e3165 digital electronic system

3.0 INTRODUCTION

Message sent by programmers (program instructions) understood by the computer and vice versa.

Language that friendly to human also understandable to computer systems.

Three level of language: Low level – machine language Middle level – assembly language High level – high-level language

3.0 INTRODUCTION

3.1 HLL,AL,ML Machine language:

Natural language of a particular computer system. Strings of numbers or binary codes (0 or 1). Machine-dependant (differ from one µP to other µP. Programmers need to know specifically the

architecture of CPU. Assembly language:

Using english-like abbreviations (MUL). Overcome problems – writing machine language is

too slow and tedious. Assembler as translator. Programmer need to know the basic architecture.

3.1 HLL,AL,ML

High-level language: Using everyday English and common

mathematical notation. (x = I + j) Overcome problems : assembly language

require many instruction to accomplish a simple task.

Single instruction in HLL = several AL instructions.

Compiler as translator. Programmers do not need to know the

architecture of CPU.

3.1 HLL,AL,ML

3.2 ASSEMBLY LANGUAGE PROGRAMING

ASM68K

3.2 ASSEMBLY LANGUAGE PROGRAMING

3.2.1 MACHINE INSTRUCTION / CODE FORMAT

3.3 INSTRUCTION FORMAT There are many MC68000 instructions.

However, theses instructions comply to a standard instruction format:

Label – a user-defined symbol representing the address associated with the instruction. Up to 8 alphanumeric, begin with letter, terminated by a space.

SLIDE 48

3.3 INSTRUCTION FORMAT Operator .

a mnemonic code for MC68000 instructions . An assembler directive / pseudo-instruction /

pseudo-operation. Macro call (repetitive instructions in a program

which arecodified only once and can be used as many times as necessary)

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

3.4.1 DATA TRANSFER / MOVE GROUP Transfer / move data between registers and

memory. Instructions of data transfer group.

MOVE.B D1,D2MOVE.B D6,CCRMOVE.W SR,D6MOVE.L USP,A1MOVE.L A2,USPMOVEA.W #$4321,A0MOVEQ.B #01,D4MOVEM.W D2/D4-D6, $4000MOVEM.L $4000,D0-D5LEA START,A0

EXG D1,D2SWAP D3CLR.B $3000

3.4.1 MOVE Byte(B), Word(W) and Long word(L). R R , R M , M M , EA SR , USP An Example 1:

MOVE.W #$72,D1 Before : D1 = $00200500 After : D1 = $00200072

Example 2: MOVE.B D0,D1 Before : D1 = $00200500 , D0 = $00002222 After : D1 = $00200522 , D0 = $00002222

Example 3: MOVE.B $3000,D1 Before : D1 = $00200500 , After : D1 = $00200532 ,

# menunjukkan data sebenar.

$ menunjukkan nombor heksa

$3000 32

$3001 43

$3002 98

3.4.1 MOVE(CONT.) Example 4:

MOVE.W D6,$4000 Before: D6 = $AB206541 , After : D6 = $AB206541 ,

Example 5: MOVEA.W D6,A2 Before : D6 = $AB206541 , A2 = $ABCD1234 After : D6 = $AB206541 , A2 = $00006541

Example 6: MOVEA.W D6,A2 Before : D6 = $AB20A541 , A2 = $ABCD1234 After : D6 = $AB206541 , A2 = $FFFFA541

$4000 3254

$4002 4377

$4004 9868

$4000 32

$4001 54

$4002 43

@

$4000 6541

$4002 4377

$4004 9868

@$4000 65

$4001 41

$4002 43


MOVEA.L D6,A2 Before : D6 = $AB206541 , A2 = $ABCD1234 After : D6 = $AB206541 , A2 = $AB206541

Example 8: MOVEQ #$04,D3 (MOVEQ #XXX,Dn) Before : D3 = $AB206541 After : D3 = $00000004

Example 9: MOVEQ #$80,D3 Before : D3 = $AB206541 After : D3 = $FFFFFF80

-128 (80) hingga 127 (7F)


MOVEM.W D2/D5/D0,$200 Before : D0 = $11112222 ,

D2 = $33334444 ,

D5 = $55556666 , After : D0 = $11112222 ,

D2 = $33334444 ,

D5 = $55556666 ,

Example 11: 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 : – 4Dlm contoh ada 3 daftar maka 3x4=12 , 50C – C = 500


MOVEM.W $200,D2/D5/D0 Before : D0 = $11112222 ,

D2 = $33334444 ,

D5 = $55556666 , After : D0 = $FFFF9868 ,

D2 = $00006541 ,

D5 = $00004377 ,

Example 13: 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 : + 4Dlm contoh ada 3 daftar maka 3x4=12 , 500 + C = 50C


MOVEP.W D2,1(A1) Before : D2 = $AB206541 ,

A1 = $00000501 After : D2 = $AB206541 ,

A1 = $00000501

Example 15: EXG D1,D5 Before : D1 = $11223344 , D5 = $66778899 After : D1 = $66778899 , D5 = $11223344

Example 16: SWAP D1 Before : D1 = $11223344 After : D1 = $33441122

$501 FF

$502 65

$503 41

$501 FF

$502 FF

$503 FF $502 dan $501

adalah alamat I/O


After : A1 = $00002001 MOVEA.L $2001,A1 ≠ LEA $2001,A1

Example 18: CLR.B D1 Before : D1 = $11223344 After : D1 = $ 11223300

OPER EQU $2001MOVEA.L #OPER,A1

MOVEA.L #$2001,A1

OPER EQU $2001LEA OPER,A1

LEA $2001,A1

3.4.2 ARITHMETIC

ADD.W $300,D1ADD.L D2,$218ADDI.B #$44,D3ADDQ.W #$32,D2ADDX D1,D2ADDX -(A1),-(A3)ADDA $100,A6SUB.W $300,D1SUB.L D2,$218SUBI.B #$44,D3SUBQ.W #$32,D2SUBX D1,D2SUBX -(A1),-(A3)SUBA $100,A6NEG D1NEGX A4MULS $32,D5MULU #$04,D5DIVS $32,D5DIVU #$04,D5EXT.W D3EXT.L D2

3.4.2 ARITHMETIC(CONT.) Byte(B), Word(W) and Long word(L). Data via register, memory, absolute data and I/O

port. Example 1:

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

Example 2: ADD.L D0,D1 Before : D1 = $00200500 , D0 = $00002222 After : D1 = $00202722 , D0 = $00002222

Example 3: ADD.W $3000,D1 Before : D1 = $00200500 , After : D1 = $00203743 ,

$3000 32

$3001 43

$3002 98

3.4.2 ARITHMETIC(CONT.) Example 4:

ADD.W D6,$4000 Before: D6 = $AB206541 , After : D6 = $AB206541 ,

Example 5: ADDA.W D6,A2 Before : D6 = $AB206541 , A2 = $ABCD1234 After : D6 = $AB206541 , A2 = $ABCD7775

Example 6: ADDA.L D6,A2 Before : D6 = $AB20A541 , A2 = $ABCD1234 After : D6 = $AB206541 , A2 = $56EDB775 , C = 1

$4000 3254

$4002 4377

$4004 9868

$4000 32

$4001 54

$4002 43

@

$4000 9795

$4002 4377

$4004 9868

@$4000 97

$4001 95

$4002 43


ADDX.B D2,D3 Before : D2 = $AB206541 , D3 = $ABCD1234 , X = 1 After : D2 = $AB206541 , D3 = $AB206576 , X = 1

Example 8: ADDQ.B #$04,D3 (ADDQ #XXX,Dn) Before : D3 = $AB206541 After : D3 = $AB206545

Example 9: SUB.W #$80,D3 Before : D3 = $AB206541 After : D3 = $AB2064C1

-128 (80) hingga 127 (7F)


SUBI.L #10,D2 Before : D2 = $AB206541 After : D2 = $AB206537

Example 11: SUBQ.W #@11,D3 (SUBQ #XXX,Dn) Before : D3 = $AB206541 After : D3 = $AB200038

Example 12: NEG.W D3 Before : D3 = $AB206541 After : D3 = $AB209ABF

@ menunjukkan nombor octal

% menunjukkan nombor binari

-128 (80) hingga 127 (7F)

NEG = Pelengkap 2

$41 = 01000001Songsangkan 10111110

+ 1$BF = 10111111


MULU #2,D2 ;16 bit x 16 bit = 32 bit Before : D2 = $AB206541 After : D2 = $0000CA82

Example 11: MULU #2,D2 Before : D2 = $AB20FFFF After : D2 = $0001FFFE

Example 12: DIVU #2,D3 ;32 bit / 16 bit = 16 bit lower (result)

16 bit upper ( remainder)

Before : D3 = $00006541 After : D3 = $000132A0

@ menunjukkan nombor octal

% menunjukkan nombor binari

6541F = 2592132A016

= 1296010

3.4.3 LOGICAL OPERATION

AND.W $300,D1AND.L D2,$218ANDI.B #$44,D3ANDI #%11011,CCRADDI #$25,SR

OR.W $300,D1OR.L D2,$218ORI.B #$44,D3ORI #%11011,CCRORI #$25,SR

EOR.W $300,D1EORI.B #$44,D3EORI #%11011,CCREORI #$25,SR

NOT.W D2

3.4.3 LOGICAL OPERATION(CONT.) Example 1:

AND.B #$3E,D1 Before : D1 = $12345674 After : D1 = $12345634

Example 2: OR.B D0,D1 Before : D1 = $1234563E , D0 = $98765474 After : D1 = $1234567E , D0 = $98765474

Example 3: NOT.B D1 Before : D1 = $12345655 , After : D1 = $123456AA ,

3.4.4 SHIFT AND ROTATE Example 4: B : 0 to 255 | $00 - $FF

3.4.4 SHIFT AND ROTATE (CONT.) Example 5:

B : 0 to 255 | $00 - $FF

3.4.4 SHIFT AND ROTATE (CONT.) Example 6: B : 0 to 255 | $00 to $FF

3.4.4 SHIFT AND ROTATE (CONT.) Example 7: B : 0 to 255 | $00 to $FF

3.4.4 SHIFT AND ROTATE (CONT.) Example 8: B : 0 to 127 | $00 - $7F

-1 to -128 | $FF to $8F

3.4.4 SHIFT AND ROTATE (CONT.) Example 9: B : 0 to 127 | $00 - $7F

-1 to -128 | $FF to $8F

3.4.4 SHIFT AND ROTATE (CONT.) Example 10:

3.4.5 CONTROL TRANSFER Normally : PC point to the next instruction.

PC has the address of next instruction.

Jump / branch : PC will be load with target address.

Jump : unconditional Branch : conditional Example:

JMP $8450BCS $8A50

PC is a 32 bit register.

3.4.5 CONTROL TRANSFER Conditional transfer instruction

i. Bcc instruction (Branch conditionally) Format: Bcc DST Example: START MOVE.W D3,D4

….. BNE NEXT ;(Branch if

Z=0) NEG D3

….NEXT SUB.W #32, D1

3.4.5 CONTROL TRANSFERii. DBcc instruction (test condition, decrement & branch) Format: DBcc Dn,DST Operation: Test condition; i. condition met; no branch, execute next instruction. ii. condition not met; (Dn - 1)

if Dn ≠ -1; branch to DST

if Dn = -1; no branch, execute next instruction.

Example: START CLR.L D4 ; D4 = $00000000

….. DBNE D4,NEXT ; condition (Z=0) NEG D3….NEXT SUB.W #32, D1

3.4.5 CONTROL TRANSFERiii. Scc instruction (set conditionally) Format: Scc DST Operation: Test condition:

i. condition met; DST will get 1s ii. condition not met; DST will get 0s

No branch involved. Example: START CLR.L D4 ; Z = 1

SNE D5 ; D5 = $00000000

3.4.5 CONTROL TRANSFER JMP

– always jump to any address. JMP DST

BRA always branch but up to 65536 (216) address. BRA DST .S for branch to 256 (28) address BRA.S DST

NOP no operation delay in execution time.

STOP operand loads into status register and stop. STOP #$78

3.4.6 STACK CONTROL Stack is a data structure that can be used to

save data. Stack works in LIFO concept.

3.4.6 STACK CONTROL There are two (2) stack:

User Stack – point by USP & Supervisor Stack – point by SSP

3.4.6 STACK CONTROL

PUSH data into stack. 1. decrement SP

2. move data into stack MOVE.B D1,-(A0)

POP data from stack. 1. move data from stack

2. increment SP MOVE.B (A0)+,D1

3.4.7 SUBROUTINE CONTROL An approach to simplify a long and complex

program. Grouping repeated instructions into

subroutine. This subroutine will be called when needed. Operation:

Before executing subroutine, current data (registers, SR, PC)need to be save in stack first.

Subroutine is carried out After the subroutine finish,

data in stack will be restored back.

Instruction : JSR, BSR, RTS, RTR

3.4.7 SUBROUTINE CONTROL JSR

JSR EA – the subroutine maybe anywhere in the memory.

Operation: PC -(SP) EA PC

BSR BSR EA – the subroutine must be

within 65536 address Operation:

PC -(SP) PC + d PC

3.4.7 SUBROUTINE CONTROL RTS

RTS – return from subroutine Operation:

(SP)+ PC

RTR RTR – return and restore CCR Operation:

(SP)+ CCR (SP)+ PC

REVISION68000 INSTRUCTIONS: DATA TRANSFER/ MOVE GROUP

move,movem,lea,exg,swap,clr ARITHMETIC OPERATION

add,sub,neg,mul,div,ext LOGICAL OPERATION

and,or,not,eor SHIFT AND ROTATE (x)

lsl,lsr,asl,asr,rol,ror,roxl,roxr (x) CONTROL TRANSFER GROUP

Jmp,bcc,dbcc,scc,nop,stop STACK CONTROL INSTRUCTION SUBROUTINE CONTROL INSTRUCTION

jsr,bsr,rts,rtr

3.4.8 ASSEMBLER DIRECTIVES Assembler directives are not executed during program execution. ORG :

– instruct the assembler to assemble the program. – initiate the address of the first instruction.

– ORG $1000 END

– terminate the assembly process. EQU

– assign value – COUNTER EQU $30

DC – reserve memory with initial values. – DATA DC.B$32,$65,$21,$33

DS – reserve memory with no initial values– SUM DS.B 6 TOTAL DS.W 1

SLIDE 11

3.4.8 ASSEMBLER DIRECTIVES

3.5 FLAG/STATUS REGISTER

3.5 FLAG/STATUS REGISTER (CONT.) Carry, C

3.5 FLAG/STATUS REGISTER (CONT.) oVerflow, V

For operation with 2’s complement. (+ve) + (+ve) = (-ve) V = 1 (salah) (+ve) + (+ve) = (+ve) V = 0 (betul)


For operation with 2’s complement. (-ve) - (+ve) = (+ve) V = 1


For operation with 2’s complement. (-ve) - (+ve) = (-ve) V = 0 (-ve) - (-ve) = (+ve) V = 0

3.5 FLAG / STATUS REGISTER Zero, Z

Result zero; Z = 1 Result not zero; Z = 0

3.5 FLAG / STATUS REGISTER Negative, N

Result negative; N = 1 Result positive; N = 0

Extend (X) Usually copy C flag. Use in multiple precision arithmetic (aritmetik titik

apungan).

3.5 FLAG / STATUS REGISTER Interrupt mask bits (I2,I1,I0)

Trace, T T = 1, Instructions will be execute step by step. T = 0, Instructions will be execute till the end.

Supervisor, S S = 1, processor in supervisor state, all instruction can

be executed. S = 0, processor in user state, executing privileged

instruction will cause trap.

e3165 digital electronic system

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