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Assembly Language

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Assembly Language

Addressing Modes

Types of InstructionArithmeticLogical

Data Transfer

Boolean Variables

Program Branching

1.2.

3.

4.

5.

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AddressingModes

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1. Immediate Addressing

Value to be stored in memory immediately follows theoperation code in memoryOr

The instruction itself dictatesmemory

ExampleMOV A,

the value to be storedin

#20h Accumulator will be loaded with the value that

immediately follows; in this case 20h

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1. Immediate Addressing

It is very fast

Value to be loaded is includedin instruction

The value tobeloadedisfixedatcompiletimeitsnotveryflexible

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1. ImmediateAddressing

Examples MOV MOV MOV

A,#25HR4,#62H

B,#40H

COUNT EQU 30 MOVA,#COUNT MOVP1,#34

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Important

MOVA,#FH ; giveserror

MOVA,#0FH ;OKAY

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2. Direct Addressing

Value to be stored in memory is obtained bydirectly

retrievingitfromanother memorylocation

ExampleMOV A, 30H

The above instruction will read30H and put that data in A.

datafromRAMaddress

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2. Direct Addressing

Generally fast, although the value to be stored is notincluded in the instruction

Its quickly accessible sinceInternal RAM

value is stored in 8051s

Much more flexible than immediate addressing. Valuetobe loaded is whatever it is found at that location whichmay be variable

Entire 128 bytes of RAM (0 to FF) can be accessed

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2 . Direct Addressing

Examples MOVR0, 40H

MOV56H, A MOVR4,7FH

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3. Indirect Addressing Also called indirect register addressing

Very powerful and flexible addressing

ExampleMOV A,@R0

This instruction analyzes the value of R0.A will be loaded withvalue from RAM whose address is found in R0

For example if R0 holds value 40H and RAM address 40h holdsvalue of 67H. Then A will have value of 67H

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3. Indirect Addressing Only R0 and R1

can

be

used

for

this

purpose

MOV A,@R0 MOV @R1,B

Note: @signaddressing

iscompulsoryinordertouseIndirect

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3. Indirect Addressing This mode makes data access dynamic

(as in case of direct addressing)instead

of

static

Only R0 and R1 can be used

Since theseregistersare8bit,useisrestrictedtointernalRAMonly

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4. Register Addressing In register addressing mode, we

stored in registerdirectly use the value

8051 has 8 working registers (ofthecurrentlyactiveregister bank) R0 toR7

ExamplesMOV A, R7

MOV A, R5 Contents of R7 and R5 will be moved in above

instructions

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4. Register Addressing Easy to use register name instead of address

Source and destination registers should matchinsize

MovebetweenA and registers is allowedMOV A, R4 ; OKAY

Movebetweenregisters is not allowedMOV R7, R4 ; gives error

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5. Relative Addressing mode Used with certain jump instructions

8 bit signed offset

127 locationisused,rangeforjumpingis-128to

SJMP[label]

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6. Absolute Addressing

Used only with ACALL and AJMP instruction

Itallowsbranchingwithin2Klimit(11bit)

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7. Long Addressing

Used only with LCALL and LJMP instruction

Allowsbranchingwithin64Klimit(16bit)

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8. Index Addressing Used with JMP and MOVC instruction

Use base register (either PC or DPTR) and an offset (theaccumulator) in formingtheeffectiveaddress

MOVC A, @A+ DPTR JMP @A+ DPTR

It is used for look up tablePreviously we studied how its

Wellcover its usage later on

stored

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Instruction Types

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Types of Instructions

ArithmeticLogical

Data Transfer

Boolean Variable Manipulation

1.

2.

3.

4.

5. ProgramBranching

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MOV Summary

MOV [Destination],[Source]

Examples

MOVA,#55H ; A will be loadedwith55H

MOVR2, A ; Load value of A in R2

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1. ArithmeticInstructions

Addition Subtraction Multiplication Division Increment Decrement

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1. Arithmetic Instructions - AdditionADD

Addition Add A, source Add A, #data

; Add source to A; Add data to A

Destination operand is always register A

Examplemov a,#5mov b,#10

Add a,b

;;

;

Set a = 5Set b = 10

what will be contents

of a?

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1. Arithmetic Instructions - AdditionADDC

Addition with carry Used in case of adding 16 bit numbers

High byte and Low Byte are added separatelyinstructions

using 2

If there isbe added

a Carry out from addition of low bytes, it hasin sum of high bytes

to

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1. Arithmetic Instructions-AdditionADDC

Addition with carry Example

Write a program558DH

toaddtwo16bitnumbers2CE7Hand

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1. Arithmetic Instructions - AdditionADDCWrite a program toadd two 16 bit numbers 2CE7H and558DH

CLR CMOV A, #0E7H

ADD A, #8DH

;clear carry flag. CY = 01st;

;

;

;

;

;

;

;

Load lower byte of numbernumber2ndAdd lower byte of

now A = 74H and CY = 1;Move lower byte of sum in R0MOV R0,A

MOV A, #2CH

ADDC A,#55H

1stLoad higher byte ofAdd with carry

number

2C + 55 + 1 (carry) = 82 HMove higher byte of sum in R1MOV R1, A

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1. Arithmetic Instructions - SubtractionSUBBSUBtract with Borrow

Subtraction with borrow Uses carry flag to read if there is any borrow or not Format

SUBB A, Source ; A = A source CY

Before the subtraction, Carry If carry = 0, there is no carry If carry = 1, there is a borrow

flag means Borrow

Note: There is no subtractioninstruction withoutborrowin8051 (many microprocessors have it)

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1. Arithmetic Instructions-SubtractionSUBB

SUBtract with Borrow

After the operation Carry = 0 if result Carry = 1 if result

is POSITIVEis NEGATIVE

Negative result isstored in formatof 2s complement

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SUBtract with Example

Borrow

clr Cmov a,#25h

mov r1,#10h

subb a,r1

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1. Arithmetic Instructions -SubtractionSUBBSUBtract with

ExampleBorrow

clr Cmov

mov

;;

;

;

set carry flag = 0A = 25h

r1 = 10h

a,#25hr1,#10h

subb a,r1 a =ar1=2510=15

What will bevalue of A?15hCarry = 0

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SUBtract with Example

Borrow

setb Cmov a,#25h

mov r1,#10h

subb a,r1

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1. Arithmetic Instructions -SubtractionSUBBSUBtract with

ExampleBorrow

setb Cmov a,#25h

mov r1,#10h

subb a,r1

;;

;

;

set carry flag = 1A = 25h

r1 = 10h

a =ar11=25101=14

What will bevalue of A?14hCarry = 0

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1. Arithmetic Instructions -SubtractionSUBBSUBtract with Example

Borrow

clr C

movmov

;

;;

set carry flag = 0

A = 10hr1 = 20ha,#10hr1,#20hsubb a,r1 ;a = F0h

What will bevalue of A?

F0h = 1111 0000. Taking 2s complement (by inverting and adding 1)= 0000 1111 + 1 = 0001 0000 = 10h

Carry = 1. Showing it is a negative number

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1. ArithmeticInstructions- MultiplicationMUL MultiplicationFormat

MUL AB Multiplies A andB Operands must be Result

inAandB

A has Low byte

B has high byte

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1. ArithmeticInstructions-MultiplicationMUL MultiplicationExample

mov a,#16mov b,#4

mul ab

What willbevalueof A and B?

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1. ArithmeticInstructions-MultiplicationMUL MultiplicationExample

mov a,#16mov b,#4mul ab

What will bevalue= 64 of A and B?A = 40hB = 0

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1. ArithmeticInstructions-DivisionDIVExample

mov a,#16mov b,#4div ab

What willbevalue of A and B?

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1. ArithmeticInstructions-DivisionDIVExample

mov a,#16mov b,#4div ab

What will bevalue of A and B?A = 4hB = 0

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1. ArithmeticInstructions - DecrementDEC DecrementFormat

DEC

source

source

Subtracts 1from Example

mov r1,#55h

dec r1

R1 = 54h

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1. Arithmetic InstructionsBCD NumbersBCD NumbersBinary Coded Decimal

Unpacked BCD Lower 4 Bits represent decimal Remaining bits are zero Example: 9 will be represented

number (0 to 9)

as 0000 1001 1 byte can represent 1 number onlyPacked BCD

Each 4 bits represent a decimal number (0 to 9)1 byte can store more 2 digits

Example: 79 will be represented as 0111 1001

Utilizes space more efficiently than unpacked BCD

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1. Arithmetic Instructions BCD NumbersBCD NumbersBinary Coded Decimal Addition of BCD numbers Consider additionoffollowingBCDnumbers

mov A,#59hadd A,#1h

Result is 5AhIs it a BCD? NO

The answer should

Solution?

have been 60

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1. ArithmeticInstructionsBCDNumbersDAADecimal Adjust

mov A,#59h

add A,#1hDAA A

Result is 60hIs it a BCD? Yes

4th 5thAuxiliary Carry (from to bit)is used to check this condition

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1. Arithmetic InstructionsImportant

Play with arithmetic instructionsand seetheresults

GoodtoreviseBinaryArithmeticfirst

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2. LogicalInstructions

AND OR Exclusive OR Clear Complement Rotate Left Right

SWAP

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2. LogicalInstructions - ANDAND Format

ANL destination,source Result

destination = destination AND source

ThedestinationisnormallyAccumulator

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2. LogicalInstructions-ANDAND Example

mov a,#74hanl a,#0Fh


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2. LogicalInstructions-ANDAND Example

mov a,#74hanl a,#0Fh

What will beA = 04h value of A?

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2. LogicalInstructions-OROR Example

mov a,#74horl a,#0Fh


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2. LogicalInstructions-OROR Example

mov a,#74horl a,#0Fh

What will beA = 7Fh value of A?

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2. Logical Instructions - XORXOR (Exclusive OR) Format

XRL

Resultdestination,source

destination = destination XOR source

ThedestinationisnormallyAccumulator

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2. LogicalInstructions-XORXOR Example

mov a,#74hxrl a,#0Fh


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2. LogicalInstructions- CLEARCLEAR Format

CLR operand Result

operand = 0

The operandPort Bit etc

maybeAccumulator,carryflag (C), any IO

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2. LogicalInstructions-CLEARCLEAR Example

mov a,#74h

clr a

WhatA = 0

willbevalueofA?

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2. LogicalInstructions - COMPLEMENTCPL Format

CPL operand Result

All bits of operand inverted

The operandPort Bit etc

maybeAccumulator, carryflag(C), any IO

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2. LogicalInstructions-COMPLEMENTCPL Example

mov a,#74h

cpl a

What

will be

value of A?

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2. LogicalInstructions-COMPLEMENTCPL Example

mov a,#74h

cpl a

What will be

A = 8Bhvalue of A?

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2. Logical Instructions RotateLeftRLC (Rotate Format

Left through Carry)

RLC A Result

RL: All bits of A shiftedtoleft

Carrybitis included inrotation

CY MSB LSB

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2. Logical Instructions RotateRightRRC (Rotate Format

Rightthrough Carry)

RRC A Result

RL: All bits of A shiftedtoleft

Carryflagisincludedinrotation

CYMSB LSB

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2. LogicalInstructions SWAPSWAP Format

SWAP A Result

Swaps lower and highernibbles

Worksonlywithaccumulator

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2. LogicalInstructionsSWAPSWAP Example

mov a,#74h

swap a

WhatwillbevalueofA?

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2. LogicalInstructionsSWAPSWAP Example

mov a,#74h

swap a

What willA = 47h

bevalueofA?

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3. Data Transfer

MOV Addressing Modes

(Already Covered)

MOVC XCH PUSH POP

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3. Data Transfer MOVCMOVC Example

Write a program which reads anumberfromport1and

sendsitssquaretop2continuously

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3. Data Transfer MOVCMOVC

Write a program which readssquare to p2 continuously

a numberfrom port1 and sends its

ORG 00hmov dptr,#300h

Label: mov a,p1movc a, @a+ dptrmov p2,a

jmp Label

; Index Addressing Mode

ORGsq:

end

300hdb 0,1,4,9,16,25,36,49,64,81,100

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3. Data Transfer XCHXCH Exchanges Format

thecontents ofoperands

XCH A,source

ExampleMov a,#15h

Mov r1,#0FEh

XCH a,r1

;;

;

;

A = 15hR1 = FEh

swaps data of A and R1A = FEh and R1 = 15h

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3. Data Transfer PUSHPUSH Used to Push or store data onto stack

Stack Pointer (SP) stores the address of current stacklocation.

At power up, SP = 07 First number will be stored at 08 and so on

When data is stored (using PUSH) command,incremented automatically

SP is

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3. Data Transfer PUSHPUSH

RAM locations 08h to 1Fhcan be used for stack

RAM locations 20h to 2Fhare bit-addressable locations

More space if SP is set to pointto30h to 7Fh i.e. general purpose RAM

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3. Data Transfer PUSHPUSH Format

PUSH expression must be Number of a register

1 for R1 2 for R2 and so on

Address

expression

Symbol referring to anaddress

Note: Accumulator cannot be used withPUSH statement

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3. Data TransferPUSHPUSH Format

PUSH expression Example

PushPush

Push

31

55h

;;

;

PushPush

Push

contents of R3contents of R1

value stored atmemory

location 55h

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3. Data Transfer POPPOP Format

POP expression expression must be Number of a register

1 for R1 2 for R2 and so on

Address Symbol referring to anaddress

Retrieves the value the last value stored Automatically decrements the SP

by PUSH

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3. Data TransferPOPPOP Examplemov r7,#85hmov 80h,#5h

; R7 = 85h; Store 5h at memory location 80h

PUSH 7PUSH 80h

; Store R7 using PUSH; Store contents of location 80h

POP 1POP 75h

; Retrieve last pushed value in R1. R1 = 5h; Retrieve the current value in stack and save

; it in memory location 75h

mov a,75h ; A = contents of location 75h. A = 85h

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4. Boolean Variable Manipulation CLR CLR bit Clears the bit Examples

CLR C CLR P0^3

i.e.makesthebit=0

SETBSETB bit

Sets the bitExamples

SETB C SETB P2^0

i.e. makes the bit =1

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4. Boolean Variable ManipulationANL ANL

bit1,bit2

Takes logical AND of both bits Example

anl c,P1^7

ORL ORL

bit1,bit2 Takes logical OR of both bits Example

orl c,P3^5

CPLCPL bitInverts the value of bit

Example

CPL P0^0

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5. Program BranchingCALL and RETUnconditional AJMP SJMP LJMPConditional

CJNEDJNZ

JZ

JNZ

JB

JNB

JC

JNC

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5. Program BranchingCALLandRETCALL and RET Used toCall Subroutines(SR)

FormatCALL name ; Calling Subroutine

Return from SRRET ;

If SR is Interrupt Subroutine (ISR), return is:

RETI ; Return from interruptSR

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5. Program Branching CALLandRETCALL and RET ExampleORG 00hCallCall

Jmp

fn1fn2

$

; Calling Subroutine fn1; Calling Subroutine fn2

; Infinite Loop

fn1: ..

RET

RET

What if infinite loop isomitted?fn2:

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5. Program Branching CALL and RETCALL and RETImportant

When SR is being executed, PSW can be changed Carry Flag Auxiliary Carry Overflow etc

All required registers are stored before calling SR and retrievedafter the SR has finished

PSW A B R0 R7

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5. Program Branching CALLandRETCALL and RETImportant

Saving Required registers

PUSH them in stack before calling SR1.2. POPthemfromstackafterreturningfromSR

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5. Program Branching Unconditional JumpsSJMP Canjump128 to 127 (8 bit signed) from currentaddress

SJMP REL; REL is 8 bit offsetAJMP Canjumptoanyaddress represented

AJMP Addr11

by11bits

LJMP Canjumptoanyaddress represented

AJMP Addr16

by16bits

JMP We can write JMP and assembler will

appropriate jumpconvert it to

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5. Program BranchingUnconditionalJumpsSJMP, AJMP, LJMP Example

Org 00h

Main:movmov

mov

a,#23ha,#25h

a, #27h

jmp main ;;

Automatically decides the appropriateJmp

end

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5. Program Branching ConditionalJumpsCJNE Compare and Jump if NOT Equal

CJNE A, expression, LabelDJNZ Decrement and Jump if NOT Zero

DJNZ, register, Label

JZ

Jump if Accumulator is ZeroJZ Label

JNZ

Jump is Accumulator is NOT ZeroJNZ Label

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5. Program BranchingConditionalJumpsJB Jump ifBitisset

JBbit, LabelJNB Jump ifBitisNotset

JNB bit, labelJC Jump ifCarry is set

JC LabelJNC Jump ifCarry NOT set

JNC Label95

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5. Program Branching Conditional JumpsCJNE Compare and Jump if NOT Equal

CJNE expression1, expression1 may be

A Register (R0 R7)

expression2,Label

expression2maybeNumber

Address

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5. ProgramBranchingConditionalJumpsCJNE Example

mov a,#01h

main: INC aCJNE A, #09h, main

Jmp $

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5. Program Branching ConditionalJumpsDJNZ Decrement and Jump if Not Zero

DJNZ register, Label

Register may be any register of the active bank

R0 to R7

First it decrements the valuethe value is zero or not

ofregisterandthenchecksif

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5. ProgramBranchingConditionalJumpsDJNZ Example

mov r7,#05h

main:DJNZ r7,main

jmp $



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5. Program Branching ConditionalJumpsJZ Jump if accumulator Format

is Zero

JZ expression

expression may beLabel8 bit signed

offset address

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5. ProgramBranchingConditionalJumpsJZ Example

mov a,#1h

main: DEC aJZ main

jmp $

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5. Program Branching ConditionalJumpsJNZ Jump if accumulator is Not Zero

JNZ expression

Expression may beLabel8bitsignedoffsetaddress

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5. ProgramBranchingConditionalJumpsJNZ Examplemov a,#07h

main: DEC aJNZ main

jmp $



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5. Program Branching ConditionalJumpsJB Jump if Bit isset

JBbit,Label Example

Mov a,#0FhLabel: DEC A

mov p0, a

JB P0^3, Label



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5. Program Branching ConditionalJumpsJNB Jump if Bit isNOTset

JNBbit,Label Example

Mov a,#01hLabel: INC A

mov P0,a

JNB P0^3, Label



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5. Program Branching ConditionalJumpsJC Jump ifCarryis set

JC Label

JNC Jump ifCarryNotset

JNCLabel

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