chapter 3_ short circuit

8/17/2019 Chapter 3_ Short Circuit

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DFT 1113

COMPUTER

ORGANISATION

CHAPTER 3INSTRUCTION SET AND

ASSEMBLY LANGUAGEPROGRAMMING NOR ZAMIRABINTI OTHMAN

JTMK


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3.1 UNDERSTAND INSTRUC TION SET AND ASSEMB LY

LANGUAGE.

3.1.1 Define Instruction set, machine and assembly language.

3.1.2 Describe the addressing modes using proper instruction

format 3.1.3 Describe various types of addressing modes with example.

3.2 APPLY ASSEMB LY LANGUAGE .

3.2.1 Write a simple program in assembly language for:

a. arithmetic operation

b. logic operation

3.2.2 Use tools in analyzing and debugging assembly language

programs

LEARNING OUTCOMES


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Capability of processor determine the capability of thecomputer system. Therefore, processor is the keyelement or heart of a computer system.

Other than PC, microprocessors are used in variouscomputerized system such as industrial automation.

INTRODUCTION


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Message sent by programmers (program

instructions) understood by the computer and

vice versa.

Language that friendly to human alsounderstandable to computer systems.

Three level of language:

Low level – machine languageMiddle level – assembly language

High level – high-level language

3.1. UNDERSTAND THE INSTRUCTION SET

AND ASSEMBLY LANGUAGE.


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INSTRUCTION SET :

The complete collection of instructions that are understoodby a CPU

• Known also as Machine Code/Machine Instruction• Binary representation

• Usually represented by assembly codes

3.1.1 DEFINE INSTRUCTION SET, MACHINE

AND ASSEMBLY LANGUAGE


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Machine language is a binary program (or binary code).

Binary code is a sequence of instruction and operand in

binary that list the exact representation instruction as

they appear in computer memory.

Natural language of a particular computer system.

Strings of numbers or binary codes (0 or 1).

MACHINE LANGUAGE:


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Machine-dependent (differ from one µP to other µP.

Program written in any other language must be

translated to binary representation of instruction

before they can be executed by computer.

Programmers need to know specifically the

architecture of CPU.


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MACHINE LANGUAGE:


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The program is difficult to

understand

The program is long

and arduous

The program did not specify the

tasks

we perform require a computer.

The program is slow to key-

in to the computer

due

to the input one bit by one bit at a time

LIMITATIONS IN PROGRAMMING USING

MACHINE LANGUAGE


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To overcome these problems, the use of

easily remembered code (mnemonic) was

introduced. It is apply with the use of

Assembly Language


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Assembly language is a symbolic representation of a

machine language of specific processor. Assembly language

is a form that is very dependent on the underlying

architecture.

Using english-like abbreviations (MUL), ( ADD), (SUB)

Assembler as translator.

Assembler - Translate ordinary mnemonics such as MOVEData, Acc, into their corresponding machine language

(the only form of instruction that computer can

executed)

ASSEMBLY LANGUAGE


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Assembly Process - Process of translating an assembly

language programs into a machine language programs.

The assembly process is straightforward (one -to-one

mapping of assembly languages statements to their

machine language counterpart) and rather simple

Programmer need to know the basic architecture.

ASSEMBLY LANGUAGE (AL)


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It clarifies the execution of instruction.

It shows how the data represent in memory.

It show how a program interacts with the operating system

(OS), processor and, input and output (I/O) system.

It clarify how a program access external devices

Sometimes difficult to access hardware drive and system

tool if using (register transfer languages) RTL

WHY ASSEMBLY LANGUAGE STUDY IS

IMPORTANT?


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Easy to make compilers, debuggers and other device tools.

Allow accessing information that is not accessible

(restricted) from high level languages.

More function library that can be used in programming

development.

Possibility to make library function that compatible withdifferent compiler and operating system

ADVANTAGES USING ASSEMBLY

LANGUAGES


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(Development) Requires longer development time

(Reliability & Security) Easy to makes error

(Debug) More possibility for errors

(Maintain) Difficult to modify because it allows

unstructured code.

(Portability) Difficult to porting to different

platforms.

DISADVANTAGES USING ASSEMBLY

LANGUAGES


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Using everyday English and common mathematical

notation. (x = I + j)

Overcome problems : assembly language require manyinstruction to accomplish a simple task.

Single instruction in HLL = several AL instructions.

Compiler as translator.

Programmers do not need to know the architectu re of CPU.

HIGH LEVEL LANGUAGE (HLL)


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THREE DIFFERENT LEVELS OF LANGUAGE


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COMPARISON OF INSTRUCTIONS OF

THREE DIFFERENT LANGUAGES


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COMPARISON OF INSTRUCTIONS OF

THREE DIFFERENT LANGUAGES


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The term microprocessor typically refers to the central processing unit

(CPU) of a microcomputer, containing the arith metic logic unit (ALU)

and the control units (CU).

Performs the main tasks for the comp uter system

data transfer between (registers) and memory or I/O

arithmetic and l ogic operations. Example: ADD, SUB, AND, OR, etc

program flow via s imple decisions. Example: Zero, Sign, Carry, etc

MICROPROCESSOR OVERVIEW


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BIT

binary digit

hold only two values: 0 or 1

Smallest unit of data

1 bit : the smallest.

4 bit : nibble.

8 bit : byte; indicated (.B) notation

16 bit : word; indicated (.W) notation

32 bit : long word; indicated (.L) notation

DATA SIZES


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For the data which has more bits, it is divided into dual-half portions upper (MSD section)

lower portions (LSB section).

1 Byte consists of two nibbles; upper nibble

lower nibble.

1 Word consists of two bytes;

upper byte lower byte.

1 Long Word consists of two word; upper word

lower word.

DATA SIZES


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Can store either

Longword

Word

Byte

DATA SIZES

1 0 0 1 1 0 0 1 1 1 0 1 0 1 1 1 0 1 0 1 0 1 1 1 0 1 0 0 0 1 0 1

0 1 0 1 0 1 1 1 0 1 0 0 0 1 0 1

0 1 0 0 0 1 0 1


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NOTATIONS FOR INFORMATION

REPRESENTATION

64 bits

8 bytes

2 words

1 doubleword

Q: How do we number these various units of information in a consistent manner?

9 6 2 1 7 6 6

Most Significant Digit (MSD) Least Significant Digit (LSD)

1 2 3 4 5 6

“Big End” -ian Numbering

6 5 4 3 2 1

“Little End” -ian Numbering


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NOTATIONS FOR INFORMATION

REPRESENTATION


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A single cell sized 1 bit can store either logic -0 or logic-1.

Thus the range of data is 0 – 1 .

Data size: n = 1

Data capacity : 2n = 21 = 2

Range : 0 – 1

A single cell sized 4 bit (Nibble) can store 16 possible

situations

Data size: n = 4Data capacity : 2n = 24 = 16

Range : 0 - 15

DATA SIZE


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DATA SIZE

Data

size

N

( in bit)

Data type Data capacity

2n

Range

1

Bit

2

0 – 1

4 Nibble 16 0 - 15

8

Byte

256

0 – 255

16

Word

65536

0 – 65535

32

Long Word

4,294,967,296 0 -

4,294,967,295


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Internal Register – theworking element of µp.

MC68000 – an internal 32-

bit processor Register Set

8 data register (D0-D7)

7address register(A0-A6)

2 stack pointer(USP&SSP) 1 program counter (PC)

1 status register (SR)

REGISTER SET OF MC68000


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DATA TYPE : BINARY


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DATA TYPE : OCTAL


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DATA TYPE : HEXADECIMAL


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DATA TYPE : DECIMAL


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ASCII CHARACTER


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DATA TYPE

%binary number

@

octal number$hexadecimal number

No sign decimal number

Character ASCI code


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3.1.2 DESCRIBEADDRESSING MODES

USING PROPER

INSTRUCTION FORMAT


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There are many MC68000 instr uctions. However, thesesinstructions comply to a standard instruction format:

INSTRUCTION FORMAT


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INSTRUCTION FORMAT

CommentsLabel>

CommentsLabel>

CommentsLabel>


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LABEL – pointer to the instruct ion’s memory location

OPERATOR -operation code ( ie;MOVE,ADD…)

OPERAND – Depends on the operator, may has more

than one operand. COMMENT – Explanation about the execution of

instruction.

Data format - Byte | Word | Longword | Sign-extended

INSTRUCTION FORMAT


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INSTRUCTION FORMAT


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Label are optional

Label – a user-defined symbol representing theaddress associated with the instruction.

Up to 8 alphanumeric, begin with letter, terminated bya space.

LABEL


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Opcode is an opera tion code that will be executed by

microprocessor

It is abbreviation of English- like word for an operation

For example MOVE – Transfer data from source to destination

ADD – Add two binary numbers stored in registers

Associates with the size of data to be execute

Can be either Byte: .B

Word: .W

Longword: .L

OPCODE/OPERATION CODE / MNEMONIC


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OPERANDS


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Comment is optional

Comment provide readability for future reference

Explain the operation of each instruction

Any line beginning with an * is a comment. Any line beginning with a ; (semi -colon) is a comment.

Example ;

COMMENT

org $1000 ; set the location counter to $1000


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mnemonics are postfixed with symbol ".B", ".W", ".L".

DATA FORMAT


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DATA FORMAT


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DATA FORMAT


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DATA FORMAT


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DATA FORMAT


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DATA FORMAT


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3.1.3 DESCRIBE

VARIOUS TYPES OFADDRESSING MODES

WITH EXAMPLE.


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Data transfer Group

Arithmetic Group

Logical Group

Shift and rotate Group

Branch & Subroutine

Stack & Queue

CLASSIFICATION OF INSTRUCTION SETS


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These instructions transfer or move data between its

internal registers, between an internal register and a

storage location in memory, or between two locationsin memory.

- memory to register

- register to memory

- register to register

- memory to memory

DATA TRANSFER GROUP


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DATA TRANSFER GROUP


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Byte(B), Word(W) and Long word(L).

R R , R M , M M , EA SR , USP An

Example 1:

MOVE.B #$72,D1

Before : D1 = $00200500 After : D1 = $00200572

Example 2:

MOVE.B D0,D1 Before : D1 = $00200500 , D0 = $00002222

After : D1 = $00200522 , D0 = $00002222 Example 3:

MOVE.W $3000,D1 Before : D1 = $00200500 , After : D1 = $00203243 ,

MOVE (SOURCE TO DESTINATION)

# indicate actual

value$ indicate

hexadecimal value

$3000 32

$3001 43

$3002 98


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Example 4 :

MOVE.W D6,$4000

Before: D6 = $AB206541 ,

After : D6 = $AB206541

Example 5:

MOVE.W D2,(A1)

Before : D2 = $AB206541 ,

A1 = $00000501 After : D2 = $AB206541 ,

A1 = $00000501

MOVE (SOURCE TO DESTINATION)

$4000 3254

$4002 4377

$4004 9868

$4000 32

$4001 54

$4002 43

$4000 6541

$4002 4377

$4004 9868

$4000 65

$4001 41

$4002 43

$501 65

$502 41

$503 FF

$501 FF

$502 FF

$503 FF

before

After

Before After


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Provides a means of initializing an address register.

Only a word or longword operand is allowed to betransferred into the specified address register.

For a word operation, the source operand is sign -extended before being loaded into the addressregister.

Example 5:

MOVEA.W D6,A2Before : D6 = $AB206541 , A2 = $ABCD1234

After : D6 = $AB206541 , A2 = $00006541

MOVEA (MOVE ADDRESS)


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Example 6:

MOVEA.W $8C00,A0

Before : A0 = $FFFFFFFF

After : A0 = $FFFF8C00

Example 7:

MOVEA.W D6,A2

Before : D6 = $AB20A541 , A2 = $ABCD1234

After : D6 = $AB20A541 , A2 = $FFFFA541

Example 8:

MOVEA.L D6,A2

Before : D6 = $AB206541 , A2 = $ABCD1234

After : D6 = $AB206541 , A2 = $AB206541

MOVEA (MOVE ADDRESS)


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• A short form of the move instruction for transferringan immediate operand to a data register .

• The immediate operand is limited to the range of

-128 to 127.

• The operation size is implied to be longword.

• Therefore the 8-bit immediate operand is sign-extended to 32 bits before being loaded into thedestination, which must be a data register.

MOVEQ (MOVE QUICK)


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MOVEQ (MOVE QUICK)


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Example 1:

MOVEQ #$04,D3 Before : D3 = $AB206541After : D3 = $00000004

Example 2:

MOVEQ #$80,D3Before : D3 = $AB206541After : D3 = $FFFFFF80

Example 3:

MOVEQ #-3,D2Before : D2 = $AB206541After : D2 = $FFFFFFFD

MOVEQ (MOVE QUICK)


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Transfer words or longwords between a register listand consecutive memory locations.

In the case of a word transfer to the registers, eachmemory word is sign-extended before being loadedinto the respective register.

MOVEM (MOVE MULTIPLE REGISTERS

MOVEM (MOVE multiplr registers)


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Example 1:

MOVEM.W D2/D5/D0,$200

Before : D0 = $11112222 ,D2 = $33334444 ,

D5 = $55556666 ,

After : D0 = $11112222 ,

D2 = $33334444 ,

D5 = $55556666 ,

Example 2 :

MOVEM.L D2/D5/D0,-(A4)

Before : D0 = $11112222 , D2 = $33334444 ,

D5 = $55556666 , A4 = $0000050C

After : D0 = $11112222 ,

D2 = $33334444 ,

D5 = $55556666 ,

A4 = $00000500

$200 6541$202 4377

$204 9868

$200 4444

$202 6666$204 2222

$500 451287A4

$504 54A2F221

$508 846AC1DD

$50C 45123211$500 33334444

$504 55556666

$508 11112222

Predecrement (tolak dulu)B : – 1 , W : – 2 , L : – 4

Dlm contoh ada 3 daftarmaka 3x4=12 ,50C – C = 500




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Example 3 :

MOVEM.W $200,D2/D5/D0

Before : D0 = $11112222 ,D2 = $33334444 ,

D5 = $55556666 ,

After : D0 = $FFFF9868 ,

D2 = $00006541 ,

D5 = $00004377 ,

Example 4 :

MOVEM.L (A4)+,D2/D5/D0

Before : D0 = $1111222 , D2 = $33334444 ,

D5 = $55556666 , A4 = $00000500

After : D0 = $846AC1DD ,

D2 = $451287A4 ,

D5 = $54A2F221 ,

A4 = $0000050C

$200 6541$202 4377

$204 9868

$200 6541

$202 4377$204 9868

$500 451287A4

$504 54A2F221

$508 846AC1DD

$50C 45123211$500 451287A4

$504 54A2F221

$508 846AC1DD

Posincrement (tmbh kmdn)B : + 1 , W : + 2 , L : + 4

Dlm contoh ada 3 daftarmaka 3x4=12 ,500 + C = 50C



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EXG (EXCHANGE)

• EXG interchanges the contents of two

registers.

• operand size for EXG is longword

Example 1:

EXG D1,D5

Before : D1 = $11223344 , D5 = $66778899After : D1 = $66778899 , D5 = $11223344


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SWAP

• SWAP exchanges the lower word of the specifieddata register with its upper word.

• the operand size for SWAP is word.

Example 1 SWAP D1 Before : D1 = $11223344 After : D1 = $33441122

Example 2

SWAP D1 Before : D1 = $1234567 After : D1 = $45670123


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Transfers the source operand address rather than its content

to the destination address register.

Therefore:

MOVEA.L OPER,A1 is equivalent toLEA OPER,A1

LEA (LOAD EFFECTIVE ADDRESS)

OPER EQU $2001

LEA OPER,A1


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Loads the target with 0

Example 1:

CLR.B D1

Before : D1 = $11223344

After : D1 = $ 11223300

CLR


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ARITHMETIC GROUP

• Similarly as the operation of mathematic,

addition (+), subtraction (-), multiplication (x),

division (÷)

• Binary arithmetic instructions performs signed

and unsigned operations.


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ARITHMETIC GROUP


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ADD (ADDITION)

• 68000 MP allows addition of 3 data size i.e 8 bit (.B),16 bit (.W), and 32 bit (.L)

• Data is accessed via register (D0 to D7), memory,absolute data and I/O port.

• < Destination> +

• Both can be data registers

• At least one must be a data register


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Example 1:

ADD.L D0,D1

Before : D1 = $00200500 , D0 = $00002222After : D1 = $00202722 , D0 = $00002222

Example 2:

ADD.W $3000,D1Before : D1 = $00200500 ,

After : D1 = $00203743 ,

ADD (ADDITION)

$3000 32

$3001 43

$3002 98


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Example 3:

ADD.B D0, D1

ADD (ADDITION)

.B

HH

22

+ 44

66


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$4000 3254

$4002 4377

$4004 9868

$4000 32

$4001 54

$4002 43

$4000 9795

$4002 4377

$4004 9868

$4000 97

$4001 95

$4002 43

ADD (ADDITION

Example 4:

ADD.W D6,$4000

Before: D6 = $AB206541 ,

After : D6 = $AB206541 ,


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“ADDI.W #$72,D1” and

“ADD.W #$72,D1” are equivalent

But “ADDI #” can add to a

memory location

ADDI (ADD IMMEDIATE)

Add an immediate value

Cannot be done using ADD as one operand must

be a data register

• Example :ADDI.W #$1234,(A0)

Example 1:

ADDI.W #$72,D1Before : D1 = $00200500After : D1 = $00200572


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• ADDA – destination of add is an address register

Example 1:

ADDA.W D6,A2 Before : D6 = $AB206541 , A2 = $ABCD1234 After : D6 = $AB206541 , A2 = $ABCD7775

Example 2:

ADDA.L D6,A2 Before : D6 = $AB20A541 , A2 = $ABCD1234 After : D6 = $AB206541 , A2 = $56EDB775 , C = 1

Example 3:

ADDA.W #$8122,A0 Before : A0 = $FFFFFFFF After : A0 = $FFFF8122

ADDA (ADDITION)


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1 until 8

ADDQ (ADDITION

• ADDQ – Add a constant (1 to 8) to a memorylocation, or register. It is executed faster.

• ADDQ #4, D1• Speed is faster than ADD #4, D1

Example 1:

ADDQ.B #$04,D3 (ADDQ #XXX,Dn)Before : D3 = $AB206541After : D3 = $AB206545


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ADDX (ADD EXTENDED)

• Add source and destination plus contents of the Xbit of the condition code register

• Both source and destination must be dataregisters

Example 1:

ADDX.B D2,D3Before : D2 = $AB206541 , D3 = $ABCD1234 , X = 1

After : D2 = $AB206541 , D3 = $AB206576 , X = 1


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SUB (SUBTRACTION)

< Destination> -

• SUB, SUBA, SUBQ, SUBI, SUBX – the

subtractions equivalents of ADD

Example 1:

SUB.W #$80,D3Before : D3 = $AB206541

After : D3 = $AB2064C1


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Example 2:

SUBI.L #10,D2

Before : D2 = $AB206541

After : D2 = $AB206537

Example 3:

SUBQ.W #@11,D3 (SUBQ #XXX,Dn)

Before : D3 = $AB206541

After : D3 = $AB200038

SUB

-128 (80) until 127 (7F)


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Example 1:

MULU #2,D2 ;16 bit x 16 bit = 32 bit

Before : D2 = $AB206541

After : D2 = $0000CA82Example 2:

MULU #2,D2

Before : D2 = $AB20FFFF

After : D2 = $0001FFFE

MULU

@ octal no.% binary no.

No sign decimal no.


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Example 1:

DIVU #2,D3 ;32 bit / 16 bit = 16 bit lower (result)

16 bit upper ( remainder)

Before : D3 = $00006541

After : D3 = $000132A0

Example 2:

DIVU #$1234,D0 ;32 bit / 16 bit = 16 bit lower(result)

16 bit upper ( remainder)

Before : D0= $00005678

After : D0 = $0DA80004

DIVU

6541F = 25921 32A016

= 1296010


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NEG – Take the 2’s complement of target

Example 1:

NEG.W D3

Before : D3 = $AB206541

After : D3 = $AB209ABF

NEG

NEG = 2’s complement

$41 = 01000001Invert 10111110

+ 1

$BF = 10111111


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Similar to the logic gate, logical operations

involve AND, OR, and NOT.

The content of the register and memory arecompared.

The operation is implemented in ALU.

LOGICAL GROUP


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LOGICAL GROUP


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Example 1:

AND.B $3E, D3 [D3(B) AND $3E D3 (B) ]

(8 bit data in data register D3 is AND-ed with 8 bit

absolute data, and the product is stored in D3)

AND


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Example 1:

OR.B D0, D1 [D1 (B) or D0 (B) D1(B) ]

8 bit data in data register D1 is OR-ed with 8 bit data in data

register D0, and the product is stored in D1)

OR

NOT


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Example 1:

NOT.B D1 [D1(B) NOT D1 ]

(content in D1 is NOT, and the product is stored back

to D1)

NOT


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chapter 3_ short circuit

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