Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
8085 Program to Subtract two multi-Byte numbers
Now let us see a program of Intel 8085 Microprocessor. This program is mainly for subtracting multi Byte numbers.
Problem Statement:
Write 8085 Assembly language program to add two multi-Byte numbers.
Discussion:
We are using 3-Byte numbers. The numbers are stored into the memory at location 8001H and 8004H. One additional information is stored at location 8000H. In this place we are storing the Byte count. The result is stored at location 8050H.
We are taking the address of first operand block address into DE register pair, and the second operand block address into HL pair. The BC register pair is storing the destination address to store the result. At first we are clearing the carry flag. In 8085 there is no instruction to set the CY flag to 0, so we need two consecutive instructions. STC and CMC, the first one is set the CY to 1, and then complement it to make it 0. We are taking the number from memory and performing subtract with borrow instruction SBB to get the result. Thus the whole result is generated.
Input:
| Address | Data |
|---|---|
| . . . | . . . |
| 8000 | 03 |
| 8001 | 73 |
| 8002 | 21 |
| 8003 | 65 |
| 8004 | 88 |
| 8005 | 51 |
| 8006 | 27 |
| . . . | . . . |
Flow Diagram:
Program:
| Address | HEX Codes | Labels | Mnemonics | Comments |
|---|---|---|---|---|
| F000 | 11, 01, 80 | LXI D,8001H | Point to first operand | |
| F003 | 21, 04, 80 | LXI H,8004H | Point to second operand | |
| F006 | 01, 50, 80 | LXI B,8050H | Point destination address | |
| F009 | 37 | STC | Set the carry | |
| F00A | 3F | CMC | Complement carry | |
| F00B | 1A | LOOP | LDAX D | Load A from memory pointed by DE |
| F00C | 9E | SBB M | Subtract memory element and borrow | |
| F00D | 02 | STAX B | Store result into memory pointed by BC | |
| F00E | 23 | INX H | Increase HL pair | |
| F00F | 13 | INX D | Increase DE pair | |
| F010 | 03 | INX B | Increase BC pair | |
| F011 | 3A, 00, 80 | LDA 8000H | Load the size into A | |
| F014 | 3D | DCR A | Decrease A by 1 | |
| F015 | 32, 00, 80 | STA 8000H | Store updated size into memory | |
| F018 | C2, 0B, F0 | JNZ LOOP | If Z = 0, jump to LOOP | |
| F01B | 76 | HLT | Terminate the program |
Output:
| Address | Data |
|---|---|
| . . . | . . . |
| 8050 | EB |
| 8051 | CF |
| 8052 | 3D |
| . . . | . . . |
1K+ Views
