Chp2introductiontothe68000microprocessor Copy 110627223604 Phpapp01

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68k Instruction Set68k Instruction Set

MOVE.B D0, D1

Opcode Operand

Source

DestinationSize



Assembly Language ProgammingAssembly Language Progamming

Machine Instruction

0011 0000 0000 0011

0011

Op-code

000000

Firstoperand

000111

Secondoperand

Ty pes of instruction i.e. addition or subtraction

Data size, byte, word or long word

The amount of the word in complete instruction

Where the data for an instruction (operand) can find



Cycle of InstructionCycle of Instruction

Collection Implement



Cycle of InstructionCycle of Instruction cont«cont«

Collection phase: Fill-in program counter with

instruction address

Collect instruction from memor y, putin the instruction register

For control the instruction, clean-up

PC for next instruction

Implement phase:

Count for the operation address Collect operation

Do the operation

Store the product of operation

Back to first step



68K Addressing Modes68K Addressing Modes

Addressing ModesAddressing Modes SyntaxSyntax

Register Direct Addressing

Data Register Direct Address Register Direct

Dn An

Absolute Data Addressing

Absolute Short Absolute Long

xxx.Wxxx.L

Program Counter Relative Addressing

Relative with offsetRelative with Index Offset

d16 (PC)d8 (PC,Xn)

Register Indirect Addressing

Register IndirectPostincrement Register IndirectPredecrement Register IndirectRegister Indirect with OffsetIndexed Register Indirect with offset

(An)(An)+-(An)

d16(An)d8 (An, Xn)

Immediate Data addressing

ImmediateQuick Immediate

#xxx#1-#8

Implied Addressing

Implied Register SR USP SP PC



R egister Direct AddressingR egister Direct Addressing

Data R egister Direct

MOVE.B D0,D3





MOVE.W D0,D3

Cont«





MOVE.L D0,D3

Cont«




Address R egister Direct

MOVEA.L A3,A0

Cont«




Address R egister Direct

MOVEA A3,A0

Cont«



Absolute Data AddressingAbsolute Data Addressing

Absolute Long Mode

CLR.B $10000

010000 42 51 p 010000 00 51

010002 55 13 010002 55 13




Absolute Long Mode

CLR.W $10000

010000 42 51 p 010000 00 00

010002 55 13 010002 55 13

Cont«




Absolute Long Mode

CLR.L $10000

010000 42 51 p 010000 00 00

010002 55 13 010002 00 00

Cont«




Absolute Short Mode

CLR.B $1000

01000 42 51 p 01000 00 51

01002 55 13 01002 55 13

Cont«



Absolute Short vsAbsolute Short vsAbsolute LongAbsolute Long

Absolute short: 16-bit (0000~7FFF, 8000~FFFF)

Sign extend the 16-bit value into 24-bit

(000000~007FFF, FF8000~FFFFFF)

Absolute Long:

24-bit(000000~FFFFFF) No sign extension

Cont«



PC R elative AddressingPC R elative Addressing

PC R elative with Offset

MOVE.W $1200(PC), D5

Assume the instruction is located at $122000

Effective address:

$00122002 + $00001200 = $00123202



PC R elative AddressingPC R elative AddressingCont«

PC R elative with Offset




PC R elative with Index and Offset

MOVE.L -20(PC,A2.L),D5

Assume the instruction is located at $00010200

Effective address:

$00010202 + $00020030 + $FFFFFFEC =$0003021E

Cont«




PC R elative with Index and Offset

Cont«



R egister Indirect AddressingR egister Indirect Addressing

MOVE.L D0,(A0)





Address R egister IndirectAddress R egister IndirectAddressing PredecrementAddressing Predecrement

MOVE.W D0,-(A6)



Address R egister IndirectAddress R egister IndirectAddressing With OffsetAddressing With Offset

MOVE.W 6(A0),D0



Address R egister IndirectAddress R egister IndirectAddressing with Index & OffsetAddressing with Index & Offset

MOVE.W $10(A0,D0.L), A1



Immediate AddressingImmediate Addressing

Immediate

MOVE.L #$1FFFF, D0

Before:D0 = 12345678

After: D0 = 0001FFFF

Note:$ = value for hexadecimal

@ = value for octal

% = value for binar y

& or blank = decimal

µAB¶ = character ASCII



Immediate AddressingImmediate Addressing

Quick Immediate

1. MOVEQ #$2C, D3

Before: D3 = 1234562C

After: D3 = 0000002C

2. MOVEQ #$8F, D3

Before: D3 = 1234568F

After: D3 = FFFFFF8F

Cont«



Implied AddressingImplied Addressing

ANDI #$27, SR

MOVE CCR, CODES

TRAPV



68K Instruction Set68K Instruction Set

Data transfer

Arithmetic

Logical

Shift and rotate Bit manipulation

BCD operations

Program control

System control



68K Instruction Set68K Instruction SetCont«

Data Transfer Instructions




Integer Arithmetic Instructions




Compare and Test Instructions




Logic Instructions




Shift Instructions




Rotate Instructions




Bit-Manipulation Instructions




Subroutine Control Instructions




Binary-Coded Decimal Arithmetic Instructions



Data Transfer InstructionData Transfer Instruction

Take Data from Memory ± Operation Load

MOVE.W $10000, D1

D1 FE ED BE EF D1 FE ED 42 51

010000 42 51 010000 42 51

010002 55 13 010002 55 13





MOVE.L $1000, D1

D1 FE ED BE EF D1 42 51 55 13

01000 42 51 01000 42 51

01002 55 13 01002 55 13

Cont«





MOVE.B $1000, D1

D1 FE ED BE EF D1 FE ED BE 42

01000 42 51 01000 42 51

01002 55 13 01002 55 13

Cont«




Store Data into Memory ± Operation Save

MOVE.W D1, $1000

D1 FE ED BE EF D1 FE ED BE EF

01000 42 51 01000 BE EF

01002 55 13 01002 55 13

Cont«





MOVE.L D1, $1000


01000 42 51 01000 FE ED

01002 55 13 01002 BE EF

Cont«





MOVE.B D1, $1000


01000 42 51 01000 EF 51

01002 55 13 01002 55 13

Cont«




Moving Memory Data To Memory

MOVE.W $1000, $1006

Cont«

001000 42 51 001000 42 51

001002 55 13 001002 55 13

001004 01 02 001004 01 02001006 90 AB 001006 42 51




Moving Memory Data To Memory

MOVE.B $1000, $1006

Cont«

001000 42 51 001000 42 51

001002 55 13 001002 55 13

001004 01 02 001004 01 02001006 90 AB 001006 42 AB




Immediate Mode

MOVE.B #$F1, D1

Cont«

D1 FE ED BE EF D1 FE ED BE F1

MOVE.B D1, #$F1




MOVEQ Instruction

MOVEQ #$F1, D1

Cont«

D1 FE ED BE EF D1 FF FF FF F1




Fill up Memory

MOVE.B #$F1, $1000

Cont«

001000 42 51 001000 F1 51

001002 55 13 001002 55 13

001004 01 02 001004 01 02

001006 90 AB 001006 90 AB




EXG Instruction

EXG D1, D5

Cont«

D1 12 12 20 30 D1 00 FF 48 7C

D5 00 FF 48 7C D5 12 12 20 30




SWAP Instruction

SWAP D1

Cont«

D1 12 34 20 30 D1 20 30 12 34




LEA Instruction ± Load Effective Address

LEA $00200001, A0

Cont«

A0 00 12 70 02 A0 00 20 00 01

Similar to:MOVEA.L #00200001, A0





LEA (A1), A0

Cont«

A0 00 12 70 02 A0 00 10 FF FF

A1 00 10 FF FF A1 00 10 FF FF

Similar to:

MOVEA.L A1, A0





LEA 8(A0), A1

Cont«

A0 00 12 70 02 A0 00 12 70 02

A1 00 00 40 00 A1 00 12 70 0A





LEA 8(A0, D4.L), A1

Cont«

D4 00 00 20 00 D4 00 00 20 00

A0 00 12 70 02 A0 00 12 70 02

A1 00 00 40 00 A1 00 12 90 0A



Arithmetic InstructionArithmetic Instruction

ADD Instruction

Destination + Source p Destination

ADD.W D1, D3D1 12 34 56 78 D1 12 34 56 78

D3 FD CC 01 23 D3 FD CC 57 9B

ADD.W $1000,$2000

ADD.W $2000,D3




Data Addition in MemoryProblem:

A = B + C, where A, B & C are memory

Solution:

- Copy the first word (B) from memory to register- Add the second word (C) to register

- Copy the addition product to the memory

Program:MOVE.W B,D1 ; Take B

ADD.W C,D1 ; Add with C

MOVE.W D1,A ; Store product in A

Cont«




Carry Production (0~255) ± Unsigned integers

Cont«

ADD.B D0, D3

No carr y, C = 0

9-bit after addition, C = 1




Overflow Production (-128~+127) ±signed integers

Cont«




ADD.B D0, D3

V = 0, C = 0

V = 0, C = 1

Cont«

V = 1, C = 0

V = 1, C = 1




ADDA Instruction

Data size: W, L

ADDA.L #$70, A1

Cont«

A1 00 00 20 30 A1 00 00 20 A0

Use ADDQ.L #$1, A1 for value 1~8,it is faster than ADDA.L #$1, A1




SUBA Instruction

Data size: W, L

SUBA.L #$70,A1

Cont«

A1 00 00 20 30 A1 00 00 1F C0

Use SUBQ.L #$1, A1 for value 1~8,it is faster than SUBA.L #$1, A1




ADDQ Instruction

ADDQ.W #3,D6

Cont«

D6 12 34 56 FF D6 12 34 57 02

1~8 only




ADDQ vs MOVEQ

Cont«

ADDQ MOVEQ

Source 1..8 -128«+127

Destination Any Dn only

Data Size B, W, L L only




SUB Instruction

Destination - Source p Destination

SUB.B D3, D1

Cont«

D1 12 34 56 78 D1 12 34 56 55

D3 FD CC 01 23 D3 FD CC 01 23

SUB.W $1000, $2000




The Effect for V Flag after ADD/SUB

ADD S, D

Cont«

s d Answer (d) Overflow

Positive Negative AnyNothing happen

(always V=0)

Negative Positive AnyNothing happen

(always V=0)

Positive PositivePositive

Negative

V=0

V=1

Negative NegativeNegative

Positive

V=0

V=1




The Effect for V Flag after ADD/SUB

SUB S, D

Cont«

s d Answer (d) Overflow

Positive Positive AnyNothing happen

(always V=0)

Negative Negative AnyNothing happen

(always V=0)

Negative PositivePositive

Negative

V=0

V=1

Positive NegativeNegative

Positive

V=0

V=1




The Effect for Flag after SUB

C = Set; 2 unsigned values

V = Set; 2 signed values

Z = Set; product is 0

N = MSB

X = C

Cont«




The Effect for Flag after SUB

SUB.W D1,D0 ;D0 D0 ± D1

Cont«

D0 00 00 30 40 D1 00 00 00 02 D0 00 00 30 3E 0 0 0 0 0

D0 00 00 00 40 D1 00 00 00 40 D0 00 00 00 00 0 0 1 0 0

D0 00 00 FF FF D1 00 00 55 AA D0 00 00 AA 55 0 1 0 0 0

D0 00 00 00 00 D1 00 00 03 00 D0 00 00 FD 00 1 1 0 0 1

D0 00 00 90 00 D1 00 00 70 00 D0 00 00 20 00 0 0 0 1 0

D0 00 00 70 00 D1 00 00 90 00 D0 00 00 E0 00 1 1 0 1 1

D0 before SUB D1 before SUB D1 before SUB

X N Z V C

After instruction




SUBQ Instruction

SUBQ.W #7,D6

Cont«

D6 12 34 56 FF D6 12 34 56 F8

1~8 only




SUBQ vs MOVEQ

Cont«

SUBQ MOVEQ

Source 1..8 -128«+127

Destination Any Dn only

Data Size B, W, L L only




Multiplication Instruction

1. Unsigned number

MULU.W source, Dn

2. Signed number

MULS.W source, Dn

Cont«

Note: Source: 16-bit of any address mode

Both operand: 32-bit data register

Multiply product: 32-bit stores in destination data register




MULU

1. Multiply 256 with 2

MOVE.W #256, D3

MULU.W #2, D3

Cont«

D3 FD CC 01 00 D3 00 00 02 00

16-bit 32-bit




MULU

2. Multiply #$FFFF with #$FFFF

MULU.W #$FFFF, D3

Cont«

D3 FD CC FF FF D3 FF FE 00 01

Note:FFFF = 65535

65535 * 65535 = 4294836225 = FFFE0001




MULS

Multiply #$FFFF with #$FFFF

MULS.W #-1, D3

Cont«

D3 FD CC FF FF D3 00 00 00 01

Note:D3 = -1

(-1) * (-1) = 1

i h i ii h i i




MULS & MULU

Cont«

Content in hex As unsigned decimal As signed decimal

Location XX 0003 3 3

Location YY B000 45056 -20480

Register D0 00A0 160 160

Register D1 FF00 65280 -256

MULS XX, D0 D0=000001E0 (3 x 160 = 48010)

MULU XX, D0 D0=000001E0 (3 x 160 = 48010)MULS XX, D1 D1=FFFFFD00 (3 x (-256) = -76810)

MULU XX, D1 D1=0002FD00 (3x 65280 = 19584010)

MULS YY, D1 D1=00500000 (-20480 x (-256) = 524288010)

MULU YY, D1 D1=AF500000 (45056 x 65280 = 294125568010)

i h i ii h i i




Division Instruction

1. Unsigned number

DIVU.W source, Dn

2. Signed number

DIVS.W source, Dn

Cont«

Note: Source: 16-bit of any address mode

Both operand: 32-bit data register

Division product: lower word destination register

Remainder: upper word destination register

A i h i I iA i h i I i





Example 1:

DIVU.W D0, D3

Cont«

D0 FE CC 01 00 D0 FE CC 01 00

D3 00 10 00 01 D5 00 01 10 00

Remainder Quotient

A ith ti I t tiA ith ti I t ti





Example 2: (Overflow)

MOVE.L #$40000, D3

DIVU.W #2, D3

Result D3 = 100 0000 0000 0000 0000 (19-bit)

Destination can hold 16-bit data only

Operand remains unchanged and a divisionoverflow sets the V = 1

Cont«






Example 3: (Case divide with zero)

MOVE.L #$122, D3

CLR D0

DIV D0, D3

Cont«







Cont«

Content in hex DecimalLocation XX 0012 Decimal 18

Location YY FFAE Decimal ±82

Location ZZ FF00 Unsigned decimal 65280

Register D0 00000308 Decimal 776

Register D1 FFFFFE00 Decimal -512







Cont«

Instruction Result in hex Quotient Remainder

DIVU XX,D0 D0=0002002B Decimal 43 Decimal 2

DIVS XX,D1 D1=FFF8FFE4 Decimal -28 Decimal ±8

DIVS YY,D0 0026FFF7 -9 38

DIVS YY,D1 FFEC0006 6 -20DIVU ZZ,D0 03080000 0 308

DIVU ZZ,D1 FFFFFE00 65792 65024





NEG Instruction (Negate)

Form 2¶s complement

X n 0 - X

NEG.W D5

Cont«

D5 34 67 00 F0 D5 34 67 FF 10





EXT Instruction (Sign Extend) Convert byte @ word operand to word @ longword

by extending the sign bit of the operand

EXT.W D5

Cont«

D5 34 67 00 F0 D5 34 67 FF F0

D5 34 27 20 65 D5 34 27 00 65





EXT Instruction (Sign Extend)

EXT.L D5

Cont«

D5 34 67 30 F0 D5 00 00 30 F0

D5 34 27 99 65 D5 FF FF 99 65

P ti l E lP ti l E l



Practical ExamplePractical Example

Write a sequence of program to computeX = 5 * Y + Z/W

where Y, Z and W are 16-bit signed integers and the resultis stored as longword X.Solution:

MOVE.W Y, D0MULS #5, D0MOVE.W Z, D1EXT.L D1DIVS W, D1

EXT.L D1ADD.L D1, D0

MOVE.L D0, X

Location andregister

AsHexadecimal

Location X 0012

Location Y FF AE

Location ZFF00

Register D0 00000308

Register D1 FFFFFE00

C & T t I t tiC & T t I t ti



Compare & Test InstructionCompare & Test Instruction

Compare Instruction ± CMP

CMP.W D1, D0

D0-D1X N Z V C1 1 1 1 1

D0 01 20 30 40 D1 01 20 30 40 0000 1 0 1 0 0

D0 01 20 30 41 D1 01 20 30 40 0001 1 0 0 0 0

D0 01 20 30 40 D1 01 20 30 41 FFFF 1 1 0 0 1

D0 01 20 00 00 D1 01 20 FF FF 0001 1 0 0 0 1

D0 01 20 FF FF D1 01 20 00 00 FFFF 1 1 0 0 0

D0 01 20 10 00 D1 01 20 20 00 F000 1 1 0 0 1

D0 01 20 70 00 D1 01 20 90 00 E000 1 1 0 1 1

D0 01 20 90 00 D1 01 20 70 00 2000 1 0 0 1 0

Logical InstructionLogical Instruction




A B AB AB AB A B

00

11

010

1

00

0

1

0111

0110

110

0

10

10





AND.s <sea>, <dea> Either one is Dn operand

ANDI.s #data, <dea> The 1st immediate operand

OR.s <sea>, <dea> Either one is Dn operand

ORI.s #data, <dea> The 1st immediate operandEOR.s <sea>, <dea> Either one is Dn operand

EORI.s #data, <dea> The 1st immediate operand

NOT.s <ea>

Cont«





AND Instruction To clear more than 1 bit in operand

ADNI.L #$0F, D0 If D0 = $0000B2DA

Cont«

D0 0000 0000 0000 0000 1011 0010 1101 1010

$0F 0000 0000 0000 0000 0000 0000 0000 1111

AB 0000 0000 0000 0000 0000 0000 0000 1010

@ D0 = $0000000A





OR Instruction To set more than 1 bit in operand

ORI.W #$00E0, D0 If D0 = $A22D

Cont«

D0 1010 0010 0010 1101

$E0 0000 0000 1110 0000

AB 1010 0010 1110 1101

@ D0 = $A2ED





EOR Instruction To invert more than 1 bit in operand

EORI.W #$00FF, D0 If D0 = $A22D

Cont«

D0 1010 0010 0010 1101

$FF 0000 0000 1111 1111

AB 1010 0010 1101 0010

@ D0 = $A2D2





NOT Instruction To get 1¶s complement of the

operand

NOT.B D0

If D0 = $A22D

Cont«

D0.B 0000 0000 1111 1111 D0.B 1010 0010 1101 0010

@ D0 = $A2D2

Shift InstructionShift Instruction




4 instructions of shift operations: ASL (arithmetic shift left) ASR (arithmetic shift right) LSL (logical shift left) LSR (logical shift right)

Size:B, W, L

Affected flag: X & C = depend to the bit shift out N & Z = depend to the product V = 1 if sign operand changes by shifting, but 0 for rotate





Arithmetic Shift Instruction

Example:

ASR.B #2, D1

Cont«

D1 12 34 56 F0 D1 12 34 56 FC

(-16) (-4)





Logical Shift Instruction Unsigned value

Example: LSR.B #2, D1

Cont«

D1 12 34 56 F0 D1 12 34 56 3C

(240) (60)

0

Rotate InstructionRotate Instruction



R otate InstructionR otate Instruction

4 instructions of rotate operations: ROL (rotate left) ROR (rotate right) ROXL (rotate left through X flag) ROXR (rotate right through X flag)

Size:B, W, L

Affected flag: X & C = depend to the bit shift out N & Z = depend to the product V = 1 if sign operand changes by shifting, but 0 for rotate





Example: ROR.B #2, D1

Cont«

D1 12 34 56 A1 D1 12 34 56 68

(10100001) (01101000)





R otate Instruction through X

Example: ROXR.B #2, D1

Cont«

D1 12 34 56 A1 D1 12 34 56 A8

(10100001) (10101000)

ExampleExample



ExampleExample

Running light in address $80000

MOVE.B #1,D0

REPEAT MOVE.B D0,$800000

JSR DELAY ;Delay

ROL.B #1,D0

BRA REPEAT

Bit Manipulation InstructionBit Manipulation Instruction




Test the specified bit in the destination

Data size = B or only 1.

The bit manipulation instruction included:

a. BCHG (bit test and change) instructionb. BCLR (bit test and clear) instruction

c. BSET (bit test and set) instruction

d. BTST (bit test) instruction





Example 1:In address $FF8000 contains $AA. What isthe result after BTST.B #3, $FF8000?

Solution: ($FF8000) = 101010102 where the 3rd bit

is 1.

Therefore Z = 0

Cont«





Example 2:Data register D5 contains 2C3459A7.What is the state of the Z flag and whatare the contents of D5 after BCHG #6, D5?

Solution: lower byte of D5 = 101001112. Bit 6, 1<0>100111, is a 0. This will set the Z flag & complement bit 6. This result in D5 = 2C3459E716.

Cont«





Example 3:Data registers D6 and D7 contain 0000000C and75793290, respectively. What is the result of BCLR D6, D7?

Solution: Register D6 specifies that the 12th bit position

001<1>001010010000) should be tested andcleared.

Since the 12th bit is a 1, Z = 0 bit 12 of D7 will be cleared D7 = 75792290.

Cont«





Example 4:

What is the result of BSET #2,(A3)?

Solution:The bit in position 2 of the memory locationpointed to by A3 is tested and then set.

Cont«

Subroutine ControlSubroutine Control



InstructionInstruction

Flow Chart Symbol





Program Control Structure

Cont«





Jump Instruction ± JMP

Unconditional transfer to target destination

PC n effective address

JMP <ea> Type of effective address:

a. Absolute short

b. Absolute long

c. Address register indirect

Cont«





Jump Instruction ± JMP

JMP START ; START = $1000

JMP $FF8000

Cont«

0100 1110 1111 10000001 0000 0000 0000

0100 1110 1111 1001

0000 0000 1111 11111000 0000 0000 0000

Subroutine ControlSubroutine Controlii




Bcc (branch condition) InstructionBcc<ea>

Cont«

Instruction Meaning Arithmetic If the test is true

BEQ EQual to zero U Z=1

B NE Not Equal to zero U Z=0

BMI Minus U N=1BPL Plus U N=0

BCS/LO Carr y Set/LOwer U C=1

BCC/HS Carr y Clear/Higher or Same U C=0

BVS oVerflow Set S V=1

BVC oVerflow Clear S V=0

BGT GreaTer than S Z+( NV)=0

BLT Less Than S NV=1

BGE Greater than or Equal S NV=0

BLE Less than or Equal S Z+( NV)=0

BHI Higher U C+Z=0

BLS Lower than or Same U C+Z=1

Subroutine ControlSubroutine Controlii




Branch Always Instruction - BR APC n PC + offsetBRA.s <ea>

Short branch: -128 e PC e +127 Syntax: BRA.S

Long branch:

-32k e PC e 32k Syntax: BRA.L

Cont«

Subroutine ControlSubroutine ControlI iI i




Control Constructions1. IF condition

THEN action 1;

2. IF condition

THEN action 1

ELSE action 2;

3. FOR counter = initial TO final value

DO action 1;

4. R EPEA

Taction1

UNTIL condition;

Cont«

5. WHILE condition

DO action 1;

6. CASE selector OF

action 1,action 2 ,

.

.action N ;

Subroutine ControlSubroutine ControlI t tiI t ti




Example for IF - THENIF A=10 THEN Action 1

CMPI.W #10, VARA

BNE NEXT.

.

.

NEXT

Cont«

Action 1 F

T





Example for IF ± THEN ± ELSEIF A<=0 THEN Action 1; ELSE Action 2

TST.W VARA

BLE NEXT1.

.

BRA NEXTNEXT1

.

.

NEXT

Cont«

Action 1

Action 2





Example for FOR LOOP

FOR I=1 TO 10 DO Action 1

MOVE.W #9, D7

LOOP.

.

.

DBF D7, LOOP

Cont«

Action 1





Example for R EPEAT ± UNTILREPEAT Action 1 UNTIL A=0 or maximum count isreached

MOVE.W LIMIT,D7SUBQ.W #1, D7LOOP

.

.

.TST.W VARADBNE D7, LOOP

Cont«

Action 1





Example for WHILE ± DOWHILE A<=10 and maximum countis not reached DO Action 1

MOVE.W LIMIT, D7AGAIN CMPI.W #10, VARA

DBGT D7, LOOPLOOP

.

.

.BRA AGAIN

NEXT...

Cont«

Action 1

Action 2

BCD (Binary Coded Decimal)BCD (Binary Coded Decimal)I t tiI t ti




Each BCD byte contents 2 digit of 4-bit BCD. Each BCD instruction involve in X bit.

Z bit is changed if the product is not zero.

Before doing the first BCD operation, it have to

start with X=0 and Z=1. Syntax:

ABCD Dx,Dy (source)10 + (destination)10 + X p destination

ABCD- (Ay)-(Ax)

NBCD <ea> 0 ± (source)10 ± X p destination

SBCD Dx,Dy (destination)10 - (source)10 - X p destination





ABCD and SBCD InstructionsExample 1:

Add BCD number in D0 to D1.

MOVE #4, CCR ; x=0 and z=1

ABCD D0, D1

Example 2:Subtract BCD number in D0 from D1

MOVE #4, CCR ; x=0 and z=1

SBCD D0, D1

Cont«





BCD InstructionAddition for 2 BCD number 8 bit (byte).

MOVE #4,CCR ;x=0 and z=1

ABCD -(A0),-(A1)

ABCD -(A0),-(A1)ABCD -(A0),-(A1)

ABCD -(A0),-(A1)

Cont«

A1 after addition p Upper Byte Upper Middle

Lower Middle Lower Byte

A1 before addition p

A0 after addition p Upper Byte Upper Middle

Lower Middle Lower Byte

A0 before addition p

Destination

Source

BCD (Binary Coded Decimal)BCD (Binary Coded Decimal)Inst ctionInst ction




NBCD Instruction Use 9¶s @ 10¶s complement notation to represent

a negative BCD number

Example 1:What is the 10s complement for fill-in D0. Let sayD0 = 23.

MOVE #0,CCR

NBCD D0NBCD operation: 100 (0 borrow 1)

Original D0: -23

X bit: - 0

Product: 77

Cont«

BCD (Binary Coded Decimal)BCD (Binary Coded Decimal)InstructionInstruction




NBCD Instruction

Example 2:

What is the 9s complement for fill-in D0. Let say D0

= 23.MOVE #$10,CCR

NBCD D0

NBCD operation : 100 (0 borrow 1)

Original D0 : -23X bit: - 1

Product : 76

Cont«

Stack and SubroutineStack and Subroutine



Stack Data structure that is used to save and restore

information in a last-in, first-out (LIFO) fashion.

Register A7 is implicitly used as the stack pointer.

Push (save):MOVE source, -(SP) orMOVE.L source, -(SP)

Pop (retrieve):

MOVE (SP)+, destination orMOVE.L (SP)+, destination




Stack

Cont«




StackSTACKA DS.W 50

MOVEA.L #STACKA + 100, A2

MOVE.L D1, -(A2)MOVE.W VAR, -(A2)

MOVE.W #$25, -(A2)

MOVE.L (A2)+, D2

MOVE.L (A2)+, D3

Cont«




Stack

Cont«

SubroutineSubroutine






Cont«

Subroutine Concept A program structure where one part of the program is called

the main program. In addition to this, a smaller segment attached to the main

program, known as a subroutine. The subroutine is written to provide a function that must be

performed at various points in the main program. A return instruction must be included at the end of the

subroutine to initiate the return sequence to the mainprogram environment.

The instructions provided to transfer control from the mainprogram to a subroutine and return control back to the main

program are called subroutine-handling instructions.




Cont«

Subroutine Control Instructions ±JSR, BSR, R TS & R TR

InterruptInterrupt



pp

An external event which informs the CPUthat a device needs its service.

Interrupt Service Routine (ISR):a. a program which associates with interrupt.

b. also called the interrupt handlerc. there are 4 bytes of memory allocated as

interrupt vector table.

d. handled in the supervisor mode, S = 1.

InterruptInterrupt



pp

Example 1:Calculate the sum of 10 numbers in theaddress memory

$2000.

Cont«

extra

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