In the article, there are few examples of data transfer program in 8085 microprocessor in assembly language programming (ALP). You need to learn the instruction set of 8085 in detail before programming.
One program can be performed in multiple ways. Here they are mentioned as Program 1, Program 2 … Comments in each program is written after semicolon (;).
Data Transfer Program in 8085
Store the Data in 8085 Microprocessor
Store the data byte 50H into memory location 2000H.
Program 1:
MVI A, 50H ; Store 50H in the accumulator
STA 2000H ; Copy accumulator contents at memory location 2000H
HLT ; Terminate program execution.
Program 2:
MVI A, 50H ; Store 50H in the accumulator
LXI H, 2000H ; Load HL register pair with value(data) 2000H
MOV M, A ; Copy accumulator contents in memory location pointed by HL register pair (2000H)
HLT ; Terminate program execution.
Program 3:
MVI A, 50H ; Store 50H in the accumulator
LXI B, 2000H ; Load BC register pair with value(data) 2000H
STAX B ; Copy the accumulator contents in memory address indicated by BC pair
HLT ; Terminate program execution.
Sample Example:
(2000H) = F3H (or mostly 00H by default)
Result= (2000H)=50H
Copy the contents of memory location 1000H to register C.
Program:
LDA 1000H ; Get the contents of memory location 1000H into accumulator
MOV C, A ; Copy the accumulator into register C
HLT ; Terminate program execution.
Sample Example:
(1000H) = 25H
Result= (C)=25H
Copy the contents of register B to memory location 1000H.
Program:
MOV A, B ; Copy the accumulator into register C
STA 1000H ; Get the contents of memory location 1000H into accumulator
HLT ; Terminate program execution.
Sample Example:
(B) = 4AH
Result= (1000H)=4AH
Exchange the contents in 8085 Microprocessor
Exchange the contents of memory locations 2000H and 2001H.
Program 1:
LDA 2000H ; Get the contents of memory location 2000H into accumulator
MOV B, A ; Save the contents into B register
LDA 2001H ; Get the contents of memory location 2001H into accumulator
STA 2000H ; Store the contents of accumulator at address 2000H
MOV A, B ; Get the saved contents back into A register
STA 2001H ; Store the contents of accumulator at address 2001H
HLT ; Terminate program execution.
Program 2:
LXI H, 2000H ; Initialize HL register pair as a pointer to memory location 2000H
LXI D, 2001H ; Initialize DE register pair as a pointer to memory location 2001H
MOV B, M ; Get the contents of memory location 2000H into B register
LDAX D ; Get the contents of memory location 2001H into A register
MOV M, A ; Store the contents of A register into memory location 2000H
MOV A, B ; Copy the contents of B register into accumulator
STAX D ; Store the contents of A register into memory location 2001
HLT ; Terminate program execution.
Program 3:
LXI B, 2000H ; Initialize BC register pair as a pointer to memory location 2000H
LDAX B ; Get the contents of memory location 2000H into accumulator
MOV D, A ; Copy accumulator into register D
INX B ; increment the content of BC register by 1 (BC=2001)
LDAX B ; Load accumulator with content of memory indicated by BC register
DCX B ; Decrement the content of BC register by 1 (BC=2000)
STAX B ; Copy the contents of accumulator to address indicated by BC register pair
INX B ; increment the content of BC register by 1 (BC=2001)
MOV A, D ; Copy the register D into accumulator
STAX B ; Store the contents of accumulator into memory location 2001
HLT ; Terminate program execution
Program 4:
LXI B, 2000H ; Initialize BC register pair as a pointer to memory location 2000H
LDAX B ; Get the contents of memory location 2000H into accumulator
MOV D, A ; Copy accumulator into register D
INX B ; Increment the content of BC register by 1 (BC=2001)
LDAX B ; Load accumulator with content of memory indicated by BC register
MOV E, A ; Copy accumulator to register E
MOV A, D ; Copy register D to accumulator
STAX B ; Store the contents of accumulator into memory location 2001
DCX B ; Decrement the content of BC register by 1 (BC=2000)
MOV A, E ; Copy register E to accumulator
STAX B ; Store the contents of accumulator into memory location 2000
HLT ; Terminate program execution
Sample Example:
(2000H) = 55H
(2001H)= 44H
Result= (2000H)=44H and (2001H)=55H
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