8085 Program to Add two 8 Bit numbers

In this program, we will see how to add two 8-bit numbers using 8085 microprocessor.

Problem Statement

Write8085 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 the addition is the 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		MVIC,00H	Clear C register
F002	21,00, 80		LXIH,8000H	Load initial address to get operand
F005	7E		MOVA, M	Load Acc with a memory element
F006	23		INX H	Point to next location
F007	46		MOVB, M	Load B with the 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	LXIH,8050H	Load the destination address
F010	77		MOVM, A	Store the result
F011	23		INX H	Point to next location
F012	71		MOVM, C	Store the carry
F013	76		HLT	Terminate the program