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8086 Assembly Language Programming I Week 4 – Overview of the x86 assembly language prgoramming

Assembly language programming • Learning assembly language programming will help understanding the operations of the microprocessor • To learn: – Need to know the functions of various s – Need to know how external memory is organized and how it is addressed to obtain instructions and data (different addressing modes) – Need to know what operations (or the instruction set) are ed by the U. For example, powerful Us floating-point operations but simple Us only integer operations

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How to learn programming • • • •

C –Concept L – Logic thinking P – Practice Concept – we must learn the basic syntax, such as how a program statement is written • Logic thinking – programming is problem solving so we must think logically in order to derive a solution • Practice – write programs

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

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Flow of program development

Program .asm

Assembl e

Object file .obj

link

Executable file .exe

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Example • Machine code for mov AL, 00H • B4 00 (2 bytes) • After assembled, the value B400 will be stored in the memory • When the program is executed, then the value B400 is read from memory, decoded and carry out the task

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Assembly Program • Each instruction is represented by one assembly language 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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Example • In c++, you can do A = (B+C)*100 • In assembly language, only one instruction per statement A = B ; only one instruction - MOVE A = A+C ; only one instruction - ADD A = A*100 ; only one instruction - Multiply

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Format of Assembly language • General format for an assembly language statement • Label Instruction Comment • Start: Mov AX, BX ; copy BX into AX Start is a 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 9

8086 Software Model

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Software model • 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 an 8086, theoretically you should define the different segments!!! • To access the active segments, it is via the segment : CS (code), SS (stack), DS (data), ES (extra) • So when writing assembly language program, you must make use of the proper segment or index when you want to access the memory

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s • In assembly programming, you cannot operate on two memory locations in the same instruction • So you usually need to store (move) value of one location into a 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 s!!!!!

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Example • In C++ A = B+C ; A, B, C are variables • In assembly language A,B, C representing memory locations so you cannot do A = B+C – MOV AL, B ; move value of B into AL – ADD, AL, C ; do the add AL = AL +C – MOV A, AL ; put the result to A

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Data s • AX, BX, CX,and DX – these are the general purpose s but each of the s also has special function Example – AX is called the accumulator – to store result in arithmetic operations

• s are 16-bit but can be used as 2 8-bit storage • Each of the 4 data s 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 output operations

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Data • In based addressing mode, base BX is used as a pointer to an operand in the current data 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 be rotated or shifted by multiple-bit rotate • DX, data , is used in all multiplication and division, it also contains an input/output port address for some types of input/output operations

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Pointer and index s • 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 is employed when using the based addressing mode and is commonly used by instructions in a subroutine that reference parameters that were ed by using the stack

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Pointer and Index • Source index (SI) and Destination index (DI) are used to hold offset addresses for use in indexed addressing of operands in memory • When indexed type of addressing is used, then SI refers to the current data segment and DI refers to the current extra segment • The index s can also be used as source or destination s in arithmetic and logical operations. But must be used in 16-bit mode 17

Data types • Data can be in three forms: 8-bit, 16-bit, or 32-bit (double word) • Integer could be signed or unsigned and in bytewide or word-wide • For signed integer (2’s complement format), the MSB 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 64-bit or even 128-bit data • In 8086, only integer operations are ed!!!

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A sample program .code ; indicate start of code segment .startup ; indicate start of program mov 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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Assembly programming 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 Example DSEG 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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Assembly language programming Example CSEG segment ‘code’ START PROC FAR ; define the start of a program (procedure) RET START ENDP CSEG ends End start

; return ; define the end of a procedure ; end of everything

Different assembler may have different syntax for the definition of the key words !!!!! Start is just a name it could be my_prog, ABC etc

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More sample Stacksg segment para ‘stack’ …. ; define the stack segment Stacksg ends Datasg segment para …… ; declare data inside the data segment Datasg ends Codesg segment para ‘code’ Main proc far ; assume ss:stacksg, ds: datasg, cs:codesg mov ax, datasg mov ds, ax …. mov ax, 4c00H int 21H Main endp Codesg ends end main

End of everything

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Definitions To declare a segment, the syntax is: segment_name SEGMENT alignment class Example Stacksg segment PARA (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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Definition • ‘data’, ‘code’ – class entry. Is used to group related segments when linking. The linker automatically groups segments of the same class in memory • PROC – define procedures (similar to a function) inside the code segment. Each procedure must be identified by an unique name. At the end of the procedure, you must include the ENDP

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Definitions 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 In some assembler, you need to move the base address of a segment directly into the segment !!! 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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Syntax of a simple assembly language program • If you are doing something simple then you do not 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 CL mov 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 = AL –op1 inc 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 27

Assembler for 8086 Emu8086 (http:// www.emu8086.com) – there is a trial version but it does not all the features such as interrupt The emu8086 consists of a tutorial and the reference for a complete instruction set Keil – www.keil.com

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Defining data in a program 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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Examples for data 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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Example for data definition DUP – duplicate DUP can be used to define multiple storages DB 10 DUP (?) ; defines 10 bytes not initialize DB 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 defines a 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 Program

• Assembly language should be more effective and it will take up less memory space and run faster • In real-time application, the use of assembly program is required because program that is written in a high-level language probably could not respond quickly enough • You can also put assembly codes into your C++ program in order to reduce the execution time!!!!

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Assembly language programming • The syntax for different microprocessor may be different but the concept is the same so once you learn the assembly programming for one microprocessor, you can easily program other kinds of system • For example, programming the 8051 series is very similar to the 8086 33

Addressing Modes • Function of the addressing modes is to access the operands • Available modes (9 modes): addressing, immediate addressing, direct addressing, 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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Why addressing mode is important?

• 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 language programming? This is via different addressing modes!!!!

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addressing mode • The operand to be accessed is specified as residing in an internal 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 s

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

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Immediate addressing mode

Source operand is part of the instruction Usually immediate operands represent constan The 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, #15 The immediate operand is stored in program st memory (i.e the code segment) This value is also fetched into the instruction qu in the BIU No external memory bus cycle is initiated! 39

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Direct addressing mode • Move a byte or word between a memory location and a • the locations following the instruction opcode 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 current value in the data segment • Physcial address = DS + offset • The instruction set does not a memory-to-memory transfer! 42

Direct addressing • Data is assumed to be stored in the data segment so DS is used in calculating the physical address!!! • External memory bus cycle is needed to do the read • Example of direct addressing: mov AL, var1 • Where Var1 can be regarded as a variable

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indirect addressing mode • Transfer a byte or word between a and a memory location addressed by an index or base – Example MOV AL, [SI] – SI – index – The symbol [] always refer to an indirect addressing

• The effective address (EA) is stored either in a pointer or an index • The pointer can be either base BX or base pointer BP • The index can be source index SI, or destination index DI • The default segment is either DS or ES

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indirect addressing • Eg MOV AX, [SI] • Value stored in the SI is used as the offset address • The segment is DS in this example • Meaning of the above is to move the data stored in the memory location : DS + SI to the AX • In indirect addressing mode, the EA (effective address) is a variable and depends on the index, or base value • Eg mov [BX], CL • Which segment will be used for the above operation

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Address (in  HEX)

Content

01236

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01235

18

01234

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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

01233

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