1. Instruction set Architecture
DEPARTMENT OF COMPUTER SCIENCE & TECHNOLOGY
FACULTY OF SCIENCE & TECHNOLOGY
UNIVERSITY OF UWA WELLASSA 1 1

2. Introduction
The instruction set of a microprocessor is a list of all the software instructions that the processor can execute.
When an instruction expressed in machine code, it is encoding using 0’s and 1’s(in hexadecimal format)
A single machine instruction can take anywhere from 1 to 6 byte code. 2 2 2

3. Instruction Format Format of a typical microprocessor instruction
E.g. MOV AX,BX
ADD AX,BX
Op-code are usually written in the form called mnemonic.
E.g. MOVE MOV
ADDITION ADD
INCREASE INC
Op-code
Operands
Identifies the action to be taken
Identifies the data to be operated 3 3 3

4. Instruction Format An instruction may have 0-3 number of operands E.g.
No of operands
Instruction
Meaning
HLT
Halt the processor 1
INC AX
Add 1 to the value of register AX 2
MOV AX,BX
Copy the value of BX to AX 3
SHLD DX,AX,4
Shift register AX 4 bits left into DX 4 4 4

5. Classes of 8086 instruction set
Data movement
MOV,PUSH,POP
Conversion
CBW,CWD
Arithmetic
ADD,SUB,MUL,DIV,CMP
Logical,shift,rotate and bit instructions
AND,OR,SHL,SHR,RCL,RCR
Input/output instructions
IN,OUT
String instructions
MOVS ,STOS,LODS
Program flow control
JMP,CALL,RET,Conditional JUMP
Miscellaneous
STC,CMC 5 5 5

6. Addressing modes of 8086
Addressing mode refers to valid form of the operand or operands of an instruction.
The operands can be registers, memory locations(memory variables) or constant values.
There are 3 types of addressing modes
Register addressing mode
Immediate addressing mode
Memory Addressing mode 6 6 6

7. Register Addressing mode
For an instruction if all the operands are internal registers of 8086 microprocessor then it is in register addressing mode.
E.g.
ADD AX,BX
MOV AX,BX
INC AX
Not all register conditions are valid addressing mode. There are some conditions.
Conditions
The size of the operands should match.
Byte registers AL,BL,CL,DL,AH,BH,….(8 bits)
Word registers AX,BX,CX,DX….(16 bits) 7 7 7

8. Register Addressing mode…
ADD AL,BL
ADD AX,BL
ADD AX,BX
ADD AL,BX
Valid
Invalid
Valid
Invalid
Both operands cannot be in segment registers
e.g.
MOV DS,CS
If you need to move the value stored in CS register to DS register have to use data registers in between.
MOV AX,CS
MOV DS,AX 8 8 8

9. Immediate Addressing mode
MOV AX,100
Source is a constant value. It is a part of this instruction. Not stored in a register or memory location. So operand is called as an immediate operand.
To accesses immediate operand immediate addressing mode is used.
e.g.
MOV AX,100
MOV a,100 where a is a byte in memory
The source value is immediately available in ALU and the processor does not need to access any other register or a memory location to fetch the data
Therefore the operation will perform much faster 9 9 9

10. Immediate Addressing mode…
Conditions
The size of two operands should be match.
Byte register can store only
MOV AL,300
MOV AX,300
We can define a constant and then use it as source operand.
This is also an immediate addressing mode.
But the size of defined constant and the destination register or memory location should be matched.
e.g.
c EQU 20
MOV AL,c
Invalid
Valid 10 10 10

11. Immediate Addressing mode…
When defining a memory variable
a db 100
a dw 1000
a dd 11 11 11

12. Memory Addressing mode
If a memory location is used for an operand then it is in memory addressing mode.
There are 5 types of memory addressing modes.
Direct memory addressing mode
Register indirect memory addressing mode
Based memory addressing mode
Indexed memory addressing mode
Based Indexed addressing mode 12 12 12

13. Memory Addressing mode…
Direct memory addressing mode
one operand is refers to a memory location directly and other operand refers to a register
e.g. a db 10 a allocate a 1 byte
MOV AL,a
ADD AL,a
SUB AL,a
All these instructions are directly accessing the memory location(without using register)
Size of memory location and size of registers must match. 10 13 13 13

14. Memory Addressing mode…
Example
Suppose we define a double word memory variable with the value 169,772,160(=0A1E8480H)
a dd a
(refers to the 1st byte)
Highest addressed word
Lowest addressed word 0A 80 1E 84 14 14 14

15. Memory Addressing mode…
MOV AL,a AL=80H(AL store only the lowest byte of a)
MOV AX, a AX=8480H(AX store only the lowest word of a) If
value <=255; Store in a byte register(8 bit)
value <=65535; Store in a word register(16 bit)
value> Store in 2 word register(32 bit)
previous double word can be
store as two words
store as four bytes
in registers 15 15 15

16. Memory Addressing mode…
Register indirect memory addressing mode
offset is placed in a register and then use that register to access the item pointed by the register
Registers used BX
SI with data segment DI
BP - with stack segment
In order to obtain the offset of a variable OFFSET operator is used. To access data stored in that memory location [] operator is used. 16 16 16

17. Memory Addressing mode…
Examples
a db 100
MOV BX,OFFSET a
MOV AL,[BX]
AL=100
Physical Address =DS:BX
Use this method to access a memory location of an array or character string DS
100
1 byte a
offset 17 17 17

18. Memory Addressing mode…
ARR db 1,2,3,4,5
MOV BX,OFFSET ARR
MOV CX,5
NEXT: MOV AL,[BX]
INC BX
LOOP NEXT DS 2 1 4 3
Offset
ARR 18 18 18

19. Memory Addressing mode…
Base memory addressing mode
This is similar to register indirect memory addressing mode, except that the displacement can be added to the register value, forming the offset to be used as the memory address.
e.g.
MOV BX,OFFSET ARR
MOV AL,[BX+3]
AL=4 DS 2 1 4 3
Offset
Displacement
ARR 19 19 19

20. Memory Addressing mode…
The displacement can be specified
MOV AX,[reg+disp]
MOV AX,[reg][disp]
MOV AX,disp[reg]
MOV AX,[reg]+disp
Example
ARR db 1,2,3,4,5
MOV BX,OFFSET ARR
MOV AL,[BX+2]
MOV AL,[BX][2]
MOV AL,2[BX]
MOV AL,[BX]+2 20 20 20

21. Memory Addressing mode…
Index memory addressing mode
Index addressing works in a similar manner to base addressing except the displacement is placed in a register and the offset of the data structure specified in the name itself.
SI and DI registers are typically used for the displacement although BX and BP can be used DS 2 1 4 3
Offset
e.g.
ARR db 1,2,3,4,5
MOV SI,3
MOV AL,ARR[SI]
MOV AL,ARR[SI+1]
Displacement
ARR 21 21 21

22. Memory Addressing mode…
Calculate the sum of two byte arrays
ARR1 db 1,2,3,4,5
ARR2 db 2,4,6,8,10
MOV SI,0
MOV CX,5
NEXT:MOV AL,ARR1[SI]
ADD ARR2[SI],AL
INC SI
LOOP NEXT 22 22 22

23. Memory Addressing mode…
Base Index memory addressing mode
This addressing mode is a combination of both base and index addressing modes.
Here both offset and displacement are store in registers
offset=BX/BI
displacement=SI/DI
Either BX or BP in combination with SI or DI is used. neither BX and BP nor SI and DI can be used together
MOV DI,OFFSET ARR
MOV SI,7
MOV BX,OFFSET ARR
MOV BP,5
Invalid
Invalid 23 23 23

24. Memory Addressing mode…
e.g.
ARR db 1,2,3,4,5
MOV BX,OFFSET ARR
MOV SI,3
MOV AL,[BX+SI]
MOV AL,[BX][SI]
MOV AL,[BX]+[SI]
MOV AL,[SI][BX] 24 24 24

25. Stack Segment
The stack segment is a special area of memory designed for quick temporary storage of program data and the subroutine return addresses.
It is a LIFO (Last In First Out) memory.
SP Register – Points to the top of the stack. Contains the offset of next memory location to be access in the stack segment.
SS Register – Points to the base of the stack.
BP Register – Contains the offset from the base any other memory location in stack segment.
BP Register
SP Register
SS Register 25 25 25

26. PUSH and POP instructions
In Assembly programming data is push on to the stack by the PUSH instruction and data is pop out from the stack by POP instruction.
Operands must be 16 bit registers.
PUSH <Operand> Operands must be 16 bit registers
POP <Operand>
In the stack segment the data define type should be in word type and therefore the operands must be 16 bit registers. 26 26 26

27. PUSH and POP instructions…
Example –
MOV AX, 100
PUSH AX
POP AX 00 AH 64 AL AH AL
AH comes first
SP Register AH AL
AL comes first
SP Register 27 27 27

28. PUSH and POP instructions…
In Assembly programming we should allocate memory space to stack segment.
db 100 dup(?) ; Allocate 100 bytes for the stack segment
Example
MOV AX, 100
PUSH AX
MOV BX, 200
PUSH BX
MOV CX, 300
PUSH CX
POP CX
POP BX
After each PUSH and POP instruction show the content of the stack segment 28 28 28

29. PUSH and POP instructions…
MOV AX, 100
PUSH AX
MOV BX, 200
PUSH BX AH AL
SP Register AH AL BH BL
SP Register 29 29 29

30. PUSH and POP instructions…
MOV CX, 300
PUSH CX
POP CX AH AL BH BL CH CL
SP Register AH AL BH BL CH CL
SP Register 30 30 30

31. PUSH and POP instructions…
POP BX AH AL BH BL
SP Register 31 31 31

32. Use of stack segment Can use as a stack data structure
We can use stack segment as a normal stack when we want to perform some operations using a stack data structure.
When interrupt is requested before transferring the control to the ISR (Interrupt Service Routine), the processor save the current program addresses and data on to the stack (using PUSH operation).
When the ISR finishes its execution by using IRET instruction processor will pop the current program data. (Original values) 32 32 32

33. Use of stack segment…
In Assembly language program, before ‘INT21H’ instruction we have different values in registers (SS, CS, DS and AX).
When we execute ‘INT21H’ instruction what will happen is, it delete all the contents in these registers and use them to perform operations within the ‘INT21H’.
Suppose we have to execute the ‘INT21H’ instruction in the middle of our program. Then already kept data and instructions in SS, CS, DS and AX registers should be saved.
In a situation like this the processor uses stack segment. Processors push these values into stack segment and then free the registers.
After performing the operations within ‘INT21H’, the original data should keep in the current registers.
Processor use IRET instruction to do this. 33 33 33