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February 29, 2012

Write an assembly language program to shift an eight bit data four bits right.

Written by @
Apparatus Used: Microprocessor Kit, Keyboard
Algorithm:
  • Load the data into accumulator
  • Rotate the content of accumulator 4 times
  • Store the result at memory location.
  • End the Program
Program:
Memory Address
Op-code
Operand
Comments
2000
LDA
2500 H
Get Data into Accumulator
2003
RAR
Rotate accumulator right
2004
RAR
Rotate accumulator right
2005
RAR
Rotate accumulator right
2006
RAR
Rotate accumulator right
2007
STA
2501 H
Store output at location
200A
RST
5
Set Breakpoint
200B
END
End the program
2nd Method
Memory Address
Op-code
Operand
Comments
2000
LDA
2500 H
Get Data into Accumulator
2003
RRC
30 H
Rotate right without carry
2005
RRC
0A H
Rotate right without carry
2007
RRC
200C
Rotate right without carry
200A
RRC
07 H
Rotate right without carry
200C
STA
2501 H
Store output at location
200F
RST
5
Set Break-point
2010
END

End the program

Used Instruction:
LDA address: Load data into register A (accumulator) directly from the address given within the instruction.
RAR: Rotates the bits of accumulator right by one position, through the carry 
RRC: Rotate the bits of the accumulator right by one position 
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: 
2500-58 
Output:
2051-85 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 (M2500)
  • Press ENTER and Provide your Input at location (2500:58)
  • 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 (M2501) and press ENTER
  • You will get your desired output.

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