Program to find the HCF of two given bytes in 8085 Microprocessor
Here we will see how to find the HCF or GCD of two given bytes using 8085. The numbers are 8-bit numbers, nor larger than that.
Problem Statement−
Write an 8085 Assembly language program to find the HCF or GCD of two numbers stored at memory location 8000H and 8001H.
Discussion−
Here we will use the Euclidean algorithm to find HCF. This algorithm is very simple. We have to follow these steps−
- If first number and second number are same, then
a) go to step 3.
Else if first number < second number, the
b) exchange no1 and no2.
- first-number <- first-number – second-number; go to step 1
- Display result as first-number
The values are stored at location 8000H and 8001H, and the result will be stored at location 8050H.
Input
First Input
Address
| Data
|
|---|
…
| …
|
8000
| 2D
|
8001
| 69
|
…
| …
|
8050
| 0F
|
…
| …
|
Second Input
Address
| Data
|
|---|
…
| …
|
8000
| 25
|
8001
| 35
|
…
| …
|
8050
| 01
|
…
| …
|
Flow Diagram
Program
Address
| HEX Codes
| Labels
| Mnemonics
| Comments
|
|---|
F000
| 21, 00, 80
|
| LXI H,8000H
| Point to the first number
|
F003
| 7E
|
| MOV A,M
| Load the first number into Acc
|
F004
| 23
|
| INX H
| Point to next location
|
F005
| 46
|
| MOV B,M
| Load the second number
|
F006
| B8
| LOOP
| CMP B
| Compare B with A
|
F007
| CA, 17, F0
|
| JZ STORE
| When A and B are same, store the result
|
F00A
| DA, 11, F0
|
| JC EXG
| If B > A, then exchange B and A
|
F00D
| 90
|
| SUB B
| if B < A, subtract B from A
|
F00E
| C3, 06, F0
|
| JMP LOOP
| Jump to LOOP
|
F011
| 48
| EXG
| MOV C,B
| Load C with B
|
F012
| 47
|
| MOV B,A
| Move A to B
|
F013
| 79
|
| MOV A,C
| Move C to A
|
F014
| C3, 06, F0
|
| JMP LOOP
| Jump to LOOP
|
F017
| 32, 50, 80
| STORE
| STA 8050H
| Store the value into memory
|
F01A
| 76
|
| HLT
| Terminate the program
|
|
|
|
|
|
|
Output
First Output
Second Output
Published on 04-Oct-2019 14:59:04
