Divide Two 8-Bit Numbers in 8085 Microprocessor

Here we will see 8085 program. This program will divide two 8-bit numbers using 8085 microprocessor.

Problem Statement −

Write an 8085 Assembly language program to divide two 8-bit numbers and store the result at locations 8020H and 8021H.

Discussion −

The 8085 has no division operation. To get the result of division, we should use the repetitive subtraction method.

By using this program, we will get the quotient and the remainder. 8020H will hold the quotient, and 8021H will hold remainder.

We are saving the data at location 8000H and 8001H. The result is storing at location 8050H and 8051H.

Input

The Dividend: 0EH

The Divisor 04H

The Quotient will be 3, and remainder will be 2

Address	HEX Codes	Labels	Mnemonics	Comments
F000	21, 0E, 00	START	LXI H,0CH	Load 8-bit dividend in HL register pair
F003	06, 04		MVI B,04H	Load divisor in B to perform num1 / num2
F005	0E, 08		MVI C, 08	Initialize the counter
F007	29	UP	DAD H	Shifting left by 1 bit HL = HL + HL
F008	7C		MOV A, H	Load H in A
F009	90		SUB B	perform A = A – B
F00A	DA, 0F, F0		JC DOWN	If MSB<divisor then shift to left
F00D	67		MOV H,A	If MSB>divisor, store the current value of A in H
F00E	2C		INR L	Tracking quotient
F00F	0D	DOWN	DCR C	Decrement the counter
F010	C2, 07, F0		JNZ UP	If not exhausted then go again
F013	22, 20, 80		SHLD 8020	Store the result at 8020 H
F016	76		HLT	Stop

Arnab Chakraborty

Updated on: 2019-10-09T07:49:58+05:30

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