessage ;display the final message 
						;on the screen
     mov ah,09h     ;using function 09h
     int 21h        ;call dos service
     mov AX,4C00h   ;terminate program and return to DOS 
     int 21h        ;using interrupt 21h function 4CH

     mov dx,offset ErrorMessage ;display an error message 
						  ;on the screen
     mov ah,09h      ;using function 09h
     int 21h         ;call dos service
     jmp ReturnToDOS ;lets end this now

StartMessage    DB "This program creates a file called, 
			    NEW.TXT in the C: directory.$"
EndMessage      DB 0Ah,0Dh,"File create OK, look at, 	
			    file to be sure.$"
Handle          DW ? ;variable to store file handle   
ErrorMessage    DB "An error has occurred!$"
WriteMe         DB "HELLO, THIS IS A TEST, HAS IT, 	
			    WORKED?",0 ;ASCIIZ 
FileName        DB "C:\new.txt",0 					

How to find out the DOS version:

In many programs it is necessary to find out what the DOS version is. This could be 
because you are using a DOS function that needs the revision to be over a certain 

Firstly this method simply finds out what the version is.

	mov	ah,30h 	;function 30h - get MS-DOS version
	int 	21h		;call DOS function

This function returns the major version number in AL and the minor version number in 
AH. For example if it was version 4.01, AL would be 4 and AH would be 01. The 
problem is that if on DOS 5 and higher SETVER can change the version that is 
returned. The way to get round this is to use this method.

	mov	ah,33h    ;function 33h - actual DOS version
 	mov	al,06h	;subfunction 06h
	int	21h	  	;call interrupt 21h	

This will only work on DOS version 5 and above so you need to check using the 
former method. This will return the actual version of DOS even if SETVER has 
changed the version. This returns the major version in BL and the minor version in BH.

Fast string print:

We have been using a DOS service, function 9 of interrupt 21h to print a string on the 
screen. This isnt too fast nor does it allow us to use different colours or position the 
text. There is another way to print a string to the screen - direct to memory. This is 
harder as you have to set up everything manually but it has a lot of benifits mainly 

TextAttribute db 7 	;contains the character attribute
				;default is grey on black 
FastTextPrint PROC            
.286    ;need this for shift instructions. Take out if 
        ;than 286
;INPUT: AH - Row
;       AL - Column
;       CX - Length of string
;       DS:DX - The string
;       TextAttribute - the colour of the text
;OUTPUT: none
        push ax bx cx dx bp si di es ;save registers
        mov bl,ah     	;move row into bl
        xor bh,bh        ;clear bh
        shl bx,5         ;shift bx 5 places to the left
        mov si,bx        ;move bx into si
        shl bx,2         ;shift bx 2 places to the left
        add bx,si        ;add si to bx
        xor ah,ah        ;clear ah
        shl ax,1         ;shift ax 1 place to the left
        add bx,ax        ;add ax onto bx
        mov ax,0b800h    ;ax contains text video memory
        mov es,ax        ;move ax into es
        mov si,dx        ;mov dx into si
       mov ah,ds:[si]	;put the char at ds[si] into ah
       mov es:[bx],ah    ;move the char in ah to es[bx]
       inc si           	;increment si (si+1)
       inc bx            ;increment bx (bx+1)
       mov ah,TextAttribute ;put the attribute into ah
       mov es:[bx],ah    ;put ah into es position at bx
       inc bx            ;increment bx (bx+1)
       loop FastTextPrintLoop 	;loop CX times
       pop es di si bp dx cx bx ax ;restore registers
       ret			;return
FastTextPrint ENDP

Explanation  of new terms in this procedure:

In this procedure there was several things that you have not come across before. Firsly 
the lines:

	push ax bx cx dx bp si di es 	;save registers
	pop es di si bp dx cx bx ax 	;restore registers

This is just an easier way of pushing and popping more than one register. When TASM 
(or A86) compiles these lines it translates it into separate pushes an pops. This way 
just saves you time typing and makes it easier to understand.

Note: To make these lines compile in A86 you need to put commas (,) in between the 

This line might cause difficulty to you at first but they are quite easy to understand. 

	mov ah,ds:[si] 	;put the char at ds[si] into ah

What this does is to move the number stored in DS at the location stored in SI into 
AH. It is easier to think of DS being like an array in this command. It is the same as 
this line in C.

	ah = ds[si]; 


There are four different ways of shifting numbers either left or right one binary 

SHL Unsigned multiple by two
SHR Unsigned devide by two
SAR Signed devide by two
SAL same as SHL

The syntax for all four is the same.

SHL operand1,operand2

Note: The 8086 cannot have the value of opperand2 other than 1. 286/386 cannot 
have operand2 higher than 31.

Using shifts is a lot faster than using MULs and DIVs.


Using Loop is a better way of making a loop then using JMPs. You place the amount 
of times you want it to loop in the CX register and every time it reackes the loop 
statement it decrements CX (CX-1) and then does a short jump to the label indicated. 
A short jum means that it can only 128 bytes before or 127 bytes after the LOOP 

Loop Label

Using graphics in mode 13h:

Mode 13h is only availible on VGA, MCGA cards and above. The reason that I am 
talking about this card is that it is very easy to use for graphics because of how the 
memory is arranged.

First check that mode 13h is possible:

It would be polite to tell the user if his computer cannot support mode 13h instead of 
just crashing his computer without warning. This is how it is done.

;Returns: DX=0 Not supported, DX=1 supported
	mov ax,1A00h      	;Request video info for VGA
	int 10h        	;Get Display Combination Code
	cmp al,1Ah    		;Is VGA or MCGA present?
	je Mode13hSupported ;mode 13h is supported
	xor dx,dx			;mode 13h isnt supported dx=0
	mov dx,1			;return mode13h supported 

Just use this to check if mode 13h is supported at the beginning of your program to 
make sure that you can go into that mode.

Note: I have not tested this on a computer that doesnt hav VGA as I dont have any. 
In theory this should work but you should test this on computers that dont have VGA 
and see if it works this out.

Setting the Video Mode:

It is very simple to set the mode. This is how it is done.

 	mov  ax,13h		;set mode 13h
 	int  10h			;call bios service

Once we are in mode 13h and have finished what we are doing we need to we need to 
set it to the video mode that it was in previously. This is done in two stages. Firstly we 
need to save the video mode and then reset it to that mode.

VideoMode db ?
	mov ah,0Fh		;function 0Fh - get current mode
	int 10h			;Bios video service call
	mov VideoMode,al	;save current mode

	;program code here

	mov al,VideoMode	;set previous video mode
	xor ah,ah			;clear ah - set mode
	int 10h			;call bios service
	mov ax,4C00h		;exit to dos
	int 21h			;call dos function

Now that we can get into mode 13h lets do something. Firstly lets put some pixels on 
the screen. 

Function 0Ch - Write Graphics Pixel

Makes a color dot on the screen at the specified graphics coordinates.

	AH = 0Ch
	AL = Color of the dot
	CX = Screen column (x coordinate)
	DX = Screen row (y coordinate)

	Nothing except pixel on screen.

Note: This function performes exclusive OR (XOR) with the new color value and the 
current context of the pixel of bit 7 of AL is set.

	mov ah,0Ch	;function 0Ch
	mov al,7		;color 7
	mov cx,160	;x position -160
	mov dx,100	;y position -100
	int 10h		;call bios service

This example puts a pixel into the middle of the screen in a the color grey. The 
problem with this method is that calling interrupts is really slow and should be avoided 
in speed critical areas. With pixel plotting if you wanted to display a picture the size of 
the screen you would have to call this procedure 64,000 times (320 x 200). 

Some optimizations:

This method isnt too fast and we could make it a lot faster. How? By writing direct to 
video memory. This is done quite easily. 

The VGA memory starts at 0A000h. To work out where each pixel goes you use this 
simple formula:

	Location = 0A000h + Xposition + (Yposition x 320)

Location is the memory location which we want to put the pixel.

This procedure is quite a fast way to put a pixel onto the screen. Thanks go to Denthor 
of Asphyxia as I based this on his code. 

PutPixel PROC
.286		;enable 286 instructions for shifts remove if 
		;you have less than an 286.	
;INPUT: 	BX=X postion
;		DX=Y position
;		CL=colour
;OUTPUT:  None
;this can be optimized by not pushing ax if you dont 
;need to save it. For A86 change push ds es ax to push 
;ds,es,ax and do the same thing with pop.
	push ds es ax		;save ds,es and ax 
	mov ax,0A000h		;ax contains address of video
	mov es,ax			;es contains address of video
	mov di,bx			;move x position into di
	mov bx,dx			;mov y postion into bx
	shl dx,8			;shift dx 8 places to the left
	shl bx,6			;shift bx 6 places to the left
	add dx,bx			;add dx and bx together
	add di,bx			;add di and bx together
	mov al,cl			;put colour in al
	stosb			;transfer to video memory
	pop ax es ds		;restore ds,es and ax
PutPixel ENDP	

Thank you for reading. I hope that you have learnt something from this. If you 
need any more help then email me.

Gavin Estey 19/2/95

Gavins Introduction to Assembly		Page 1

[Back to TUTOR SWAG index]  [Back to Main SWAG index]  [Original]