Al Balqa Applied University
Al-Huson College University
Electrical Engineering Department
Microprocessor 8086 Lab Manual
Eng. Buthayna AL-Sharaa
Table of Contents
No
1
Lab Title
Introduction to Assembly Language
Page
3
2
Assembly Language Fundamentals
9
3
Data Transfer & Arithmetic
11
4
Loop Instruction
20
5
Addressing Modes
22
6
7
8
Boolean Instructions & Conditional Structured
Shift, Rotate, Multiplication and Division Instructions
Stacks and Subroutines
25
31
38
9
Debug Program
43
10
Keyboard Input with int 16h & Macros
47
Appendix ASCII CODE Tables
A
50
Lab1
Introduction to Assembly Language
1. Objective:
To be familiar with Assembly Language.
2. Introduction
Machine language (computer's native language) is a system of impartible instructions
executed directly by a computer's central processing unit (CPU). Instructions consist of
binary code: 1s and 0s.
Machine language can be made directly from java code using interpreter.
Assembly Language is a programming language that is very similar to machine language,
but uses symbols instead of binary numbers. It is converted by the assembler (e.g. Tasm and
Masm) into executable machine-language programs.
The difference between compiling and interpreting is as follows. Compiling translates the
high-level code into a target language code as a single unit. Interpreting translates the
individual steps in a high-level program one at a time rather than the whole program as a
single unit. Each step is executed immediately after it is translated.
C, C++ code is executed faster than Java code, because they transferred to assembly
language before machine language.
Assembly language is machine-dependent; an assembly program can only be executed on a
particular machine.
In Our Lab we will learn the assembly language of the 8086intel Microprocessor.
To make programs in assembly language, you must know some information about the 8086
microprocessor. The 8086 contains 14 registers. Each register is 16 bits long. See Figure (1)
Each register has different usage as shown in Table (1) below. The general purpose registers
can be "split". AH contains the high byte of AX and AL contains the low byte. You also
have: BH, BL, CH, CL, DL, DH. So if for example, DX contains the value 1234h DH would
be 12h and DL would be 34h.
And a 16-bit FLAG Register. The FLAGS Register consists of 9 status bits. These bits are
also called flags, because they can either be SET (1) or NOT SET (0). All these flags have a
name and purpose.
op-code destination operand, source operand
Segments:
The 8086 microprocessor has a memory of size 1MB. This memory is divided into
segments, each segment has a size of 64KB. The segments are: Data Segment (DS), Code
Segment(CS), Stack Segment (SS), Extra Segment(ES).
❖ the address of any location within a segment consist of 2 parts:
a. segment address: 16-bit selector defining the beginning address of any 64KB (216)
memory segment
b. offset address: selecting any location within the 64KB (216) memory segment
3. Running your first example: Hello Program!
1. Click Start → All Programs → Emu8086
2. Click New → Com Template. Where a white page opens.
3. Enter the code shown in the following screen
4. Click Compile
press
- Note:
. If your program does not contain any errors → save program →
. then click
from the screen that shows.
is used to execute the program one instruction at a time.
Analysis of the Hello program:
Homework:
Write an assembly language program to print all letters as follows: AB......MN
Note: To print a character on the screen you have to use the int 21h with the service 2, the
character to be printed have to be in dL register.
For Example, the following code print A on the screen.
mov ah, 2
mov dl, 41h
int 21h
Note: Don’t use loop in your solution
Lab2
Assembly Language Fundamentals
Objective:
To be familiar with Assembly Language Fundamentals.
Our compiler supports two types of data types: Byte, Word
Variables
A variable is a memory location that has an address and a value.
The syntax for variable declaration is:
Varname DB value
Varname DW value
DB : stands for Define Byte
DW : stands for Define Word
Value: is the numerical value of variable in any of the supported number systems(
Hexadecimal, decimal, binary, octal). The letter ? can be used to leave the content of
variable not changed.
The characters used for each numeric system are :
h – hexadecimal
d – decimal
b – binary
o – octal
Examples: 30d, 6Ah, 42, 1101b, 777o
- Hexadecimal beginning with letter: 0A5h
- If no radix is given, the integer constant is decimal
- A hexadecimal beginning with a letter must have a leading
- the definition of an array is similar to variable definition , except that a list of values are
specified.
Example: defining array named list1 of 4 elements each element is a byte, with initial values
1,2,3,4
List1
db
1,2,3,4
Character Constants
A character constant is a single character enclosed in either single or double quotes. The
assembler converts it to the binary ASCII code matching the character.
Examples are:
'A'
"x"
ASCII character = 1 byte.
String Constants
String constant is a string of characters enclosed in either single or double quotes:
'ABC'
"xyz"
'Say "Goodnight," Gracie'
-Each character occupies a single byte.
-it is preferred to end String with the symbol $
Comments
All single line comments start with the symbol
;
Lab3
Data Transfer & Arithmetic
❖ Direct-Offset Operands:
a direct-offset operand: (also called Direct Addressing Mode ) is adding a displacement( a number ) to the
name of a variable or array. Let’s begin with an array of bytes named arrayB:
.data
arrayB db 10h,20h,30h,40h,50h
.code
; the following instruction with will move the first byte in the array into AL:
mov
AL , arrayB
; AL = 10h
; the following instruction with will move the second byte in the array into AL:
Mov
AL , [arrayB+1]
; AL = 20h
Note:
An expression such as arrayB+1 produces what is called an offset address by adding a
constant to the variable’s name. Surrounding an offset address with brackets indicates the
expression is used to obtain the contents of memory at that address.
Lab4
Loop Instruction
Example:
LEA BX, List ;Load the offset address of List into register Bx
Note: The operator Offset performs the same as LEA
Example:
Mov BX, OFFSET List ; this instruction is the same as the above.
This operator specifies the size to which an address points to in memory, as shown:
• WORD PTR : address or register points to a word
• BYTE PTR : address or register points to a Byte
Example:
INC WORD PTR var1 ; increment the content of the word in memory addressed by var1
Lab Work:
Exercise 1:
Write the code to print your name 5 times on screen
Exercise 2:
Write the code to find the square of the number 4
Exercise 3:
Lab5
Addressing Modes
Objective:
To know more about Assembly language, such as how CPU can access data in memory.
Addressing Modes:
The addressing modes are different ways to find the memory address (how CPU can access
data in memory). There are five addressing modes used in assembly programming with
80x86 family, these modes are:
1. Immediate addressing mode:
One of the operands is an immediate number
Ex mov al,5
2. Register addressing mode
Both operands exist in registers.
Ex add cl,bl
3. Direct Addressing : Mode In this mode the 16 bit Offset is taken directly from the
instruction.
Examples:
mov [START+1],CL
; copy CL into location START + 1
mov DX, [200]
; copy the word at address 200 into register DX
4. Register Indirect: [bx], [si], [di], [bp]
Addressing data at any location through an offset address held in BP, BX, DI, and SI.
Examples :
mov ax,[bx] ; copies the word in data segment memory location addressed by bx to ax
mov [BP],dl ; copies dl into stack segment memory at the address stored in BP
5. Relative register indirect: Register added to a constant to form the offset
MOV AX, [BX+5] ↔ MOV AX,5[BX]
↔ MOV AX, [BX]+5
6. Base-Plus-Index Addressing:
In this mode one base register (BP or BX) + 1 of index register (DI or SI)
– base register: hold beginning location of a memory array
– index register: hold relative position of an element in the array
-if base register BP is used➔ data addressed is in stack segment
- if base register BX is used➔ data addressed is in data segment
Examples:
MOV DX, [BX+DI] ; move data(16 Bit ) is data segment at address BX+DI to register DX
mov ax,[bp+bx]
mov ax,[si+di]
; error , two base registers
; error , two index registers
7. Base-Indexed with displacement (relative base plus index)
MOV AX, [bx+si+2]
Homework:
1. Write an assembly program that allows the user to enter two strings and
determines if they are equal or not. (Each string must end with $).
2. Write an assembly program that allows the user to enter 5-digits Number, and
then prints it in reversed-order digits.
Lab 6/Boolean Instructions & Conditional Structured
OTHER logical instructions: XOR,OR. They have the same syntax a the AND
instruction
Truth table OR
0
0
0
0
1
1
1
0
1
1
1
1
Example: mov
al,0C5H
OR al, 0FH
Final value of al=0CFh
Truth table XOR
0
0
0
0
1
1
1
0
1
1
1
0
note 0 xor A = A
1 xor A = A’ (complement)
Example : write code to flip(complement) bit 0 in al and keep the other bits
unchanged
Xor al,1
Exercise 4
Lab 7/ Shift, Rotate, Multiplication and Division
Instructions
Lab 8
Stacks and Subroutines
Write a program that finds the largest unsigned number in an array of 5 numbers , each
number is of one byte. Use subroutines
Lab 9
Debug Program
Objective:
Debugging and writing assembly programs using Debug program under windows.
Debug is available on every Windows box through the DOS command prompt we will it to
explore the computer organization of the 80x86 machine.
To open the program:
Click to open
Write debug then press enter
-enter ? then press enter. The following shows
This is a menu that contains the set of all commands that can performed in this program.
Some of the Debugger’s Commands:
1. Register ( r ) : to display the content of CPU registers.
To change the content of register AX
Old value of AX
new value of AX
2. Dump (d ): to display the content of data segment.
The following example dumps the content of data segment at offset 0000
address
Content of memory in
hexadecimal
Content of memory in
ASCII
1. Enter (e) : to fill or change the contents of the data segments.
Example: to change the content at ds:0000, enter
Old value at address
0000
New entered value at
address 0000
- Press Enter to exit this command
- Intel uses Little Indian organization to store data in memory( where most significant
byte of operand is stored in highest address in memory
2. Assemble (a): to enter our code to the code segment
Example: to enter instructions at CS:100
3. Trace ( t ): to execute the code one instruction at a time. The instruction executed is the
one whose address is stored in register IP.
Enter :
-t
4. Go (g ): used to execute the set of instructions between two addresses.
The statement g=100 10b runs the instructions between these addresses.
5. HEX (H) : to find the length of our code
syntax:
H end address of code
start address of code
The output of this command is:
Sum of two address
difference between two addresses
6. Name (n): used to specify the name of file
7. Write (w ): to write our code to the file we specified using command ( n). the size of
code to write in bytes should be stored in register CX
Example:
Address of the string
To exit
Lab 10
Keyboard Input with int 16h & Macros
Example:
ORG 100h
MyMacro2 MACRO
LOCAL label1, label2
CMP AX, 2
JE label1
CMP AX, 3
JE label2
label1:
INC AX
label2:
ADD AX, 2
ENDM
;;;;;main program
ORG 100h
MyMacro2 ; first call to macro
MyMacro2 ;second call to macro
RET
Exercise 2.
Write a program that finds the largest unsigned number in an array of 5 numbers , each
number is of one byte. First Use subroutines. Second use macros
ASCII CODE Table