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