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February 24, 2012

Write an assembly language program to convert given hexadecimal number into equivalent BCD number.

Written by @
Apparatus Used: Microprocessor Kit, Keyboard
Assumption: In this program we have to convert given hex number into its equivalent BCD number. If a hex number is FF then its equivalent BCD number will be 0255.
Algorithm:
  • Initialize HL pair as memory pointer
  • Get number into any register as C
  • Perform repeated addition for C number of times.
  • Adjust for BCD at each step
  • Store the BCD number at memory location.
  • End the Program
Program:
Memory Address
Op-code
Operand
Comments
2000
LXI
H,2050 H
Memory pointer Initialization
2003
MVI
D,00 H
For MSD
2005
XRA
A
Clear the accumulator
2006
MOV
C,M
Get hex data into A
2007
ADI
01
Count no. one by one
2009
DAA

Adjust for BCD
200A
JNC
200E
Jump for no carry
200D
INR
D
Increment D
200E
DCR
C
Decrement counter C
200F
JNZ
2007
Jump for no zero
2012
STA
2051 H
Store LSD
2015
MOV
A,D

2016
STA
2052 H
Store MSD
2019
RST
5
Set Break-point
201A
END

End the program

Used Instruction:
LXI: This instruction is used to store the 16-bit data in the register pair designated in the operand.
MVI: This instruction is used to store the 8 bit data in specified register.
MOV: This instruction is used to copy the content from source register to destination register.
JNC: Jump if not carry at specified memory address.
JNZ: Jump if value of specified register is not zero.
DAA: If the value of low order four bits in the accumulator is greater then 9 or if auxiliary carry flag is set, the instruction adds 6 to the low order four bits. If the value of high order four bits in the accumulator is greater then 9 or if auxiliary carry flag is set, the instruction adds 6 to the high order four bits
INR: It will increment the content of specified register by one.
ADI: This instruction is used to add the 8-bit data to the content of accumulator and store output in accumulator.
XRA: The contents of the register are logically EX-OR ed with the contents of accumulator.
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 H
Output:
2051-55 (LSB)
2052-02 (MSB)
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 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 (M2051) and press ENTER
  • You will get your desired output.

3 comments:

  1. hanieshjohari26 February, 2012

    information on site is very helpful. But it will be more helpful if u add some project and related information to it like how any project is made, how it will work, coding etc....
    otherwise it was excellent.

    ReplyDelete
  2. Pradip Yadav28 February, 2012

    Thanks for your valuable comment... I am trying to do that but it takes some time.....so after some gap you can also find details regarding project

    ReplyDelete
  3. DHARANI17 January, 2013

    very very useful for me ,as a beginner..,,,,,,,,thank u :)

    ReplyDelete

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