8085 Program to Exchange 10 bytes

In this program we will see how to exchange a block of 10-byte data using 8085.

Problem Statement

Write 8085 Assembly language program to exchange a block of data, where block size is 10.

Discussion

The data are stored at location 8010H to 8019H and 9010H to 9019H. The location 8000H is holding the number of bytes to exchange. In this case number of bytes are 10D so it will be 0AH.

The logic is very simple, The HL and DE register pair is pointing the first and second data block respectively. By taking the data we are just swapping the values of each memory locations. Then repeating this process to swap two blocks completely.

Input

Flow Diagram

Program

Output

Anvi Jain

Published on 21-Jan-2019 10:12:26

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