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Sıra DOSYA ADI Format Bağlantı
01. Transfer Microprocessor Contents Locations ppt Sunumu İndir

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Assembly Language Programming of 8085BYProf. U. V. THETEDept. of Computer ScienceYMA
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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 0’s and 1’s only• This language is called Machine Language • 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
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A Machine language program to add two numbers 00111110 ;Copy value 2H in register A0000001000000110 ;Copy value 4H in register B0000010010000000 ;A = A + B
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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
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Assembly language program to add two numbersMVI A, 2H ;Copy value 2H in register AMVI B, 4H ;Copy value 4H in register BADD B ;A = A + BNote:• Assembly language is specific to a given processor• For e.g. assembly language of 8085 is different than that of Motorola 6800 microprocessor
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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 languageAssemblyLanguageProgramAssemblerProgramMachine Language Code
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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
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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 as1. Data Transfer (Copy) 2. Arithmetic3. Logical and Bit manipulation4. Branch5. Machine Control
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1. Data Transfer (Copy) Operations1. Load a 8-bit number in a Register2. Copy from Register to Register3. Copy between Register and Memory4. Copy between Input/Output Port and Accumulator5. Load a 16-bit number in a Register pair6. Copy between Register pair and Stack memory
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Example Data Transfer (Copy) Operations / Instructions1. Load a 8-bit number 4F in register B2. Copy from Register B to Register A3. Load a 16-bit number 2050 in Register pair HL4. Copy from Register B to Memory Address 20505. Copy between Input/Output Port and AccumulatorMVI B, 4FHMOV A,BLXI H, 2050HMOV M,BOUT 01HIN 07H
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2. Arithmetic Operations1. Addition of two 8-bit numbers2. Subtraction of two 8-bit numbers3. Increment/ Decrement a 8-bit number
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Example Arithmetic Operations / Instructions1. Add a 8-bit number 32H to Accumulator2. Add contents of Register B to Accumulator3. Subtract a 8-bit number 32H from Accumulator4. Subtract contents of Register C from Accumulator5. Increment the contents of Register D by 16. Decrement the contents of Register E by 1ADI 32HADD BSUI 32HSUB C INR DDCR E
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3. Logical & Bit Manipulation Operations1. AND two 8-bit numbers2. OR two 8-bit numbers3. Exclusive-OR two 8-bit numbers4. Compare two 8-bit numbers5. Complement 6. Rotate Left/Right Accumulator bits
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Example Logical & Bit Manipulation Operations / Instructions1. Logically AND Register H with Accumulator2. Logically OR Register L with Accumulator3. Logically XOR Register B with Accumulator4. Compare contents of Register C with Accumulator5. Complement Accumulator6. Rotate Accumulator LeftANA HORA LXRA BCMP C CMARAL
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4. Branching OperationsThese operations are used to control the flow of program execution1.Jumps• Conditional jumps• Unconditional jumps2.Call & Return• Conditional Call & Return• Unconditional Call & Return
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Example Branching Operations / Instructions1. Jump to a 16-bit Address 2080H if Carry flag is SET2. Unconditional Jump3. Call a subroutine with its 16-bit Address 4. Return back from the Call5. Call a subroutine with its 16-bit Address if Carry flag is RESET6. Return if Zero flag is SETJC 2080HJMP 2050HCALL 3050H RETCNC 3050HRZ
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5. Machine Control InstructionsThese instructions affect the operation of the processor. For e.g.HLT Stop program executionNOP Do not perform any operation
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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
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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 done2. Program Logic– Add two numbers– Store result in register C– Example 10011001 (99H) A+00111001 (39H) D 11010010 (D2H) C
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1. Get two numbers 2. Add them3. Store result4. Stop• Load 1st no. in register D• Load 2nd no. in register E3. Algorithm Translation to 8085 operations • Copy register D to A• Add register E to A• Copy A to register C• Stop processing
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4. Make a FlowchartStartLoad Registers D, E Copy D to AAdd A and ECopy A to CStop• Load 1st no. in register D• Load 2nd no. in register E• Copy register D to A• Add register E to A• Copy A to register C• Stop processing
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5. Assembly Language Program1. Get two numbers2. Add them3. Store result4. Stopa) Load 1st no. in register Db) Load 2nd no. in register Ea) Copy register D to Ab) Add register E to Aa) Copy A to register Ca) Stop processingMVI D, 2HMVI E, 3HMOV A, DADD EMOV C, AHLT
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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) B100110010 (132H)– The 9th bit in the result is called CARRY bit.
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0• How 8085 does it?– Adds register A and B– Stores 8-bit result in A– SETS carry flag (CY) to indicate carry bit1001100110011001AB+99H99H10011001 A1CY0 1 0 10 99H32
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• Storing result in Register memory10011001A32H1CYRegister CRegister BStep-1 Copy A to CStep-2 a) Clear register Bb) Increment B by 1
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2. Program Logic1. Add two numbers2. Copy 8-bit result in A to C3. If CARRY is generated– Handle it4. Result is in register pair BC
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1. Load two numbers in registers D, E2. Add them3. Store 8 bit result in C4. Check CARRY flag5. If CARRY flag is SET• Store CARRY in register B6. Stop• Load registers D, E3. AlgorithmTranslation 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• t r i t r
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4. Make a FlowchartStartLoad Registers D, E Copy D to AAdd A and ECopy A to CStopIf CARRYNOT SETClear BIncrement BFalseTrue
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5. Assembly Language ProgramMVI D, 2HMVI E, 3HMOV A, DADD EMOV C, AHLT• 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• t r i t r JNC ENDMVI B, 0HINR BEND:
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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
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• The various formats of specifying operands are called addressing modes• Addressing modes of 80851. Register Addressing2. Immediate Addressing3. Memory Addressing4. Input/Output Addressing
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1. Register Addressing• Operands are one of the internal registers of 8085• Examples-MOV A, BADD C
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2. Immediate Addressing• Value of the operand is given in the instruction itself• Example-MVI A, 20HLXI H, 2050HADI 30HSUI 10H
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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
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3(a) Direct Addressing• 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 2050HSTA 3050H ;store A with contents of memory location with address 3050H
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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 HL30HA 20HH50HL30H2050H
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4. Input/Output Addressing• 8-bit address of the port is directly specified in the instruction• Examples-IN 07HOUT 21H
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5. Instruction & Data Formats8085 Instruction set can be classified according to size (in bytes) as1. 1-byte Instructions2. 2-byte Instructions3. 3-byte Instructions
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1. One-byte Instructions• Includes Opcode and Operand in the same byte• Examples-Opcode Operand Binary Code Hex CodeMOV C, A 0100 1111 4FHADD B 1000 0000 80HHLT 0111 0110 76H
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2. Two-byte Instructions• First byte specifies Operation Code• Second byte specifies Operand• Examples-Opcode Operand Binary Code Hex CodeMVI A, 32H 0011 11100011 00103EH32HMVI B, F2H 0000 01101111 001006HF2H
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3. Three-byte Instructions• First byte specifies Operation Code• Second & Third byte specifies Operand• Examples-Opcode Operand Binary Code Hex CodeLXI H, 2050H 0010 00010101 00000010 000021H50H20HLDA 3070H 0011 10100111 00000011 00003AH70H30H
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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 451111 0000 F0---------------0100 0000 40• RRC• RRC• RRC ; Adjust higher digit as a lower digit.• RRC 0000 0100 after 4 rotations
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Contd.• STA 2300H ; Store the partial result• LDA 2200H ; Get the original BCD no.• ANI 0FH ; Mask higher nibble0100 0100 450000 1111 0F---------------0000 0100 05• STA 2301H ; Store the result• HLT ; Terminate program execution
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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 pointerBK: 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•
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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