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Introduction to Assembly Language Programming

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High Level Language

Compiler

Assembly Language

Assembler

Machine Code

Microprocessor Hardware

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8085A Instruction Set Data Transfer Instruction Move data between registers or between memory

locations and registers. Includes moves, loads, stores and exchanges.

Arithmetic Instruction Adds, Subtracts, Increments, Decrements data in

registers or memory.

Logic Instruction ANDs, ORs, XORs, compares, rotates or

complements data in registers or between memory and registers.

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Branch/Jump Instruction Initiates conditional or unconditional jumps,

calls, returns and restart.

Stack, I/O and Machine Control Instruction Includes instructions for maintaining stack,

reading from input port, writing to output port, setting and reading interrupt mask and clearing flags.

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Programming Model0000H

0003H

0001H0002H

0005H0004H

FFFDH

0006H

FFFFHFFFEH

MEMORY

ABD EH L

CFLAG

SPPC

CPU

8 bit

8 bit 00H

03H

01H 02H

05H 04H

FDH

06H

FFH FEH

I/O

8 bit

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Data TransferIMMEDIATE DATA TRANSFER• MVI reg , data8 ;data8 (reg)• LXI rp ,data16;data16 (rp)

REGISTER DATA TRANSFER • MOV reg1 , reg2 ;(reg2) (reg1)

Reg (Register) : A,B,C,D,E,H,LRp (Register Pair) : BC,DE,HL & SP

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ExampleMVI A ,10 ;A=0AHMVI B ,10010001B ;B=91HMVI D ,7FH ;D=7FH

LXI B ,3 ;B=00H , C=03HLXI H ,2345H ;H=23H , L=45HLXI D ,100 ;D=00H , E=64HLXI SP,3FF0H ;SPH=3FH,SPL=F0H

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Example

MVI B, 55HMOV A , B MOV C , AMOV H , CMOV L , AMOV E , LHLT

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DIRECT DATA TRANSFER • LDA address16• STA address16• LHLD address16• SHLD address16

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Example

• LDA 3000H(3000H) (A)

• STA 2100H(A) (2100H)

x1

0000H

2100H

0001H..

2102H2101H

3000H

..

3509H

y1

..

..

Ax1

Ay1

STORE

LOAD

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Example

LHLD 8000H(8000H) (L)(8000H + 1) (H)

SHLD 3500H(L) (3500H) (H) (3500H + 1)

x1

y2

0000H

3500H

0001H..

3502H3501H

8000H

..

3509H

y1

..

8001H

Hx1

L y1

STORE

LOAD

x2

y2

x2

L

H

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INDIRECT DATA TRANSFER• LDAX B ;pointer is BC register• LDAX D ;pointer is DE register• STAX B ;pointer is BC register• STAX D ;pointer is DE register• MOV reg , M ;pointer is HL register• MOV M , reg ;pointer is HL register• MVI M , data8 ;pointer is HL register

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ExampleLXI B , 2020HMVI A , 88HSTAX BINX BLDAX BLXI H , 3000HMOV D , MMOV M , A 88H

AAH

0000H

0003H

0001H0002H

0005H0004H

2020H

0006H

3000H

B

D88H

00H30H

20H 20HAH L

C 2021H

FFH

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88HAAH

0000H

0003H

0001H0002H

0005H0004H

2020H

0006H

3000H

B

DAAH

00H30HFFH

20H 21HAH L

C 2021H

AAH

• Instruction INX – increment Register pair

• BC = 2021H

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• Transfer 10 byte data from memory location 3000h• To memory location 3500h using LDA & STA

LDA 3000HSTA 3500H

.

.LDA 3009HSTA 3509H

x1

x2

..x10

..

0000H

3000H

0001H..

..3001H

3500H

3009H

3509H

x1

x10

..

..

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• Transfer 10 byte data from memory location 3000h• To memory location 3500h

MVI H,10LXI B , 3000HLXI D , 3500H

LOOP: LDAX BSTAX DINX BINX DDCR HJNZ LOOPHLT

x1

x2

..x10

..

0000H

3000H

0001H..

..3001H

3500H

3009H

3509H

x1

x10

..

..

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Arithmetic Operation

• ALU • FLAG• CPU REGISTER

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Arithmetic Instruction• ADDITION

• ADI data8 (A) + data8 (A)• ADD reg (A) + (reg) (A) • ACI data8 (A) + data8 + CY (A)• ADC reg (A) + (reg) + CY (A)• DAD rp (HL) + (rp) (HL)

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EXAMPLE

• ADI 99H ; A contains 88 (H)• register A 1 0 0 0 1 0 0 0 1 3 6 decimal constant 1 0

0 1 1 0 0 1 1 5 3 decimal _____________ _____ register A 1 0 0 1 0 0 0 0 1 2 8 9 decimal

• S = 0 Bit D7 = 0 after addition• Z = 0 The accumulator contains other than zero after addition• AC = 1 There is a carry out of bit D3 to bit D4 during addition• P = 1 The accumulator contains an even number of ‘1’s after

addition• CY = 1 There is an overflow as a result of the addition

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EXAMPLE

• ADC B ; A contains 88 (H) B contains 99 (H); CY =1

CY 1register A 1 0 0 0 1 0 0 0register B 1 0 0 1 1 0 0 1

_____________register A 1 0 0 1 0 0 0 1 0

• Flag : S = 0 , Z = 0, AC = 1, P = 1,CY = 1

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SUBTRACTION

• SUI data8(A) - data8 (A)• SUB reg (A) - (reg) (A)

• SBI data8(A) - data8 - CY (A)• SBB reg (A) - (reg) - CY (A)

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INCREMENT/DECREMENT

• INR reg (reg) + 1 (reg)• DCR reg (reg) - 1 (reg)

• INX rp (rp) + 1 (rp)• DCX rp (rp) - 1 (rp)Note : No Flag Effected for INX & DCX

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Logic InstructionAND

AND Immediate With Accumulator • ANI data8 (A) Λ Data8 (A)

AND Register/Memory With Accumulator• ANA reg (A) Λ (Reg) (A)

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OR

OR Immediate With Accumulator • ORI data8 (A) V Data8 (A)

OR Register/Memory With Accumulator• ORA reg (A) V (Reg) (A)

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EXCLUSIVE-OR

EX-OR Immediate With Accumulator • XRI data8 (A) ⊕ Data8 (A)

EX-OR Register/Memory With Accumulator• XRA reg (A) ⊕ (Reg) (A)

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COMPLEMENT THE ACCUMULATOR

• CMA ( ) (A)

COMPLEMENT THE CARRY STATUS

• CMC ( ) (CY)

A

A

CY

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COMPARE

Compare Accumulator With Immediate Data

• CPI data8 (A) – data8

Compare Accumulator With Register/Memory

• CMP reg (A) – (reg)

Note: Only flag affected

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Rotate

Rotate Accumulator Right Through Carry• RAR (A0) (CY)

(An+1) (An)

(CY) (A7)A0A1A2A3A4A5A6A7CY

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Rotate

Rotate Accumulator Left Through Carry• RAL (A7) (CY)

(An) (An+1)

(CY) (A0)A0A1A2A3A4A5A6A7CY

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Rotate

Rotate Accumulator Right• RRC (A0) (A7)

(An+1) (An)

(A0) (CY)A0A1A2A3A4A5A6A7CY

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Rotate

Rotate Accumulator Left• RLC (A7) (A0)

(An) (An+1)

(A7) (CY)A0A1A2A3A4A5A6A7CY

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Branch InstructionUnconditional Jump• JMP address16(Byte 3) (Byte 2) (PC)

Conditional Jump• J Condition address16If (Condition= true)

(Byte 3) (Byte 2) (PC)

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Condition• JZ Z=1 Jump if Zero flag SET• JNZ Z=0 Jump if Zero flag NOT SET• JC CY=1 Jump if Carry flag SET• JNC CY=0 Jump if Carry flag NOT SET• JM S=1 Jump if Sign flag SET• JP S=0 Jump if Sign flag NOT SET• JPE P=1 Jump if Parity flag SET• JPO P=0 Jump if Parity flag NOT SET

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Example 1Check Zero Flag

MVI B, 255LOOP:DCR B

JNZ LOOP ;if Z == 0 then goto ;LOOP

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Example 2 – Find the smallest value between two number

(A) = x1 ; (B) = x2

LOOP: CMP B ;(A) – (B)JNC EXIT;if CY == 0 then EXITJMP STORE

EXIT: MOV A, BSTORE:STA 2050H

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Unconditional Call Subroutine • CALL address16

(PCH) ((SP) –1)(PCL) ((SP) –2)(SP) – 2 (SP) (Byte 3)(Byte 2) (PC)

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Conditional Call Subroutine C Condition address16

If (Condition = True)(PCH) ((SP) –1)(PCL) ((SP) –2)(SP) – 2 (SP) (Byte 3)(Byte 2) (PC)

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• CZ Z=1 Call if Zero flag SET• CNZ Z=0 Call if Zero flag NOT SET• CC CY=1 Call if Carry flag SET• CNCCY=0 Call if Carry flag NOT SET• CM S=1 Call if Sign flag SET• CP S=0 Call if Sign flag NOT SET• CPE P=1 Call if Parity flag SET• CPOP=0 Call if Parity flag NOT SET

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Return From Subroutine• RET

((SP)) (PCL)((SP) + 1) (PCH)(SP) + 2 (SP)

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Return From Subroutine (Conditional)

• R ConditionIf (Condition = True)((SP)) (PCL)((SP) + 1) (PCH)(SP) + 2 (SP)

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• RZ Z=1 Return if Zero flag SET• RNZ Z=0 Return if Zero flag NOT SET• RC CY=1 Return if Carry flag SET• RNCCY=0 Return if Carry flag NOT SET• RM S=1 Return if Sign flag SET• RP S=0 Return if Sign flag NOT SET• RPE P=1 Return if Parity flag SET• RPOP=0 Return if Parity flag NOT SET

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ExampleLXI SP, 3FF0H ;init Stack PointerMVI A, 80HOUT 83H ;Init 8255, all port as output

REPEAT: MVI A,0OUT 80HCALL DELAY ;Call subroutine MVI A,1OUT 80HCALL DELAY ;Call subroutine JMP REPEAT

DELAY: MVI B, 0 ;SubroutineLOOP: DCR B

JNZ LOOPRETEND

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I/O,Stack, Machine Control Instruction

Stack OperationWrite The Content of Register Pair Onto The Stack• PUSH rp(reg high) ((SP) –1)(reg low) ((SP) –2)(SP) – 2 (SP)

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Write The Content of Accumulator & Flag Status Onto The Stack

• PUSH PSW(A) ((SP) –1)(Flag) ((SP) –2)(SP) – 2 (SP)

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Retreive The Content of Register Pair From The Stack

• POP rp((SP)) (reg low)((SP) + 1) (reg high)(SP) + 2 (SP)

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Retreive The Content of Accumulator & Flag Status From The Stack

• POP PSW((SP)) (Flag)((SP) + 1) (A)(SP) + 2 (SP)

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STACK OPERATIONLecture 2

(Revision)

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How the Stack Works• The stack is a reserved area of memory. It operates as a last-in

first-out bank of registers.

• The memory locations, which constitute the stack, are used to store binary information temporarily during program execution.

• The stack can be located anywhere in read/write memory, but is usually defined such that it neither interferes with the program memory space or the data memory space.

• The start address of the stack is specified at the initialisation stage of the program by loading the 16-bit CPU register, called the stack pointer, with the desired address of the start of the stack.

– e.g LXI SP, data 16

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How the Stack Works• Data from CPU register pairs are stored in the stack area of

memory when the processor executes a push rp instruction.

• The contents of the program counter is automatically stored in the stack area of memory whenever the processor executes a call or restart (rst n) instruction.

• Data stored in the stack area of memory are returned to processor register pairs when the processor executes a pop rp instruction.

• Data is automatically transferred from the stack area of memory to the program counter whenever the processor executes a return (ret) instruction.

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Writing to the Stack• To execute the instruction push HL

assuming initial sp contents is 2099 H.

• The stack pointer is decremented by 1 (sp=2098) and the contents of H are written to this location.

• The stack pointer is decremented by 1 (sp=2097) and the contents of L are written to this location.

• Note : When data is written to the stack the stack pointer is first decremented and then the data is written

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Reading from the Stack• To execute the instruction pop BC

assuming initial sp contents is 2097 H.

• The contents of the memory location at the address specified by the contents of sp is moved to register C and sp is incremented.

• The contents of the memory location at the address specified by the contents of sp is moved to register B and sp is incremented.

• Note : When data is read from the stack the data is read first and then the stack pointer incremented.

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ExampleWrite a program to exchange the contents of BC register with DE register

Program 1 Program 2

MOV H,B PUSH BMOV L,C PUSH DMOV B,D POP BMOV C,E POP D MOV D,HMOV E,L

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Input/Output Operation Input From The Port• IN Port_Address

(port) (A)

Output To Port• OUT Port_Address

(A) (Port)

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Example

Input From The Port

IN 80H ;Read from Port 80HSTA 2100H ;Store to Memory

Output To PortMVI A, 01HOUT 81H ;Write 01H to Port 81H

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Interrupt

RIM Read interrupt maskSIM Set Interrupt maskDI Disable InterruptEI Enable Interrupt

(Detail discussion in interrupt topic)

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ASSEMBLY LANGUAGE PROGRAMMING