Program to Add two 8 Bit numbers in 8085 Microprocessor


Here we will see one 8085 assembly language program. In this program we will see how to add two 8-bit numbers.

Problem Statement 

Write an 8085 Assembly language program to add two 8-bit numbers and store the result at locations 8050H and 8051H.

Discussion 

To perform this task, we are using the ADD operation of 8085 Microprocessor. When the result of addition is 1-byte result, then the carry flag will not be enabled. When the result is exceeding the 1-byte range, then the carry flag will be 1

We are using two numbers at location 8000H and 8001H. When the numbers are 6CH and 24H, then the result will be (6C + 24 = 90) and when the numbers are FCH and 2FH, then the result will be (FC + 2F = 12B) Here the result is exceeding the range of 1-byte.

Input

first input

Address
Data


8000
6C
8001
24


second input

Address
Data


8000
FC
8001
2F


  

Flow Diagram

Program

Address
HEX Codes
Labels
Mnemonics
Comments
F000
0E, 00


MVI C,00H
Clear C register
F002
21, 00, 80


LXI H,8000H
Load initial address to get operand
F005
7E


MOV A,M
Load Acc with memory element
F006
23


INX H
Point to next location
F007
46


MOV B,M
Load B with second operand
F008
80


SUB B
Add B with A
F009
D2, 0D, F0


JNC STORE
When CY = 0, go to STORE
F00C
0C


INR C
Increase C by 1
F00D
21, 50, 80
STORE
LXI H,8050H
Load the destination address
F010
77


MOV M,A
Store the result
F011
23


INX H
Point to next location
F012
71


MOV M,C
Store the carry
F013
76


HLT
Terminate the program

 

Output

first output

Address
Data


8050
90
8051
00


second output

Address
Data


8050
2B
8051
01


raja
Published on 09-Oct-2019 07:42:27
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