1. UNDERSTANDING
ASSEMBLY LANGUAGE

2. Instruction A microprocessor executes instructions given by the user.
Instructions should be in a language known to the microprocessor.
Microprocessor understands the language of 0’s and 1’s only.
This language is called Machine Language.
An instruction is a binary pattern designed inside a microprocessor to perform a specific function .
Intel 8085 has 246 instructions.
Each instruction is represented by an 8-bit binary value.

3. Instruction
The instruction set is a collection of pre-defined machine codes, which the CPU is designed to expect and be able to act upon when detected.
Each machine code of an instruction set consists of two separate fields:
The OPCODE is a short code which indicates what operation is expected to be performed. Each operation has a unique opcode.
The OPERAND, or operands, indicate where the data required for the operation can be found and how it can be accessed (the addressing mode, which is discussed in full later).

4. Instruction
The length of a machine code can vary - common lengths vary from one to twelve bytes in size.
Opcodes are also given mnemonics (short names) so that they can be easily referred to in code listings and similar documentation.

5. Microprocessor understands Machine Language only!
Microprocessor cannot understand a program written in Assembly language.
A program known as Assembler is used to convert an Assembly language program to machine language.
Assembly language is the language in which the mnemonics (short-hand of instructions) are used to write a program.

6. Examples
Opcode
Operand
MOV
C, A
ADD B
MVI
A, 32H
B, F2H

7. A Machine language program to add two numbers
;Copy value 2H in register A
;Copy value 4H in register B
;A = A + B

8. Assembly Language of 8085
It uses English like words to convey the action/meaning called as MNEMONICS
For e.g.
MOV to indicate data transfer
ADD to add two values
SUB to subtract two values

9. Assembly language program to add two numbers
Note:
Assembly language is specific to a given processor

10. Instruction Set of 8085 Consists of 74 operation codes, e.g. MOV
246 Instructions, e.g. MOV A,B
8085 instructions can be classified as
Data Transfer (Copy) instructions
Arithmetic instructions
Logical and Bit manipulation
Branching instructions
Machine Control instructions

11. Example Data Transfer (Copy) Operations / Instructions
Load a 8-bit number 4F in register B
Copy from Register B to Register A
Load a 16-bit number 2050 in Register pair HL
Copy from Register B to Memory Address 2050
Copy between Input/Output Port and Accumulator
MVI B, 4FH
MOV A,B
LXI H, 2050H
MOV M,B
OUT 01H
IN 07H

12. Example Arithmetic Operations / Instructions
Add a 8-bit number 32H to Accumulator
Add contents of Register B to Accumulator
Subtract a 8-bit number 32H from Accumulator
Subtract contents of Register C from Accumulator
Increment the contents of Register D by 1
Decrement the contents of Register E by 1
ADI 32H
ADD B
SUI 32H
SUB C
INR D
DCR E

13. Example Logical & Bit Manipulation Operations / Instructions
Logically AND Register H with Accumulator
Logically OR Register L with Accumulator
Logically XOR Register B with Accumulator
Compare contents of Register C with Accumulator
Complement Accumulator
Rotate Accumulator Left
ANA H
ORA L
XRA B
CMP C
CMA
RAL

14. Addressing Modes of 8085
Addressing modes specify the way in which the address of the operand is given in the instruction.
         Opcode    [Operands] [;comments]   E.g.:     MVI   B, 05   : Here MVI (Move Immediate) is the opcode part and the rest is the operand part.                 MOV  A , B   : Here MOV is the opcode part and the rest is the operand part.                 ADD   B        : Here ADD is the opcode part and
the rest is the operand part.       

15. Addressing Modes of 8085
The various formats of specifying operands are called addressing modes.
Addressing modes of 8085:
Register Addressing
Immediate Addressing
Memory Addressing
Input/Output Addressing

16. 1. Register Addressing
In this addressing mode, a register contains the operand. Depending upon the instruction, the register may be the first operand, the second operand or both.
Operands are one of the internal registers of 8085.
Examples:
MOV A, B
ADD C

17. 2. Immediate Addressing
In the immediate addressing mode, direct data is given in the operand which move the data in accumulator. It is very fast.
Value of the operand is given in the instruction itself
Examples:
MVI A, 20H
LXI H, 2050H
ADI 30H
SUI 10H

18. 3. Memory Addressing One of the operands is a memory location
Depending on how address of memory location is specified, memory addressing is of two types:
Direct addressing
Indirect addressing

19. 3(a) Direct Addressing
In the direct addressing mode, address of the operand is given in the instruction and data is available in the memory location which is provided in instruction. We will move this data in desired location.
16-bit Address of the memory location is specified in the instruction directly.
Examples:
LDA 2050H ;load A with contents of memory location with address 2050H
STA 3050H ;store A with contents of memory location with address 3050H

20. Direct Addressing Diagram
Instruction
Opcode
Address A
Memory
Operand 6

21. 3(b) Indirect Addressing
In the indirect addressing mode, the instruction specifies a register which contain the address of the operand. Both internal RAM and external RAM can be access via indirect addressing mode.
A memory pointer register is used to store the address of the memory location.
Example:
MOV M, A ;copy register A to memory location whose address is stored in register pair HL H L A
30H
20H
50H
2050H
30H

22. Indirect Addressing Diagram
Instruction
Opcode
Address A
Memory
Pointer to operand
Operand 6

23. HL Register
The HL register is the only register that can be used to specify a memory address.
It is called a pointer register because it points to a memory location.
The H register contains the high byte of the address, and the L register the low byte.
High and low bytes may be unfamiliar to you. Consider the decimal number 47. From convention we know that 4 is the number of "tens", and the 7 is the number of "units", high and then low.
For the hex number 47, 4 is the number of "sixteens", and 7 is the number of "units".
In the same way for the number 4768, 47 is the number of "two-hundred-fifty-sixes", and 68 is the number of "units".
We split it this way because the first two digits can fit in the high byte (register) and the second two can fit in the low register.

24. MOV Instruction
This instruction is used to copy the data from one place to another.
The MOV instruction takes two operands.
Syntax:
Syntax of the MOV instruction is:
The MOV instruction may have one of the following five forms:

25. MOV Instruction Examples:
MOV Rd, Rs (This instruction copies the content of Rs to Rd)
MOV M, Rs (This instruction copies the content of register Rs to memory location pointed by HL Register)
MOV Rd, M (This instruction copies the content of memory location pointed by the HL register to the register Rd.)
*Note: the contents of the source register are not altered.

26. MVI Instruction
MVI: move immediate date to a register or memory location.
Examples:
MVI Rd, #30H (30h is stored in register Rd)
MVI M, #30H (30h is stored in memory location pointed by HL Reg)

27. 4. Input/Output Addressing
8-bit address of the port is directly specified in the instruction
Examples:
IN 07H
OUT 21H

28. 5. Instruction & Data Formats
8085 Instruction set can be classified according to size (in bytes) as:
1-byte Instructions
2-byte Instructions
3-byte Instructions

29. 1. One-byte Instructions
Includes Opcode and Operand in the same byte
Examples:
Opcode
Operand
Binary Code
Hex Code
Task
MOV
C, A
4FH
Copy the contents of the accumulator in the register C.
ADD B
80H
Add the contents of register B to the contents of the accumulator.
HLT
76H

30. 2. Two-byte Instructions
First byte specifies Operation Code
Second byte specifies Operand
Examples:
Opcode
Operand
Binary Code
Hex Code
MVI
A, 32H
3EH
32H
B, F2H
06H
F2H

31. 3. Three-byte Instructions
First byte specifies Operation Code
Second & Third byte specifies Operand
Examples:
Opcode
Operand
Binary Code
Hex Code
LXI
H, 2050H
21H
50H
20H
LDA
3070H
3AH
70H
30H

32. An example assembly language program
Address Instruction
202A MVI A, 21 ;Copies 21 into accumulator
202C MVI B, 2A ;Copies 2A into B register
202E ADD B ;Adds B reg content with Acc and stores the result in Acc.
202F STA 41 FF ; Stores the Acc (the sum) into the memory location 41 FF.
2032 HLT ; Stops the program

33. Another example assembly language program
Address Instruction
2020 MVI B, 24 ;Copies 24 into accumulator
2022 INR B ;Increment B reg content by 1
2023 MOV A, B ;Copies B register into Acc.
2024 SUB B ;Subtracts B reg content from Acc and stores the result in Acc.
2025 STA 5F FF ; Stores the Acc content into the memory location 5F FF.
2028 HLT ; Stops the program