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February 13, 2012

Write a program to multiply two 8 bit numbers, output more than 8 bit number using memory location and repetitive addition method.

Written by @
Apparatus Used: Microprocessor kit, Keyboards
Assumption: Suppose in this case first number is stored at 2050 and second number is stored at 2051. Multiplication can be achieved by adding the multiplicand repeatedly as many times as the multiplier. As 05*08. 05+05+05+05+05+05+05+05=40
 
Algorithm:
  • Start the Program by loading HL register pair with address of memory location. 
  • Move data in Register B.
  • Get second number and load into accumulator.
  • Add Second number and accumulator.
  • Check for the carry
  • Increment the value of carry.
  • Check whether repeated addition is over and the store the value of product and carry at memory location. 
  • End the Program.
Program:
Memory Add
Mnemonics
Op-code
Comments
2000
MVI
D, 00 H
Initialize register D to 00
2002
MVI
A,00 H
Clear the accumulator
2004
LXI
H,2050
Store first input location
2007
MOV
B,M
Store first number into B
2008
INX
H

2009
MOV
C,M
Get Second Number in Register C
200A
ADD
B
Add Register B and Accumulator
200B
JNC
200F
Jump of no carry at 200F
200E
INR
D
Increment content of Register D
200F
DCR
C
Decrement counter by one
2010
JNZ
200A
Repeat until the value of C=0
2013
STA
2052
Store output
2016
MOV
A,D

2017
STA
2053
Store the carry at this location
201A
RST
5
Set Break-point
201B
END

End the program
 
Description of used instruction:  
MVI: This instruction is used to store 8 bit data in specified register. 
LXI: This instruction is used to store the 16-bit data in the register pair designated in the operand. 
MOV rd,M: This instruction copies data from the Memory to destination register. 
INX: This instruction is used to increment the content of register pair by one. 
ADD: This instruction is used to add the content of specified register to the content of accumulator and store output in accumulator. 
DCR: This instruction is used decrement the content of specified register by 1. 
JNC: Jump if not carry at specified memory address. 
JNZ: If result is not zero it will jump to the target address.
INR: It will increment the content of specified register by one. 
STA: This instruction is used to store the content of accumulator at specified memory address. 
RST 5: This instruction is used to set break-point for the execution. 
END: This instruction is used to execute the program. 
 
Result: 
Input:
2050-FF 
2051-FF 
Output:
2052 H- 01 H 
2053 H- FE H 
 
Procedure to look output  
· After press ENTER, You will get first screen
 · Press G and Provide Initial address (as 2000)
 · Press SHIFT+4,You will get first screen again
 · Press M and Provide Input location for Input (M2050)
 · Press ENTER and Provide your Input at location (2050: FF)
 · Press ENTER and Provide your Second Input (2051 :FF)
 · Press SHIFT+4
 · Press G and Provide Initial address (as 2000)
 · Press SHIFT+4,You will get first screen again
 · Press M and Provide address for output (M2052) and press ENTER
 · You will get your desired output.

5 comments:

  1. Anonymous20 February, 2012

    Load 89H in C register and Mask higher 4 bits and store lower 4 bits in a memory location.

    Shikhar Sharma
    shikharraj@rediffmail.com

    ReplyDelete
  2. Pradip Yadav21 February, 2012

    Hi Shekar

    In days I m little busy in my exam schedule. After this definitely i will post answer of it. Just wait till 23 Feb. You will get answer of this.

    ReplyDelete
  3. Anonymous27 March, 2012

    not getting program at 1 point sir, i.e..
    why decremented counter,as we have not set any counter in the beginning..
    rply plzz
    @(rahulsingh062@gmail.com)

    ReplyDelete
  4. Pradip Yadav27 March, 2012

    HI rahul

    look at
    2009 MOV C,M Get Second Number in Register C
    you will get your answer...

    ReplyDelete

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