In the article, there are few examples of multiplication programs in8085 microprocessorin assembly language programming (ALP). You need to learn the instruction set of 8085 in detail before programming.
One program can be performed in multiple ways. Here they are mentioned as Program 1, Program 2 … Comments in each program is written after semicolon (;).
Multiplication Programs in 8085 Microprocessor
Multiplication of two 8 bit numbers in 8085 microprocessor
Multiply two 8-bit numbers stored in memory locations 2200H and 2201H by repetitive addition and store the result in memory locations 2202H and 2203H.
Program 1:
LDA 2200H ; Copy the contents from memory location 2200H to accumulator
MOV E, A ; Copy accumulator to register E
MVI D, 00H ; Get the first number in DE register pair
LDA 2201H ; Copy the contents from memory location 2201H to accumulator
MOV C, A ; Initialize counter by value of multiplier
LXI H, 0000H ; Result = 0
BACK:DAD D ; Result = result + first number
DCR C ; Decrement counter
JNZ BACK ; If count≠0 (i.e. Zero flag≠0) repeat the loop
SHLD 2202H ; Store value of H at memory location 2202H and L at 2203H
HLT ; Terminate program execution.
Program 2:
LHLD 2200H ; Loads content of 2200H in H and content of 2201H in L
XCHG ; Exchange contents of H with D and contents of L with E
MOV C, D ; Copy contents of register D in C
MVI D, 00H ; Copy 00H into register D
LXI H, 0000H ; assigns 00 to H and 00 to L
BACK:DAD D ; Result = result + first number
DCR C ; Decrement counter
JNZ BACK ; If count≠0 (i.e. Zero flag≠0) repeat the loop
SHLD 2202H ; stores value of H at memory location 2203 and L at 2202
HLT ; Terminate program execution.
Sample Example:
(2200H) = 03H
(2201H) = B2H
Result = B2H + B2H + B2H = 216H
= 216H
(2202H) = 16H
(2203H) = 02H
Above two programs generate 16-bit result and stores it onto successive memory location. Multiple times addition can be performed using ADD instruction as follows which is useful for 8-bit result only.
Program 3:
LXI H, 2200H ; H-L register pair is loaded with 2200H
MOV B, M ; Copy the contents of memory 2200H into register B
INX H ; Increment the address of HL pair by one and make it 2201H
MOV C, M ; Copy the content of memory into register C (Get second number)
MVI A, 00H ; Assign 00H to accumulator.
BACK:ADD B ; Add the content of accumulator with register B and store the result in accumulator.
DCR C ; Decrement the register C (counter)
JNZ BACK ; If count≠0 (i.e. Zero flag≠0) repeat the loop
INX H ; Increment the address of HL pair by one and make it 2202H
MOV M, A ; Copy the content of accumulator (answer) to register M.
HLT ; Terminate program execution.
Sample Example 1:
(2200H) = 15H
(2201H) = 08H
Result = (2202H) = A8H
This is the correct answer.
Sample Example 2:
(2200H) = 15H
(2201H) = A2H
Result = (2202H) = 4AH
Note that, the answer for above calculation (sample example 2) should be D4AH but it will only show A2H because accumulator can only save data of 8-bit. Hence for the numbers generating 16-bit answer above program will show incomplete (/wrong) answer.
multiplication of two BCD numbers in 8085
Write an assembly language program to multiply 2 BCD numbers
Program:
MVI C, Multiplier ; Load BCD multiplier
MVI B 00 ; Initialize counter
LXI H, 0000H ; Result = 0000
MVI E, multiplicand ; Load multiplicand
MVI D 00H ; Extend to 16-bits
BACK: DAD D ; Result = Result + Multiplicand
MOV A, L ; Get the lower byte of the result
ADI 00H
DAA ; Adjust the lower byte of result to BCD.
MOV L, A ; Store the lower byte of result
MOV A, H ; Get the higher byte of the result
ACI 00H
DAA ; Adjust the higher byte of the result to BCD
MOV H, A ; Store the higher byte of result.
MOV A, B ; [Increment
ADI 01H ; counter
DAA ; adjust it to BCD and
MOV B, A ; store it]
CMP C ; Compare if count = multiplier
JNZ BACK ; if not equal repeat
HLT ; Stop
