assembly8086 tutorial

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Tutorial

by: GLENN VON C. POSADAS

Presented by: Glenn Von C. Posadas


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INTRODUCTION

Assembly Language is a low-levelprogramming language in which eachstatement corresponds directly to a single

machine instruction. It gives the computer-detailed command, thus it is specific to a givenprocessor.

Assembly Language might be used instead of ahigh-level language for any of the three majorreasons:



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INTRODUCTION: 3 MAJOR REASONS

SPEED

CONTROL

PREFERENCES

It provides precise control of the computer. Theused of assembly language lets a programmerinteract directly with the hardware (processor,

memory, display and input / output ports), thus itexecutes faster than those generated by thecompilers.



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HOW TO: assemble, compile and run

Run the DOSbox.

Mount the hard drive where your TASM folder islocated. By typing: Mount DriveLetter: Driveletter:\

Locate your TASM folder: type cd TASM To assemble your program with a file extension .asm,

type TASM

To link, type: tlink \t (this applies only to

.COM programs. tlink to link .EXE programs Type to run the program



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2 STRUCTURES OF ASSEMBLY

LANGUAGE

The two kinds/structure of Assembly Program

are:

1. The .COM (Read as dot com or COM)

program

and the second one is the

2. The .EXE (Read as dot ekse or EXE) program.



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First of all, the .COM is a program that requires anassembler to run. And compared to .EXE

programs, .COM is smaller in size and its structurecodes are easy to memorize. So basically, I preferto use this during exams in our theoretical as wellas in our laboratory exams. HAHA!

On the other hand, the .EXE program DOESN'TREQUIRE an assembler ANYMORE!And it is bigger in size since it is a stand-alone

program. You can run it anytime you wantwithout using other program(Assembler)



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.COM PROGRAM

cseg segment para 'code'

assume cs:cseg, ds:cseg, ss:cseg, es:cseg

org 100h

start: jmp begin;VARIABLE DECLARATION

begin:

;Codes

int 20h

cseg ends

end start



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.EXE PROGRAM

sseg segment para stack 'stack'

dw 200h

sseg ends

dseg segment para 'data'

dseg ends


;...CONTINUATION1


;CONTINUATION HERE

main proc far

assume cs:cseg,ds:dseg,ss:sseg

mov ax, dseg

mov ds, ax

mov es, ax

mov ah, 4ch

int 21h

main endp

cseg ends

end main


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Assembly Language 8086 Basic

Instructions

These are the most used instructions in 8086assembly language:

MOV instruction

Format: MOV {destination}, {source}This instruction can ONLY be used in the followingways:

REGISTER TO REGISTER

REGISTER TO MEMORY LOCATION IMMEDIATE TO REGISTER

IMMEDIATE TO MEMORY LOCATION



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LEA (load effective address) instruction

Format: LEA {register}, {memory location}

This means that we can only use such

instruction in that way. This instruction is

usually used for printing a string.

CMP(compare) instruction

Compare Destination to Source.

We can use this to create and to use jumpinstructions.



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Declaring Variables

For string:

VARIABLE_NAME db STRING HERE$

-note: put a $ in the end of the string toterminate. This serves as a null.

For Character:

VARIABLE_NAME db character

For a variable with null (or to initiate):

VARIABLE_NAME db ?



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CONTROL CHARACTERS (from ASCII

TABLE)

HEX DECIMAL NAME FUNCTIO

07 7 BELL Emits a one-second

beep

08 8 BACKSPACE Moves the cursor

one column to the

left.

0A 10 Line Feed Moves the cursor

down one row

0D 13 Carriage Return Moves the cursor to

the beginning of

the line



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Data Organization: DB, DW, and EQU

Bytes are allocated by define bytes DB.

Words are allocated by define words DW.

Both allow more than one byte or word to be allocated.

Question marks specify uninitialized data.

Strings allocate multiple bytes.

Labels in front of the directives remember offsets from thebeginning of the segment which accommodates the directive.

DUP allows to allocate multiple bytes. The following two linesproduce identical results:

DB ?, ?, ?, ?, ? DB 5 DUP(?)

Note that EQU directive does not allocate any memory: it creates aconstant value to be used by Assembler:

CR EQU 13 DB CR . mov al, CR



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Complete Instruction Set



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Some Important Interrupts for

8086/8088 microprocessors

INT 21h / AH=2 - write character to standard output.entry: DL = character to write, after execution AL = DL.

example: mov ah, 2 mov dl, 'a' int 21h

INT 21h / AH=1 - read character from standard input,with echo, result is stored in AL.if there is no character in the keyboard buffer, the

function waits until any key is pressed.

example: mov ah, 1 int 21h



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INT 10h / AH = 2 - set cursor position.input:DH = row.DL = column.BH = page number (0..7).example: mov dh, 10 mov dl, 20 mov bh, 0 mov ah, 2 int 10h

INT 10h / AH = 09h - write character and attribute at cursor position.

input:AL = character to display.BH = page number.BL = attribute.CX = number of times to write character.INT 10h / AH = 0Ah - write character only at cursor position.

input:AL = character to display.BH = page number.CX = number of times to write character.

http://www.computing.dcu.ie/~ray/teaching/CA296/notes/8086_bios_and_dos_interrupts.htmlhttp://www.computing.dcu.ie/~ray/teaching/CA296/notes/8086_bios_and_dos_interrupts.htmlhttp://www.computing.dcu.ie/~ray/teaching/CA296/notes/8086_bios_and_dos_interrupts.htmlhttp://www.computing.dcu.ie/~ray/teaching/CA296/notes/8086_bios_and_dos_interrupts.htmlhttp://www.computing.dcu.ie/~ray/teaching/CA296/notes/8086_bios_and_dos_interrupts.html


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INT 10h / AH = 0Ch - change color for a single pixel.

input:

AL = pixel colorCX = column.DX = row.example: mov al, 13h mov ah, 0 int 10h ; set graphics video mode.mov al, 1100b mov cx, 10 mov dx, 20 mov ah, 0ch int 10h ; set pixel.

INT 10h / AH = 2 - set cursor position.input:DH = row.DL = column.BH = page number (0..7).example: mov dh, 10 mov dl, 20 mov bh, 0 mov ah, 2 int 10h

INT 20h - exit to operating system.



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bit color table:


note: ; use this code for compatibility with dos/cmd

prompt full screen mode:

mov ax, 1003h

mov bx, 0 ; disable blinking.

int 10h


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Attributes:

Blink --- Red --- Green --- Blue || Intens --- Red --- Green --- Blue

BACKGROUND TEXT

1 denotes on, 0 denotes off

Example:

Blinking character, Green Foreground, Violet

background (Blinking G/V)

1 1 0 1 || 0 0 1 0 ==> in hex ==> 0D2h



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Answer:


assume cs:cseg;ds:cseg;es:cseg;ss:cseg

org 100h

start: jmp begin

begin:

mov ax, 0003h

int 10h

mov ah,02hmov dh, 12

mov dl, 37

int 10h


mov ah,09h

mov al, 03h ;

mov bl, 84h

mov cx,1

int 10h

int 20h

cseg ends

end start


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Dealing with STRINGS


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