Program to multiply two 8-bit numbers (shift and add method) in 8085 Microprocessor

Let us see one 8085 Microprocessor problem. In this problem we will see how to multiply two numbers using shift and add methods, not by using additive approach.

Problem Statement −

Write an 8085 Assembly language program to multiply two 8-bit numbers using shift and add method.

Discussion −

The shift and add method is an efficient process. In this program, we are taking the numbers from memory location 8000H and 8001H. The 16 bit results are storing into location 8050H onwards.

In this method we are putting the first number into DE register pair. The actual number is placed at E register, and D is holding 00H. The second number is taken into A. As the numbers are 8-bit numbers, then we are shifting the Accumulator contents eight times. When the carry flag is set while rotating, then the DE content is added with HL. Initially HL pair will hold 0000H. Then HL is also added with HL itself. Thus the result will be generated.

Input

Address	Data
…	…
8000	25
8001	2A
…	…

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	21, 00, 80		LXI H,8000H	Point to first operand
F003	5E		MOV E,M	Load the first operand to E
F004	16, 00		MVI D,00H	Clear the register D
F006	23		INX H	Point to next location
F007	7E		MOV A,M	Get the next operand
F008	0E, 08		MVI C,08H	Initialize counter with 08H
F00A	21, 00, 00		LXI H, 0000H	Clear the HL pair
F00D	0F	LOOP	RRC	Rotate the acc content to right
F00E	D2, 12, F0		JNC SKIP	If carry flag is 0, jump to skip
F011	19		DAD D	Add DE with HL
F012	EB	SKIP	XCHG	Exchange DE and HL
F013	29		DAD H	Add HL with HL itself
F014	EB		XCHG	Exchange again the contents of DE and HL
F015	0D		DCR C	Decrease C register
F016	C2, 0D, F0		JNZ LOOP	if Z = 0, jump to LOOP
F019	22, 50, 80		SHLD 8050H	Store the result
F01C	76		HLT	Terminate the program