Chapter 3 ASSEMBLY LANGUAGE PROGRAMMING

7/28/2019 Chapter 3 ASSEMBLY LANGUAGE PROGRAMMING

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TOPIC 3

ASSEMBLY LANGUAGE PROGRAMMING


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At the end of the class youshould :

Understand assembly language programming

Compare high, assembly and machine language

Describe important terminologies

Determine four fields of an assembly languageinstruction


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Differentiate between Machine, Assemblyand High Level Languages

Machine Language Assembly Language High Level Language

Basic programminglanguage

Used for thecommunication betweencomputer and human

Used for thecommunication betweencomputer and human

Were developed inwhich single statementscould be written toaccomplish substantialtasks

The translator programsthat convert middle levelprograms into machinelanguage are calledassembler

The translator programsthat convert high levelprograms into machinelanguage are calledcompilers


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Differentiate between Machine, Assemblyand High Level Languages

Machine Language Middle Language High Level Language

Machine language,programmer s writeinstructions using binarycode

Middle language allowprogrammers to writeinstructions usingmnemonic

High level languagesallow programmers towrite instructions thatlook like everyday

English and containcommonly usedmathematical notations

Binary (0,1) Assembly language Ex : Basic, Pascal, C etc

Programmer need to

know specifically thearchitecture of CPU

Programmer need to

know basic architectureof CPU such as registers

Programmer do not

need to know thearchitecture of CPU torealize the high levellanguage programming


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Important Terminologies :

Machine code

the binary language understood by themicroprocessor

Mnemonic Source files

contains all the instruction mnemonics needed to

execute a program


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Object files, list file and hex file object file The source file is converted into an object file containing

the actual binary information the machine will understand

by a special program called an assembler list file contains all the original source file text plus the additional

code generated by the assembler

hex file contains printable object code in a text format

Hex files (called S-record files by Motorola) are useful fordownloading over serial lines.


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Assembler

Single-line assembler

Work with one source line at a time and arerestricted in operation.


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Cross-assembler

Cross-assemblers are programs written in onelanguage, such as C, that translate source

statements into a second language: the machinecode of the desired processor

Pseudo-opcode


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Assembly Language Format

Syntax : [label[:]] [mnemonic[operands]][;comments]

LABEL Labels (or symbols) are optional and are used to

identify a particular statement so that it can bereferred to from other parts of the program

MNEMONIC The mnemonic defines the function of a particular line

and it can either be one of the 68000 instructionmnemonics or it may be an assembler directive or

pseudo-instruction.

LABEL MNEMONIC OPERANDS COMMENTS


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OPERANDS

Operand may be a 68000 register name, anaddress or data or a label referring to either an

address or data When two operands are required, the first

operand specifies the destination for the result ofthe operation and the second operand specifies

the source of the data for the operation


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COMMENTS

Comments are used to describe the function ofthe program

Any text preceded by a semi-colon(;) or asterisk(*) is ignored by the assembler but will beincluded as a comment in the listing of theprogram


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Assembler Directives

Assembler directives statements give theassembler information that affects the assemblyprocess, but do not necessarily cause object codeto be generated

ORG sets the ORIGIN address this is the firstmemory location to be used for storage of theprogram

EQU used to define (or EQUATE) the value of alabel

DC Define Constant used to store a byte (orb tes of data in memor


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Assembler Directives

DS Define Storage used to reserve memorylocations for use by the program

ALIGN make the program counter even

EVEN - make the program counter even

TCALL allows calls to the operating systemusing the 68000 TRAP instruction


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Understand instruction set and addressingmodes

Identify groups of instruction sets of the 68000

microprocessor. Write assembly language using each group of

instruction set.

Use "DEBUG" program to debug assembly

language programs.


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4 groups of instruction sets of68000 microprocessor

Data Movement/Data Transferinstruction set

Arithmetic instruction setLogic instruction set

Shift and Rotate instruction set


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Move instru ct ion

Use to copy/transfer data from source todestination

Transfer can be done between register andmemory

Instruction formatMOVE.s source, destination

s is data size. For MOVE the size are: B(byte)

W(word)

L(longword)


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Move instru ct ion

Use to copy/transfer data from source todestination

Transfer can be done between register andmemory

Instruction formatMOVE.s source, destination

s is data size. For MOVE the size are: B(byte)

W(word)

L(longword)


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Data Movement/Data Transfer

instruction set


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instruction set


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instruction set


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instruction set


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instruction set


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instruction set


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instruction set


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/ f


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D M /D T f


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D t M t/D t T f


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D t M t/D t T f


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D t M t/D t T f


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D t M t/D t T f


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D t M t/D t T f


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D t M t/D t T f


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Addressing Modes


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Addressing Modes


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Arithmetic instruction set

ADDITION ADD EA, Dn (EA) + Dn Dn

allow Byte, Word and Long Word

data is accessed via register D0 D7, memory,

absolute data and I/O Port


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SUBTRACTION SUB EA, Dn Dn - (EA) Dn


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MULTIPLICATION (MULU, MULS) S (16 bit) x D (16 bit) D (32 bit)


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DIVISION (DIVU, DIVS) D (32 bit) / S (16 bit) D [16 bit / 16 bit]

= D [Remainder / Quotient]


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Logic instruction set

similar to the logic gate, logical operationsinvolve AND, OR and NOT

the content of the register and memory arecompared

the operation is implemented in ALU


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Logic instruction set


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Shift Instructions :Arithmetic (signed) (0 to 127, -1 to -128)

Logical (unsigned) (0 to 255)

Rotate Instructions :

Ordinary Rotate

Rotate through X Flag


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Shif d R i i


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Shift d R t t i t ti t


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At the end of the class you


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List types of addressing modes.

Code assembly language instructions usingeach addressing mode.


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What is Addressing Modes?

Addressing Modes is :

is the techniqueused to fetch the desiredoperand during the execution of an instruction.

the location of the operand(s) on which toperform the function

The MC68000 supports 14


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The MC68000 supports 14different addressing modes Data register direct Address register direct

Absolute short

Absolute long

Register indirect

Post-increment register indirect Pre-decrement register indirect

Register indirect with offset

Register indirect with index and offset

PC-relative with offset

PC-relative with index and offset Immediate

Immediate quick

Implied register


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Addressing Mode summary


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Data reg ister d irect

Description

data register (D0-D7) is the source or destination ofdata

Example since .B is appended to MOVE, only the low byte

of the destination data register is affectedINSTRUCTION MOVE.B D0,D3MEMORY REGISTER

ADDRESS CONTENTS NAME CONTENTS

BEFORED0

D3

1020 4FFF

1034 F88A

AFTERD0

D3

1020 4FFF

1034 F8FF


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Add ress register direct

Description

address register (A0-A7) is the destination of data

only word or longword operands may be specified

A word operand is sign-extended to fit the register Example

The contents of A3 are copied onto A0INSTRUCTION MOVEA.B A3,A0

MEMORY REGISTER


BEFOREA0

A3

0020 0000

0004 F88A

AFTERA0

A3

0004 F88A

0004 F88A


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Absolute sho r t

Description source or destination is a memory location whose address is

specified in one extension word of the instruction

bits 16-23 of the full address are obtained by sign-extension of the16-bit short address

Example

The source is immediate, destination is absolute short address

Since operation is .w, source is sign-extended to two bytes

INSTRUCTION MOVE.W #$1E, $800

MEMORY REGISTER


BEFORE000800

000801

12

34

AFTER

000800

00080100

1E


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Absolute long

Description

source or destination is a memory location whose address is specified in twoextension word of the instruction

bits 16-23 of the full address are obtained by sign-extension of the 16-bit short address

Example

The source is immediate, destination is absolute long address Since operation is .b, only one byte of memory is changed

short address

Note

Often, the distinction between abs-short and abs-long is transparent to the

programmer due to the use of labelsINSTRUCTION MOVE.B #$1E, $8F00

MEMORY REGISTER


BEFORE 08F000 FF

AFTER08F000

1E


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Reg ister ind irect

DescriptionAn address registercontains the address of the

address of the source or destination operand

Example The instruction moves a longword stored in D0 to

the memory location specified by he address in A0INSTRUCTION MOVE. L D0, (A0)

MEMORY REGISTER


BEFORE

001000

001001

001002

001003

55

02

3F

00

A0

D0

00001000

1043834F

AFTER

001000

001001

001002

001003

10

43

83

4F

A0

D0

00001000

1043834F


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Pos t-inc rement reg ister ind irect

Description Indicated by a + sign after (Ai)

afterreading or writing data the address register is incrementedby the number of bytes transferred

Byte : [Ai] [Ai] + 1 Word : [Ai] [Ai] + 2 Longword : [Ai] [Ai] + 4

INSTRUCTION MOVE. W (A5)+, D0

MEMORY REGISTER


BEFORE

001000001001

001002

001003

4567

89

AB

A5

D0

00001000

0000FFFF

AFTER

001000

001001

001002

001003

45

67

89

AB

A5

D0

00001002

00004567


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Pre-decrement regis ter ind irect

Description Indicated by a - sign before (Ai)

before reading or writing data the address register isdecremented by the number of bytes transferred

Byte : [Ai] [Ai] - 1 Word : [Ai] [Ai] - 2 Longword : [Ai] [Ai] - 4

INSTRUCTION MOVE. W D0,-(A7)

MEMORY REGISTER


BEFORE

001000001001

001002

001003

1012

83

47

A7

D0

00001002

00000143

AFTER

001000

001001

001002

001003

01

43

83

47

A7

D0

00001002

00000143


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Reg ister ind irect w ith o ffset

Description A variation of register indirect that includes a 16-bit signed offset

(displacement) as an extension word in the instruction

The sign-extended offset is added to the address register to form

the effective address of the source or destination Example

Effective address is 6 plus address register

Value stored in the address register does not changeINSTRUCTION MOVE. W 6 (A0),D0

MEMORY REGISTERADDRESS CONTENTS NAME CONTENTS

BEFORE001026

001027

07

BF

A0

D0

00001020

00000000

AFTER001026

001027

07

BF

A0

D0

00001020

000007BF

Register indi rect w i th index and


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Register indirect w i th index and

offset

Description Another variation of register indirect. An index is used as well as

an 8-bit signed offset

The effective address is formed by adding the sign-extended

offset, the contents of the index register and the contents of theaddress register

Example

=$10+$100A+$2=$101CINSTRUCTION MOVEA $10(A0,D0.L),A1


BEFORE00101C

00101D

EF

10

A0

A1

D0

0000100A

00000000

00000002

AFTER00101C

00101D

EF

10

A0

A1

D0

0000100A

FFFFEF10

00000002

PC l t i i th f f t


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PC-relat ive w ith o ffset

Description a 16-bit offset is added to PC to form effective address

only source operand can be addressed this way

this mode provides position-independent code

assembler computes offset by subtracting PC from label

INSTRUCTION MOVE.W COUNT(PC),D5

MEMORY REGISTER


BEFORE001026

001027

AB

CD

PC

D5

00001000

12345678

AFTER001026

001027

AB

CD

PC

D5

00001004

1234ABCD

00001000 1 ORG $1000

00001000 3A3A 0FFE 2 MOVE.W COUNT (PC), D5

00002000 3 ORG $2000

00002000 ABCD 4 COUNT DC.W $ABCD

00002002 5 END

$0FFE = $2000-$1000-$2

PC l t i i th i d d ff t


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PC-relat ive w ith index and o ffset

Description an 8-bit signed offset plus and index register are used to

compute the address relative to the PC

INSTRUCTION MOVE.B TABLE(PC,D0.W),D0

MEMORY REGISTER


BEFORE 001015 19PC

D0

0000100A

ABCD0005

AFTER 001015 19PC

D0

0000100E

00000019

00001000 1 ORG $1000

00001000 303C 0005 2 MOVE.W #5,D0

00001004 6100 0004 3 BSR SQUARE00001008 4E75 4 RTS

0000100A 103B 0004 5 SQUARE MOVE.B TABLE (PC,D0.W),D0

0000100E 4E75 6 RTS

00001010 000104091019 7 TABLE DC.B 0,1,4,9,16,25

00001016 8 END

I d i t


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Immediate

Description Immediate address uses two extension words to hold the source

operand

Data may be expressed in :

decimal (& prefix or none) hexadecimal ($ prefix)

octal (@ prefix)

binary (% prefix)

ASCII (string within )INSTRUCTION MOVE.B #$1FFFF,D0

MEMORY REGISTER


BEFORE D0 12345678

AFTER D0 0001FFFF

I d i t i k


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Immed iate qu ick

Description Immediate quick addressing is an optimized case of immediate

addressing whose binary code fits in one word (including theoperand)

Immediate operand is sign-extended to fit the 32-bit destination Available with the following instructions

MOVEQ (operand must be a 8-bit signed integer)

ADDQ (operand must be in the range 1 to 8)

SUBQ (operand must be in the range 1 to 8)INSTRUCTION MOVEQ #$1F,D0


BEFORE D0 12345678

AFTER D0 000001F



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yshould :

Explain the operation of stacks and queuesaccording to LIFO and FIFO concept.

Flag Register / Stat s register


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Flag Register / Status register

One of the registers in the MC68000microprocessor is status register.

It is a 16 bits register. Each bit has it own functionwith some of the bits unused.

The status register has 16 bits and is divided intothe system byte and user byte.

The user byte contains five condition flags. Theremaining 3 bits in the user byte are not used and

remain zero. The condition flags contain information on the result

of the last processor operation.



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Carry Flag (C)


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Carry Flag (C)

C = 1 when out o f the MSBC = 1 when bo rrow to MSB

C = 0 when no ou t or borrow to MSB

Ex1 : move.b #$EE,d0move.b #$70,d1

add.b d0,d1

Ex2 : move.b #$F0,d0move.b #$EE,d1sub.b d0,d1

Negative Flag (N)


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Negative Flag (N)

N = 1 when MSB = 1N = 0 when MSB = 0

Ex1 : move.w #$7000,d0

move.w #$4000,d1add.w d0,d1

Ex2 : move.w #$2000,d0move.w #$4000,d1add.w d0,d1

Zero Flag (Z)


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Zero Flag (Z)

Z = 1 when resu lt is zeroZ = 0 when result is non -zero

Ex1 : move.b #$40,d0

move.b #$70,d1sub.b d0,d1

Ex2 : move.b #$70,d0move.b #$70,d1sub.b d0,d1

Overflow Flag (V)


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Overflow Flag (V)

V = 1 when abnormalV = 0 when normal


move.b #$40,d1

add.b d0,d1


move.b #$40,d1

add.b d0,d1


move.b #$D0,d1

sub.b d0,d1

Extend Flag (X)


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Extend Flag (X)

X = C

except for : i) movement (move)

ii) logic (AND)

iii) comparison (CMP)

Example :


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Example :

If initial data for Status Register (SR) = $A71Fand

D2 = 6677 8899, what are the end data in SR

andD2 after ADDI.B #$5E,D2 instruction been

executed.

Show how each bit of flags obtainedBits SR T 0 S 0 0 I2 I1 I0 0 0 0 X N Z V CInitial A71F 1 1 1 0 0 1 1 1 0 0 0 1 1 1 1 1End ? 1 1 1 0 0 1 1 1 0 0 0 ? ? ? ? ?



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yshould :

Explain the operation of stacks and queuesaccording to LIFO and FIFO concept.

What is STACK ?


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What is STACK ?

Stack is a memory-based structures

a STACK is an area in memory reserved forreading and writing special data items such as

return addresses and register values.

Ex :

i) move.l D3,-(A7)

Entire contents of D3 are written into the stack areapointed to by A7. A7 is automatically decremented by4 during execution.

Item previously pushed onto the stack

Cont


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Cont

Ex : ii) move.l (A7)+,D3

Stack memory is read out into D3 and A7 isautomatically incremented by 4 during execution.

items popped off the stack (read out of memor)

STACK Characteristic


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STACK Characteristic

One characteristic of a stack is that the last itempushed is always the first item popped.

Stack are commonly referred to as a LIFO (last

in, first out) structure.

LIFO ILLUSTRATION


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LIFO ILLUSTRATION

A stack of trays.

The last tray to beplaced on top is

also the first to betaken off the top.

QUEUE


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QUEUE

Stack is a memory-based structures

In a QUEQU, the first item loaded is the firstitem to be removed.

Queues are referred to as FIFO (first in, firstout).

FIFO ILLUSTRATION


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FIFO ILLUSTRATION

Personsstanding inline.

Personsleave thequeue in theorder they

arrive.

Stack Pointer


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Stack Pointer

Stack PointerA stack pointer is used to point to thelast stack location used.

Cont


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Cont

PUSH (input) Data PUSH into stack (Stack SAVE) as data are pushed into the stack, the stack grows

towards the lower address

POP (output) Data POP from stack

moving towards higher address

Examples :


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Examples :

Question 1 :

Stack Pointer contains 00003000H

move.w #$1234, -(SP)

SOLUTION :

location 0000 2FFEH contains12H

location 0000 2FFFH contains34H

Stack Pointer = 0000 2FFEH

SP Address Stack

SP-2 00002FFE

12

SP-1 0000 2FFF 34

SP 0000 3000 X

Examples :


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Examples :

Question 1 :

Stack Pointer (A7)=0000 4200H,

D0 = 1122 3344

move.w (SP)+,D0

Has the following contents

SOLUTION :

After D0 = 1122 5634 Stack Pointer = 0000 4202H

SP Address Stack

0000 41FE 9A

0000 41FF 78

0000 4200 56

0000 4201 34

0000 4202 12

SP Address Stack

0000 41FE 9A

0000 41FF 78

0000 4200 56

SP 0000 4201 34

SP + 1 0000 4202 12

Exercise :


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Exercise :

Question 1 :

Stack Pointer contains 00001000H

move.b #$AABB CC DD, -(SP)

ANSWER

Exercise :


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Exercise :

Question 2 :

Stack Pointer (A7)=0000 1000H

D0 = AABB CCDD

move.l (SP)+,D0

Has the following contents

SP Address Stack

0000 0FFF DE

0000 1000 BC

0000 1001 9A

0000 1002 78

0000 1003 56

0000 1004 34

ANSWER



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should :

Draw the flowchart for a simple problemusing branching and subroutines.

Apply assembly-level structured

programming techniques usingbranching and subroutines.

Concept of Programming


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Concept of Programming

Definition of problem Logical design

Programming

Test run and Analysis the program Documentation of the program

Steps :


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Steps :

1. Definition ofproblem

Inpu t data of two num bers. These two numbersare added, and the produ ct is stored in a

memory locat ion.

2. Logical designSTART

Data1 D0

Data2 D1

D0 + D1 D1

D1Memory

END

Steps :


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Steps :

3. Programming ORG $7000

MOVE.B #12,D0MOVE.B #34,D1ADD.B D0,D1MOVE.B D1,$7050

END

Steps :ORG $7000 ; The program initiled at address$7000

This is a assembler directive to store the opcodes of this program in the memory


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Steps :

4. Test run andAnalysis theprogram

ADD.B D0,D1 ;D0 + D1 D1

From instruction list, the operation of this

instruction:

ADD.B EA,Dn >> (EA) + Dn Dn

ADD.B D0,D1 >> D0 + D1 D1

D0 + D1 = $0C + $22 = $2E

D1 = $xxxxxx2E

s s a asse b e d ect e to sto e t e opcodes o t s p og a t e e o y

started at address $7000.MOVE.B #12,D0 ; decimal data 12stored in D0; end program execution.

From instruction list, the operation of this

instruction:

MOVE.B SRC,DST >> (SRC EA) DST EA

MOVE.B #12,D0 >> #12 D0Since: 1210 = 0CH; thus, D0 = $xxxxxx0C

MOVE.B #34,D1 ; decimal data 34stored in D1

MOVE.B #34,D1 >> #34 D1Since: 3410 = 22H; thus, D1 = $xxxxxx22

D0 XXXXXXXX XXXXXX0C

Before After

D1 XXXXXXXX XXXXXX22

Before After

D0 XXXXXX0C XXXXXX0CBefore After

D1 XXXXXX22 XXXXXX2E

MOVE.B D1,$7050 ; product ofaddition stored in memory at location $7050

From instruction list, the operation of thisinstruction:

MOVE.B SRC,DST >> (SRC EA) DST EA

MOVE.B D1,$7050 >> D1 M($7050)

$2EM($7050)

D1 XXXXXX2E XXXXXX2E

Before After

7050 XXXXXXXX XXXXXX2E

END ; end program execution.

Before After

D0 XXXXXXXX XXXXXX0C

Steps :


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Steps :

5. Documentation ofthe program

ORG $7000 ; The program initiled at address$7000

MOVE.B #12,D0 ; decimal data 12 stored in D0MOVE.B #34,D1 ; decimal data 34 stored in D1ADD.B D0,D1 ; D0 + D1 D1MOVE.B D1,$7050 ; product of addition stored in

memory atlocation $7050

END ; end program execution.

Flowchart


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Flowchart

A flowchart is really a shorthand technique fordescribing the processes within a system

Flowcharts can also be used to describecomputer programs

Flowchart basic symbols :


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Symbols Descriptions

Process

Input / Output

Decision



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Symbols Descriptions

Start / Stop

Connector

Program flow

direction

Branch Instruction


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Branch Instruction

There are 9 branch commands in the 68000instruction set

Instruction Example

Bcc BNE $5

BRA BRA $FFFE

BSR BSR $FFEE

DBcc DBEQ D1,$009

JMP JMP (A1)

JSR JSR $11010

RTE RTE

RTR RTR

RTS RTS

Branch Instruction cont


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Bcc instruction is for branching based on acondition

BRA - instruction is for the branching always

BSR - instruction is for the branching to asubroutine

DBcc - instruction is for testing a condition,decrementing and branching

JMP instruction is for jumping

JSR istruction is for jumping to a subroutine



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RTE instruction is for returning from anexception

RTS instruction is for returning from a

subroutine RTR - instruction is for returning from a

subroutine and restoring condition codes

Types of programming structure


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ypes o p og a g s uc u e

Construct of Sequential

Types of programming structuret


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cont

Construct of Loop


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Types of programming structurecont


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cont

Construct of decision making byBranch/JumpOne-Choice Two-Choices

Subroutines


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The subroutine concept is an approach tosimplify a very long or complex program.

There are several instructions in a programtend to be repeated to perform the similarfunction, or with a little alteration. Thus it ismore practical to group these few instructions

into a sub-program, and to be called whenevernecessary by calling its name or label.

Subroutines cont


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This sub-program is normally referred assubrout inein M68000. In other words, asubroutine is a special segment of programthat can be called for execution from any pointin a program.

Subroutine concept


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p

Nested Subroutine concept


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p

Ex :


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Ex :


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Subroutines Instructions


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Jump to Subroutine (JSR) The JSR instruction transfers program execution

to a subroutine

Return From Subroutine (RTS)

The RTS instruction restores the program counterfrom the point at which it left the main program

Subroutines Instructions


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Branch to Subroutine Instruction (BSR) This instruction has a similar function to JSR

except that the address of the subroutine to becalled is specified by a relative address (offset)

rather than an absolute address. Like the BRAinstruction, BSR is restricted to 8-bit or 16-bitoffsets

Chapter 3 ASSEMBLY LANGUAGE PROGRAMMING

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