8086 assembly language programming i

7/30/2019 8086 Assembly Language Programming I

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Learning assembly language programming

will help understanding the operations ofthe microprocessor To learn:

Need to know the functions of various registers Need to know how external memory is organized

and how it is addressed to obtain instructionsand data (different addressing modes) Need to know what operations (or the

instruction set) are supported by the CPU. Forexample, powerful CPUs support floating-pointoperations but simple CPUs only support integer

operations

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CConcept

L Logic thinking P Practice

Concept we must learn the basic syntax,such as how a program statement is written

Logic thinking programming is problemsolving so we must think logically in orderto derive a solution

Practice write programs

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The native language is machine language

(using 0,1 to represent the operation) A single machine instruction can take up one or

more bytes of code Assembly language is used to write the

program using alphanumeric symbols (or

mnemonic), eg ADD, MOV, PUSH etc. The program will then be assembled (similar to

compiled) and linked into an executableprogram.

The executable program could be .com, .exe,or .bin files

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Program

.asm

Object file

.obj

Executable file

.exe

Assemble link


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Machine code for mov AL, 00H

B4 00 (2 bytes)

After assembled, the value B400 will bestored in the memory

When the program is executed, then thevalue B400 is read from memory, decodedand carry out the task

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Each instruction is represented by one assemblylanguage statement

The statement must specify which operation(opcode) is to be performed and the operands Eg ADD AX, BX ADD is the operation AX is called the destination operand BX is called the source operand The result is AX = AX + BX When writing assembly language program, you

need to think in the instruction level

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In c++, you can do A = (B+C)*100

In assembly language, only one instructionper statement

A = B ; only one instruction - MOVE

A = A+C ; only one instruction - ADDA = A*100 ; only one instruction - Multiply

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General format for an assembly languagestatement

Label Instruction Comment

Start: Mov AX, BX ; copy BX into AX

Start is a user defined name and you only put in a

label in your statement when necessary!!!!

The symbol : is used to indicate that it is a label

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In 8086, memory is divided into segments

Only 4 64K-byte segments are active and these are:code, stack, data, and extra

When you write your assembly language program for an8086, theoretically you should define the differentsegments!!!

To access the active segments, it is via the segmentregister: CS (code), SS (stack), DS (data), ES (extra)

So when writing assembly language program, youmust make use of the proper segment register orindex register when you want to access the memory

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In assembly programming, you cannotoperate on two memory locations in thesame instruction

So you usually need to store (move) value ofone location into a register and then perform

your operation After the operation, you then put the result

back to the memory location Therefore, one form of operation that you

will use very frequent is the store (move)

operation!!! And using registers!!!!!

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In C++ A = B+C ; A, B, C are variables

In assembly language A,B, C representingmemory locations so you cannot do A =B+C

MOV AL, B ; move value of B into AL register ADD, AL, C ; do the add AL = AL +C

MOV A, AL ; put the result to A

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AX, BX, CX,and DX these are the general purposeregisters but each of the registers also has specialfunction

Example AX is called the accumulator to store result in arithmetic

operations

Registers are 16-bit but can be used as 2 8-bit storage Each of the 4 data registers can be used as the source or

destination of an operand during an arithmetic, logic,shift, or rotate operation.

In some operations, the use of the accumulator is

assumed, eg in I/O mapped input and outputoperations

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In based addressing mode, base register BX isused as a pointer to an operand in the currentdata segment.

CX is used as a counter in some instructions, eg.CL contains the count of the number of bits by

which the contents of the operand must berotated or shifted by multiple-bit rotate DX, data register, is used in all multiplication

and division, it also contains an input/outputport address for some types of input/output

operations

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Stack is used as a temporary storage Data can be stored by the PUSH instruction

and extracted by the POP instruction Stack is accessed via the SP (Stack Pointer)

and BP (Base Pointer) The BP contains an offset address in the

current stack segment. This offset address isemployed when using the based addressingmode and is commonly used by instructions

in a subroutine that reference parametersthat were passed by using the stack

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Source index register (SI) and Destinationindex register (DI) are used to hold offsetaddresses for use in indexed addressing ofoperands in memory

When indexed type of addressing is used,then SI refers to the current data segmentand DI refers to the current extra segment

The index registers can also be used assource or destination registers in arithmeticand logical operations. But must be used in16-bit mode

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Data can be in three forms: 8-bit, 16-bit, or 32-bit (double word)

Integer could be signed or unsigned and inbyte-wide or word-wide

For signed integer (2s complement format), theMSB is used as the sign-bit (0 for positive, 1 for

negative) Signed 8-bit integer 127 to 128, For signed word 32767 to 32768 Latest microprocessors can also support 64-bit

or even 128-bit data In 8086, only integer operations are supported!!!

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.code ; indicate start ofcode segment

.startup ; indicate start of programmov AX, 0

mov BX, 0000H

mov CX, 0

mov SI, AX

mov DI, AX

mov BP, AX

END ; end of file

The flow of the program is usually top-down and

instructions are executed one by one!!!


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In general, an assembly program must include the code segment!!

Other segments, such as stack segment, data segment are not

compulsory

There are key words to indicate the beginning of a segment as

well as the end of a segment.Just like using main(){} in C++

Programming

ExampleDSEG segment data ; define the start of a data segment

DSEG ENDS ; defines the end of a data segment

Segment is the keyword DSEG is the name of the segment

Similarly key words are used to define the beginning of a program,

as well as the end.


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Example

CSEG segment codeSTART PROC FAR ; define the start of a program (procedure)

RET ; return

START ENDP ; define the end of a procedure

CSEG ends

End start ; end of everything

Different assembler may have different syntax for the definition

of the key words !!!!!

Startis just a name it could be my_prog, ABC etc


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Stacksg segmentpara stack

. ; define the stack segment

Stacksg ends

Datasg segment para ; declare data inside the data segment

Datasg ends

Codesg segmentpara code

Main proc far ;

assume ss:stacksg, ds: datasg, cs:codesg

mov ax, datasgmov ds, ax

.

mov ax, 4c00H

int 21H

Main endp

Codesg endsend main

End of everything


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To declare a segment, the syntax is:

segment_nameSEGMENTalignment class

Example StacksgsegmentPARA (this statement is used in previous slide)

PARA define the alignment of the segment base address, the segment with a starting

addressing that is evenly Divisible by 16. But the default value is also base address

divisible by 16 so the key word PARA can be ignored!


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data, code class entry. Is used to group

related segments when linking. The linkerautomatically groups segments of the sameclass in memory

PROC define procedures (similar to afunction) inside the code segment. Eachprocedure must be identified by an uniquename. At the end of the procedure, you

must include the ENDP

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FAR is related to program execution. When you request execution

of a program, the program loader uses this procedure as the entry

point for the first instruction to execute.

Assume to associate, or to assign, the name of a segment with a segment register

In some assembler, you need to move the base address of a segment directly into the

segment register!!!END ends the entire program and appears as the last statement. Usually the name of the

first or only PROC designated as FAR is put after END


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If you are doing something simple then you donot need to define the segment

Everything will be stored in the code segment

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start:

mov DL, 0H ; move 0H to DL

mov CL, op1 ; move op1 to CLmov AL, data ; move data to AL

step:

cmp AL, op1 ; compare AL and op1

jc label1 ; if carry =1 jump to label1

sub AL, op1 ; AL = ALop1inc DL ; DL = DL+1

jmp step ; jump to step

label1:

mov AH, DL ; move DL to AH

HLT ; Halt end of program

data db 45 ; define a variable called data

op1 db 6 ; define a variable called op1

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Emu8086 (http:// www.emu8086.com) there is a trial version but

it does not support all the features such as interrupt

The emu8086 consists of a tutorial and the reference for acomplete instruction set

Keil www.keil.com
http://www.emu8086.com/http://www.emu8086.com/


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Data is usually stored in the data segment

You can define constants, work areas (a chunk of memory )

Data can be defined in different length (8-bit, 16-bit)8-bit then use DB 16-bit then use DW

The definition for data:

[name] Dn expression ; Dn is either DB or DW

Name a program that references a data item by means of a name.

The name of an item is otherwise optional

Dn this is called the directives. It defines length of the data

Expression define the values (content) for the data


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FLDA DB ? ; define an uninitialized item called FLDA 8-bit

FLDB DB 25 ; initialize a data to 25

Define multiple data under the same name (like an array)

FLDC DB 21, 22, 23, 34 ; the data are stored in adjacent bytes

FLDC stores the first value

FLDC + 1 stores the second value

You can do mov AL, FLDC+3


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DUP duplicate

DUP can be used to define multiple storages

DB 10 DUP (?) ; defines 10 bytes not initializeDB 5 DUP (12) ; 5 data all initialized to 12

String :

DB this is a test

EQU this directive does not define a data item; instead, it definesa value that the assembler can use to substitute in other instructions

(similar to defining a constant in C programming or using the #define )

factor EQU 12

mov CX, factor


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Assembly language should be moreeffective and it will take up less memoryspace and run faster

In real-time application, the use of assemblyprogram is required because program that is

written in a high-level language probablycould not respond quickly enough

You can also put assembly codes into yourC++ program in order to reduce the

execution time!!!!

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The syntax for different microprocessor may bedifferent but the concept is the same so onceyou learn the assembly programming for one

microprocessor, you can easily program otherkinds of system

For example, programming the 8051 series isvery similar to the 8086

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Function of the addressing modes is toaccess the operands

Available modes (9 modes): registeraddressing, immediate addressing, directaddressing, register indirect addressing,

based addressing, indexed addressing,based indexed addressing, string addressing,and port addressing

Addressing modes provide different ways

of computing the address of an operand

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In c++, you can define an array, or a variable

int x[10], y, *z; Then to access different elements, you can do

Z = x ;

*(x+2);

x[0] = y

How this can be done using assembly languageprogramming? This is via different addressing

modes!!!!

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The operand to be accessed is specified asresiding in an internal register of the 8086

Eg MOV AX, BX

Move (MOV) contents of BX (the source

operand), to AX (the destination operand) Both operands are in the internal registers

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Pay attention to the value of IP and content of AX, BX


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Source operand is part of the instruction

Usually immediate operands represent constant dataThe operands can be either a byte or word

e.g MOV AL, 15

15 is a byte wide immediate source operand

Or it could be MOV AL, #15The immediate operand is stored in program storage

memory (i.e the code segment)

This value is also fetched into the instruction queue

in the BIUNo external memory bus cycle is initiated!


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Move a byte or word between a memorylocation and a register

the locations following the instructionopcode hold an effective memory address

(EA) instead of data The address is a 16-bit offset of the storage

location of the operand from the currentvalue in the data segment register

Physcial address = DS + offset The instruction set does not support a

memory-to-memory transfer!

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Data is assumed to be stored in the datasegment so DS is used in calculating thephysical address!!!

External memory bus cycle is needed to dothe read

Example of direct addressing: mov AL, var1

Where Var1 can be regarded as a variable

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Transfer a byte or word between a register and amemory location addressed by an index or baseregister Example MOV AL, [SI]

SI index register

The symbol [] always refer to an indirect addressing

The effective address (EA) is stored either in apointer register or an index register

The pointer register can be either base register BXor base pointer register BP

The index register can be source index register SI, or

destination index register DI The default segment is either DS or ES

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Eg MOV AX, [SI]

Value stored in the SI register is used as theoffset address

The segment register is DS in this example

Meaning of the above is to move the datastored in the memory location : DS + SI tothe AX register

In register indirect addressing mode, the EA(effective address) is a variable and dependson the index, or base register value

Eg mov [BX], CL

Which segment register will be used for the

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Address (in

HEX) Content

01236 1901235 1801234 2001233

According to the memory map

The result of the operation

Mov [BX], CL will result in what???

If CL = 88 and BX = 1233H and DS =0H

Physical address = DS + BX = 01233H

8086 assembly language programming i

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