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
…
…
Updated on: 2019-10-09T07:43:51+05:30

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