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8085 program to move blocks of bits from source location to a destination location

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Problem – Write a program to move blocks of bits from source location starting at 2500 to destination location starting from 2600 where size of blocks is 05 bytes. Example – Algorithm –

  1. Load register pair H-L with the address 2500H
  2. Load register pair D-E with the address 2600H
  3. Move the content at memory location into accumulator
  4. Store the content of accumulator into memory pointed by D-E
  5. Increment value of register pair H-L and D-E by 1
  6. Decrements value of register C by 1
  7. If zero flag not equal to 1, go to step 3
  8. Stop

Program –

Memory Mnemonics Operands Comment
2000 MVI C, 05 [C] <- 05
2002 LXI H, 2500 [H-L] <- 2500
2005 LXI D, 2600 [D-E] <- 2600
2008 MOV A, M [A] <- [[H-L]]
2009 STAX D [A] -> [[D-E]]
200A INX H [H-L] <- [H-L] + 1
200B INX D [D-E] <- [D-E] + 1
200C DCR C [C] <- [C] – 1
200D JNZ 2008 Jump if not zero to 2008
2010 HLT   Stop

Explanation – Registers A, D, E, H, L, C are used for general purpose:

  1. MOV is used to transfer the data from memory to accumulator (1 Byte)
  2. LXI is used to load register pair immediately using 16-bit address (3 Byte instruction)
  3. MVI is used to move data immediately into any of registers (2 Byte)
  4. STAX is used to store accumulator into register pair indirectly (3 Byte instruction)
  5. DCR is used to decrease register by 1 (1 Byte instruction)
  6. INX is used to increase register pair by 1 (1 Byte instruction)
  7. JNZ is used to jump if not zero to given memory location (3 Byte instruction)
  8. HLT is used to halt the program


  • The program is simple and easy to understand, making it useful for teaching purposes.
  • It can be used to efficiently move blocks of data within a memory, such as copying data from one buffer to another, or shifting bits within a data structure.
  • The program is customizable, as the source and destination addresses, as well as the block size, can be easily changed.


  • The program is not optimized for speed, as it uses multiple instructions to move each byte of data.
  • It is not suitable for moving large blocks of data, as it requires a loop that iterates over each byte individually.
  • The program does not check for errors or boundary conditions, such as when the source or destination address is outside the valid memory range.
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Last Updated : 11 Apr, 2023
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