Divide Two 16-Bit Numbers in 8085

Here we will see how to divide two 16 bit numbers using 8085.

Problem Statement

Write 8085 Assembly language program to divide two 16-bit numbers.

Discussion

8085 has no division operation. To perform division, we have to use repetitive subtraction. To perform 16-bit division, we have to do the same but for the register pairs. As the register pairs are used to hold 16-bit data.

The Divisor is stored at location FC00 and FC01, the dividend is stored at FC02 and FC03. After division, the quotient will be stored at FC04 and FC05, and the remainder will be stored at FC06 and FC07.

Input

Address	Data
FC00	8A
FC01	5C
FC02	5A
FC03	1D

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	01, 00, 00		LXI B,0000H	Clear BC register pair
F003	2A, 02, FC		LHLD FC02H	Take the Divisor into HL first
F006	EB		XCHG	Exchange DE and HL
F007	2A, 00, FC		LHLD FC00H	Take the dividend
F00A	7D	LOOP	MOV A,L	Load L into A
F00B	93		SUB E	Subtract E from A
F00C	6F		MOV L,A	Store A into L
F00D	7C		MOV A,H	Load H into A
F00E	9A		SBB D	Subtract B from A with Borrow
F00F	67		MOV H,A	Store A into H again
F010	DA, 17, F0		JC SKIP	If CY is 1, Skip
F013	03		INX B	Increase B by 1
F014	C3, 0A, F0		JMP LOOP	Jump to Loop
F017	19	SKIP	DAD D	Add HL and DE
F018	22, 06, F0		SHLD FC06H	Store remainder into FC06 and FC07
F01B	69		MOV L,C	Load C into L
F01C	60		MOV H,B	Load B into H
F01D	22, 04, FC		SHLD FC04H	Store quotient into FC04 and FC05
F020	76		HLT	Terminate the program

Output

Address	Data
FC04	03
FC05	00
FC06	7C
FC07	04

Ankith Reddy

Updated on: 2019-07-30T22:30:26+05:30

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