Saturday, April 9, 2011

8086 basics

Introduction to 8086 Programming
(The 8086 microprocessor is one of the family of
8086,80286,80386,80486,Pentium,PentiumI,II,III .... also
referred to as the X86 family.)
Learning any imperative programming language involves
mastering a number of common concepts:
Variables: declaration/definition
Assignment: assigning values to variables
Input/Output: Displaying messages
Displaying variable values
Control flow: if-then
Subprograms: Definition and Usage
Programming in assembly language involves mastering the
same concepts and a few other issues.
For the moment we will skip details of variable declaration and
simply use the 8086 registers as the variables in our programs.
Registers have predefined names and do not need to be
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The 8086 has 14 registers. Each of these is a 16-bit register.
Initially, we will use four of them – the so called the general
purpose registers:
ax, bx, cx, dx
These four 16-bit registers can also be treated as eight 8-bit
ah, al, bh, bl, ch, cl, dh, dl
In Java, assignment takes the form:
x = 42 ;
y = 24;
z = x + y;
In assembly language we carry out the same operation but we
use an instruction to denote the assignment operator (“=” in
Java). The above assignments would be carried out in 8086
assembly langauge as follows
mov x, 42
mov y, 24
add z, x
add z, y
The mov instruction carries out assignment.
It which allows us place a number in a register or in a memory
location (a variable) i.e. it assigns a value to a register or
Example: Store the ASCII code for the letter A in register
mov bx, ‘A’
The mov instruction also allows you to copy the contents of
one register into another register.
mov bx, 2
mov cx, bx
The first instruction loads the value 2 into bx where it is stored
as a binary number. [a number such as 2 is called an integer
The Mov instruction takes two operands, representing the
destination where data is to be placed and the source of that
General Form of Mov Instruction
mov destination, source
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where destination must be either a register or memory location
source may be a constant, another register or a memory
Note: The comma is essential. It is used to separate the two
A missing comma is a common syntax error.
Anything that follows semi-colon (;) is ignored by the
assembler. It is called a comment. Comments are used to
make your programs readable. You use them to explain what
you are doing in English.
More 8086 Instructions
add, inc, dec and sub instructions
The 8086 provides a variety of arithmetic instructions. For the
moment, we only consider a few of them. To carry out
arithmetic such as addition or subtraction, you use the
appropriate instruction.
In assembly language you can only carry out a single
arithmetic operation at a time. This means that if you wish to
evaluate an expression such as :
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You will have to use 3 assembly language instructions – one
for each arithmetic operation.
These instructions combine assignment with the arithmetic
mov ax, 5 ; load 5 into ax
add ax, 3 ; add 3 to the contents of ax,
; ax now contains 8
inc ax ; add 1 to ax
; ax now contains 9
dec ax ; subtract 1 from ax
; ax now contains 8
sub ax, 6 ; subtract 4 from ax
; ax now contains 2
The add instruction adds the source operand to the destination
operand, leaving the result in the destination operand.
The destination operand is always the first operand in 8086
assembly language.
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The inc instruction takes one operand and adds 1 to it. It is
provided because of the frequency of adding 1 to an operand
in programming.
The dec instruction like inc takes one operand and subtracts 1
from it. This is also a frequent operation in programming.
The sub instruction subtracts the source operand from the
destination operand leaving the result in the destination
1) Write instructions to:
Load character ? into register bx
Load space character into register cx
Load 26 (decimal) into register cx
Copy contents of ax to bx and dx
2) What errors are present in the following :
mov ax 3d
mov 23, ax
mov cx, ch
move ax, 1h
add 2, cx
add 3, 6
inc ax, 2
3) Write instructions to evaluate the arithmetic expression 5 +
(6-2) leaving the result in ax using (a) 1 register, (b) 2 registers,
(c) 3 registers
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4) Write instructions to evaluate the expressions:
a = b + c –d
z = x + y + w – v +u
5) Rewrite the expression in 4) above but using the registers
ah, al, bh, bl and so on to represent the variables: a, b, c, z, x,
y, w, u, and v.
Implementing a loop: The jmp instruction
Label_X: add ax, 2
add bx, 3
jmp Label_X
The jmp instruction causes the program to start executing from
the position in the program indicated by the label Label_X. This
is an example of an endless loop.
We could implement a while loop using a conditional jump
instruction such as JL which means jumi-if-less-than. It is used
in combination with a comparision instruction – cmp.
mov ax, 0
Label_X: add ax, 2
add bx, 3
cmp ax, 10
jl Label_X
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The above loop continues while the value of ax is less than 10.
The cmp instruction compares ax to 0 and records the result.
The jl instruction uses this result to determine whether to jump
to the point indicated by Label_X.
Each microprocessor provides instructions for I/O with the
devices that are attached to it, e.g. the keyboard and screen.
The 8086 provides the instructions in for input and out for
output. These instructions are quite complicated to use, so we
usually use the operating system to do I/O for us instead.
In assembly language we must have a mechanism to call the
operating system to carry out I/O.
In addition we must be able to tell the operating system what
kind of I/O operation we wish to carry out, e.g. to read a
character from the keyboard, to display a character or string on
the screen or to do disk I/O.
In 8086 assembly language, we do not call operating system
subprograms by name, instead, we use a software interrupt
An interrupt signals the processor to suspend its current
activity (i.e. running your program) and to pass control to an
interrupt service program (i.e. part of the operating system).
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A software interrupt is one generated by a program (as
opposed to one generated by hardware).
The 8086 int instruction generates a software interrupt.
It uses a single operand which is a number indicating which
MS-DOS subprogram is to be invoked.
For I/O and some other operations, the number used is 21h.
Thus, the instruction int 21h transfers control to the operating
system, to a subprogram that handles I/O operations.
This subprogram handles a variety of I/O operations by calling
appropriate subprograms.
This means that you must also specify which I/O operation
(e.g. read a character, display a character) you wish to carry
out. This is done by placing a specific number in a register.
The ah register is used to pass this information.
For example, the subprogram to display a character is
subprogram number 2h.
This number must be stored in the ah register. We are now in a
position to describe character output.
When the I/O operation is finished, the interrupt service
program terminates and our program will be resumed at the
instruction following int.
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3.3.1 Character Output
The task here is to display a single character on the screen.
There are three elements involved in carrying out this
operation using the int instruction:
1. We specify the character to be displayed. This is done by
storing the character’s ASCII code in a specific 8086 register.
In this case we use the dl register, i.e. we use dl to pass a
parameter to the output subprogram.
2. We specify which of MS-DOS’s I/O subprograms we wish
to use. The subprogram to display a character is subprogram
number 2h. This number is stored in the ah register.
3. We request MS-DOS to carry out the I/O operation using
the int instruction. This means that we interrupt our program
and transfer control to the MS-DOS subprogram that we have
specified using the ah register.
Example 1: Write a code fragment to display the character ’a’
on the screen:
mov dl, ‘a‘ ; dl = ‘a‘
mov ah, 2h ; character output subprogram
int 21h ; call ms-dos output character
As you can see, this simple task is quite complicated in
assembly language.
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3.3.2 Character Input
The task here is to read a single character from the keyboard.
There are also three elements involved in performing character
1. As for character output, we specify which of MS-DOS’s I/O
subprograms we wish to use, i.e. the character input from the
keyboard subprogram. This is MS-DOS subprogram number
1h. This number must be stored in the ah register.
2. We call MS-DOS to carry out the I/O operation using the
int instruction as for character output.
3. The MS-DOS subprogram uses the al register to store
the character it reads from the keyboard.
Example 2: Write a code fragment to read a character from
the keyboard:
mov ah, 1h ; keyboard input subprogram
int 21h ; character input
; character is stored in al
The following example combines the two previous ones, by
reading a character from the keyboard and displaying it.
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Example 3: Reading and displaying a character:
mov ah, 1h ; keyboard input subprogram
int 21h ; read character into al
mov dl, al ; copy character to dl
mov ah, 2h ; character output subprogram
int 21h ; display character in dl
A Complete Program
We are now in a position to write a complete 8086 program.
You must use an editor to enter the program into a file. The
process of using the editor (editing) is a basic form of word
processing. This skill has no relevance to programming.
We use Microsoft’s MASM and LINK programs for assembling
and linking 8086 assembly language programs. MASM
program files should have names with the extension (3
characters after period) asm.
We will call our first program prog1.asm, it displays the letter
‘a‘ on the screen. (You may use any name you wish. It is a
good idea to choose a meaningful file name). Having entered
and saved the program using an editor, you must then use the
MASM and LINK commands to translate it to machine code so
that it may be executed as follows:
C> masm prog1
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If you have syntax errors, you will get error messages at this
point. You then have to edit your program, correct them and
repeat the above command, otherwise proceed to the link
command, pressing Return in response to prompts for file
names from masm or link.
H:\> link prog1
To execute the program, simply enter the program name and
press the Return key:
H:\> prog1
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Example 4: A complete program to display the letter ‘a‘ on the
; prog1.asm: displays the character ‘a’ on the screen
; Author: Joe Carthy
; Date: March 1994
.model small
.stack 100h
mov dl, ‘a’ ; store ascii code of ‘a’ in dl
mov ah, 2h ; ms-dos character output function
int 21h ; displays character in dl register
mov ax, 4c00h ; return to ms-dos
int 21h
end start
The first three lines of the program are comments to give the
name of the file containing the program, explain its purpose,
give the name of the author and the date the program was
The first two are directives, .model and .stack. They are
concerned with how your program will be stored in memory
and how large a stack it requires. The third directive, .code,
indicates where the program instructions (i.e. the program
code) begin.
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For the moment, suffice it to say that you need to start all
assembly languages programs in a particular format (not
necessarily that given above. Your program must also finish in
a particular format, the end directive indicates where your
program finishes.
In the middle comes the code that you write yourself.
You must also specify where your program starts, i.e. which is
the first instruction to be executed. This is the purpose of the
label, start.
(Note: We could use any label, e.g. begin in place of start).
This same label is also used by the end directive.
When a program has finished, we return to the operating
system. Like carrying out an I/O operation, this is also
accomplished by using the int instruction. This time MS-DOS
subprogram number 4c00h is used.
It is the subprogram to terminate a program and return to MS-
DOS. Hence, the instructions:
mov ax, 4c00h ; Code for return to MS-DOS
int 21H ; Terminates program
terminate a program and return you to MS-DOS.
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Time-saving Tip
Since your programs will start and finish using the same
format, you can save yourself time entering this code for each
program. You create a template program called for example,
template.asm, which contains the standard code to start and
finish your assembly language programs. Then, when you wish
to write a new program, you copy this template program to a
new file, say for example, prog2.asm, as follows (e.g. using the
MS-DOS copy command):
H:\> copy template.asm prog2.asm
You then edit prog2.asm and enter your code in the
appropriate place.
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