8085 Program to Add N Numbers of Size 8 Bits

In this program, we will see how to add a block of data using the 8085 microprocessor.

Problem Statement

Write 8085 Assembly language program to add N 1-byte numbers. The value of N is provided.

Discussion

In this problem, we are using location 8000H to hold the length of the block. The main block is stored from address 8010H. We are storing the result at location 9000H and 9001H. The 9000H holding the lower byte, and 9001H is holding the upper byte.

Repeatedly we are taking the number from the memory, then adding it with the accumulator and increase the register E content when carry flag is set. Initially, Eis cleared.

Input

Address
 Data
.
.
.
 .
.
.
8000
 08
.
.
.
 .
.
.
8010
 AF
8011
 2E
8012
 7C
8013
 81
8014
 2C
8015
 BF
8016
 FB
8017
 1C
.
.
.
 .
.
.

Flow Diagram

Program

Address
 HEX Codes
 Labels
 Mnemonics
 Comments
F000
 21, 00, 80

LXI H, 8000H
 Load the address to get a count of numbers
F003
 4E

MOV C, M
 Load C with the count value
F004
 21,10, 80

LXI H, 8010H
 Load HL with the starting address
F007
 AF

XRA A
 Clear Accumulator
F008
 5F

MOV E, A
 Clear the E register also
F009
 86
 LOOP
 ADD M
 Add Memory content with Accumulator
F00A
 D2, 0E, F0

JNC SKIP
 When the carry is false, skip next task
F00D
 1C

INR E
 Increase E, when the C flag is set
F00E
 0D
 SKIP
 DCR C
 Decrease C register by 1
F00F
 23

INX H
 Point to next location
F010
 C2, 09, F0

JNZ LOOP
 When Zero is false, go to LOOP
F013
 21, 00, 90

LXIH, 9000H
 Load address to store the result
F016
 77

MOV M, A
 Save accumulator content
F017
 23

INX H
 Increase HL pair
F018
 73

MOV M, E
 Store carry
F019
 76

HLT
 Terminate the program

Output

Address
 Data
.
.
.
 .
.
.
9000
 DC
9001
 03
.
.
.
 .
.
.
Updated on: 2020-06-26T12:10:37+05:30

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