Programming with 8085: An Introduction to Assembly Language Programming

1. PROGRAMMING WITH 8085 BTCS-404 (MALP) B.Tech 4th SEM. IT Ajay Kumar Dogra Associate Professor, CSE BEANT COLLEGE OF ENGG. & TECH. GURDASPUR

2. Assembly Language Programming of 8085

3. Topics 1. Introduction 2. Programming model of 8085 3. Instruction set of 8085 4. Example Programs 5. Addressing modes of 8085 6. Instruction & Data Formats of 8085

4. 1. Introduction A microprocessor executes instructions given by the user Instructions should be in a language known to the microprocessor Microprocessor understands the language of 0s and 1s only This language is called Machine Language

5. For e.g. 01001111 Is a valid machine language instruction of 8085 It copies the contents of one of the internal registers of 8085 to another

6. A Machine language program to add two numbers 00111110 ;Copy value 2H in register A 00000010 00000110 ;Copy value 4H in register B 00000100 10000000 ;A = A + B

7. 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

8. Assembly language program to add two numbers MVI A , 2H ;Copy value 2H in register A MVI B , 4H ;Copy value 4H in register B ADD B ;A = A + B Note: Assembly language is specific to a given processor For e.g. assembly language of 8085 is different than that of Motorola 6800 microprocessor

9. Microprocessor understands Machine Language only! Microprocessor cannot understand a program written in Assembly language A program known as Assembler is used to convert a Assembly language program to machine language Assembly Language Program Assembler Program Machine Language Code

10. Low-level/High-level languages Machine language and Assembly language are both Microprocessor specific ( Machine dependent ) so they are called Low-level languages Machine independent languages are called High-level languages For e.g. BASIC, PASCAL,C++,C,JAVA, etc. A software called Compiler is required to convert a high-level language program to machine code

11. 2. Programming model of 8085 Accumulator ALU Flags Instruction Decoder Register Array Memory Pointer Registers Timing and Control Unit 16-bit Address Bus 8-bit Data Bus Control Bus

12. A ccumulator (8-bit) Flag Register (8-bit) B (8-bit) C (8-bit) D (8-bit) E (8-bit) H (8-bit) L (8-bit) Stack Pointer (SP) (16-bit) Program Counter (PC) (16-bit) S Z AC P CY 16- Lines Unidirectional 8- Lines Bidirectional

13. Overview: 8085 Programming model 1. Six general-purpose Registers 2. Accumulator Register 3. Flag Register 4. Program Counter Register 5. Stack Pointer Register

14. 1. Six general-purpose registers B, C, D, E, H, L Can be combined as register pairs to perform 16-bit operations ( BC, DE, HL ) 2. Accumulator identified by name A This register is a part of ALU 8-bit data storage Performs arithmetic and logical operations Result of an operation is stored in accumulator

15. 3. Flag Register This is also a part of ALU 8085 has five flags named Zero flag (Z) Carry flag (CY) Sign flag (S) Parity flag (P) Auxiliary Carry flag (AC)

16. These flags are five flip-flops in flag register Execution of an arithmetic/logic operation can set or reset these flags Condition of flags (set or reset) can be tested through software instructions 8085 uses these flags in decision-making process

17. 4. Program Counter (PC) A 16-bit memory pointer register Used to sequence execution of program instructions Stores address of a memory location where next instruction byte is to be fetched by the 8085 when 8085 gets busy to fetch current instruction from memory PC is incremented by one PC is now pointing to the address of next instruction

18. 5. Stack Pointer Register a 16-bit memory pointer register Points to a location in Stack memory Beginning of the stack is defined by loading a 16-bit address in stack pointer register

19. 3. 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 1. Data Transfer (Copy) 2. Arithmetic 3. Logical and Bit manipulation 4. Branch 5. Machine Control

20. 1. Data Transfer (Copy) Operations 1. Load a 8-bit number in a R egister 2. Copy from R egister to R egister 3. Copy between R egister and Memory 4. Copy between Input / Output Port and A ccumulator 5. Load a 16-bit number in a R egister pair 6. Copy between R egister pair and Stack memory

21. Example Data Transfer (Copy) Operations / Instructions 1. Load a 8-bit number 4F in register B 2. Copy from Register B to Register A 3. Load a 16-bit number 2050 in Register pair HL 4. Copy from Register B to Memory Address 2050 5. Copy between Input / Output Port and A ccumulator MVI B , 4FH MOV A , B LXI H , 2050H MOV M , B OUT 01H IN 07H

22. 2. Arithmetic Operations 1. Addition of two 8-bit numbers 2. Subtraction of two 8-bit numbers 3. Increment/ Decrement a 8-bit number

23. Example Arithmetic Operations / Instructions 1. Add a 8-bit number 32H to A ccumulator 2. Add contents of Register B to A ccumulator 3. Subtract a 8-bit number 32H from A ccumulator 4. Subtract contents of Register C from A ccumulator 5. Increment the contents of Register D by 1 6. Decrement the contents of Register E by 1 ADI 32H ADD B SUI 32H SUB C INR D DCR E

24. 3. Logical & Bit Manipulation Operations 1. AND two 8-bit numbers 2. OR two 8-bit numbers 3. Exclusive-OR two 8-bit numbers 4. Compare two 8-bit numbers 5. Complement 6. Rotate Left/Right Accumulator bits

25. Example Logical & Bit Manipulation Operations / Instructions 1. Logically AND Register H with A ccumulator 2. Logically OR Register L with A ccumulator 3. Logically XOR Register B with A ccumulator 4. Compare contents of Register C with A ccumulator 5. Complement A ccumulator 6. Rotate A ccumulator Left ANA H ORA L XRA B CMP C CMA RAL

26. 4. Branching Operations These operations are used to control the flow of program execution 1. Jumps Conditional jumps Unconditional jumps 2. Call & Return Conditional Call & Return Unconditional Call & Return

27. Example Branching Operations / Instructions 1. Jump to a 16-bit Address 2080H if C arry flag is SET 2. Unconditional Jump 3. Call a subroutine with its 16-bit Address 4. Return back from the Call 5. Call a subroutine with its 16-bit Address if C arry flag is RESET 6. Return if Z ero flag is SET JC 2080H JMP 2050H CALL 3050H RET CNC 3050 H RZ

28. 5. Machine Control Instructions These instructions affect the operation of the processor. For e.g. HLT Stop program execution NOP Do not perform any operation

29. 4. Writing a Assembly Language Program Steps to write a program Analyze the problem Develop program Logic Write an Algorithm Make a Flowchart Write program Instructions using Assembly language of 8085

30. Program 8085 in Assembly language to add two 8- bit numbers and store 8-bit result in register C . 1. Analyze the problem Addition of two 8-bit numbers to be done 2. Program Logic Add two numbers Store result in register C Example 10011001 (99H) A +00111001 (39H) D 11010010 (D2H) C

31. 1. Get two numbers 2. Add them 3. Store result 4. Stop Load 1 st no. in register D Load 2 nd no. in register E 3. Algorithm Translation to 8085 operations Copy register D to A Add register E to A Copy A to register C Stop processing

32. 4. Make a Flowchart Start Load Registers D, E Copy D to A Add A and E Copy A to C Stop Load 1 st no. in register D Load 2 nd no. in register E Copy register D to A Add register E to A Copy A to register C Stop processing

33. 5. Assembly Language Program 1. Get two numbers 2. Add them 3. Store result 4. Stop a) Load 1 st no. in register D b) Load 2 nd no. in register E a) Copy register D to A b) Add register E to A a) Copy A to register C a) Stop processing MVI D, 2H MVI E, 3H MOV A, D ADD E MOV C, A HLT

34. Program 8085 in Assembly language to add two 8- bit numbers. Result can be more than 8-bits. 1. Analyze the problem Result of addition of two 8-bit numbers can be 9-bit Example 10011001 (99H) A +10011001 (99H) B 1 00110010 ( 1 32H) The 9 th bit in the result is called CARRY bit.

35. 0 How 8085 does it? Adds register A and B Stores 8-bit result in A SETS carry flag (CY) to indicate carry bit 10011001 10011001 A B + 99H 99H 10011001 A 1 CY 00110010 99H 32H

36. Storing result in Register memory 10011001 A 32H 1 CY Register C Register B Step-1 Copy A to C Step-2 a) Clear register B b) Increment B by 1

37. 2. Program Logic 1. Add two numbers 2. Copy 8-bit result in A to C 3. If CARRY is generated Handle it 4. Result is in register pair BC

38. 1. Load two numbers in registers D, E 2. Add them 3. Store 8 bit result in C 4. Check CARRY flag 5. If CARRY flag is SET Store CARRY in register B 6. Stop Load registers D, E 3. Algorithm Translation to 8085 operations Copy register D to A Add register E to A Copy A to register C Stop processing Use Conditional Jump instructions Clear register B Increment B Copy A to register C

39. 4. Make a Flowchart Start Load Registers D, E Copy D to A Add A and E Copy A to C Stop If CARRY NOT SET Clear B Increment B False True

40. 5. Assembly Language Program MVI D, 2H MVI E, 3H MOV A, D ADD E MOV C, A HLT Load registers D, E Copy register D to A Add register E to A Copy A to register C Stop processing Use Conditional Jump instructions Clear register B Increment B Copy A to register C JNC END MVI B, 0H INR B END:

41. 4. Addressing Modes of 8085 Format of a typical Assembly language instruction is given below- [ Label: ] Mnemonic [ Operands ] [ ;comments ] HLT MVI A , 20H MOV M , A ;Copy A to memory location whose address is stored in register pair HL LOAD: LDA 2050H ;Load A with contents of memory location with address 2050H READ: IN 07H ;Read data from Input port with address 07H

42. The various formats of specifying operands are called addressing modes Addressing modes of 8085 1. Register Addressing 2. Immediate Addressing 3. Memory Addressing 4. Input / Output Addressing

43. 1. Register Addressing Operands are one of the internal registers of 8085 Examples- MOV A , B ADD C

44. 2. Immediate Addressing Value of the operand is given in the instruction itself Example- MVI A , 20H LXI H , 2050H ADI 30H SUI 10H

45. 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

46. 3(a) Direct Addressing 16-bit Address of the memory location is specified in the instruction directly Examples- LDA 2050 H ; load A with contents of memory location with address 2050H STA 3050 H ;store A with contents of memory location with address 3050H

47. 3(b) Indirect Addressing 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 30H A 20H H 50H L 30H 2050H

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

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

50. 1. One-byte Instructions Includes Opcode and Operand in the same byte Examples- Opcode Operand Binary Code Hex Code MOV C , A 0100 1111 4F H ADD B 1000 0000 80 H HLT 0111 0110 76 H

51. 1. Two-byte Instructions First byte specifies Operation Code Second byte specifies Operand Examples- Opcode Operand Binary Code Hex Code MVI A , 32H 0011 1110 0011 0010 3E H 32 H MVI B , F2H 0000 0110 1111 0010 06 H F2 H

52. 1. Three-byte Instructions First byte specifies Operation Code Second & Third byte specifies Operand Examples- Opcode Operand Binary Code Hex Code LXI H , 2050H 0010 0001 0101 0000 0010 0000 21 H 50 H 20 H LDA 3070H 0011 1010 0111 0000 0011 0000 3A H 70 H 30 H

53. Separate the digits of a hexadecimal numbers and store it in two different locations LDA 2200H ; Get the packed BCD number ANI F0H ; Mask lower nibble 0100 0101 45 1111 0000 F0 --------------- 0100 0000 40 RRC RRC RRC ; Adjust higher digit as a lower digit. RRC 0000 0100 after 4 rotations

54. Contd. STA 2300H ; Store the partial result LDA 2200H ; Get the original BCD no. ANI 0FH ; Mask higher nibble 0100 0100 45 0000 1111 0F --------------- 0000 0100 05 STA 2301H ; Store the result HLT ; Terminate program execution

55. Block data transfer MVI C, 0AH ; Initialize counter i.e no. of bytes Store the count in Register C, ie ten LXI H, 2200H ; Initialize source memory pointer Data Starts from 2200 location LXI D, 2300H ; Initialize destination memory pointer BK: MOV A, M ; Get byte from source memory block i.e 2200 to accumulator. STAX D ; Store byte in the destination memory block i.e 2300 as stored in D-E pair

56. Contd. INX H ; Increment source memory pointer INX D ; Increment destination memory pointer DCR C ; Decrement counter to keep track of bytes moved JNZ BK ; If counter 0 repeat steps HLT ; Terminate program