Multiplication of Two 8 bit numbers in 8085

jojo January 18, 2011 9 Comments

AIM

To prepare an assembly language program for 8085 to multiply two 8 bit numbers

PROGRAM

MVI C,00

LDA 4200

MOV B,A

LDA 4201

MOV A,D

MVI A,00

LABEL:ADD B

DCR D

JNZ LABEL

JNC LOOP

INR C

LOOP:STA 4202

MOV A,C

STA 4203

HLT

SAMPLE OUTPUT

ADDRESS

DATA

INPUT/OUTPUT

4200

04

Input

4201

02

Input

4202

08

Output

4203

00

Output

RESULT

A program to multiply two 8 bit numbers was prepared and output obtained

Comments
  • Sailesh
    February 24, 2016

    I dnt kn a single letter of program plz guide me

  • February 19, 2015

    MVI C 00
    LDA 4200
    MOV B A
    LDA 4201
    MOV A D
    MVI A 00
    LABEL:ADD B
    DCR D
    JNZ LABEL
    JNC LOOP
    INR C
    LOOP:STA 4202
    MOV A C
    STA 4203
    HLT

    Your Code is wrong on line: 5

  • Neel Terdal
    November 17, 2011

    MVI C,00
    MVI A,02
    STA 00A0
    LDA 00A0
    MOV B,A
    MVI A,04
    STA 00A1
    LDA 00A1
    MOV D,A
    MVI A,00
    LABEL:ADD B
    DCR D
    JNZ LABEL
    STA 00A2
    HLT

  • Neel Terdal
    November 17, 2011

    5th line should be
    MOV D,A

  • Qatada
    March 18, 2011

    What are the two numbers being multiplied and in what base?

  • pruthviraj mane
    March 11, 2011

    pls give simple ckt based on 8085

  • SADDY
    January 15, 2011

    THANKS BRO.. 🙂

  • arindam
    August 31, 2009

    LXI H, 2200 : Initialize the memory pointer
    MOV E, M : Get multiplicand
    MVI D, 00H : Extend to 16-bits
    INX H : Increment memory pointer
    MOV A, M : Get multiplier
    LXI H, 0000 : Product = 0
    MVI B, 08H : Initialize counter with count 8
    MULT: DAD H : Product = product x 2
    RAL
    JNC SKIP : Is carry from multiplier 1 ?
    DAD D : Yes, Product =Product + Multiplicand
    SKIP: DCR B : Is counter = zero
    JNZ MULT : no, repeat
    SHLD 2300H : Store the result
    HLT : End of program

  • arindam
    August 31, 2009

    subscribe me

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