Convert Two-Digit BCD to Binary in 8085

In this program we will see how to convert BCD numbers to binary equivalent.

Problem Statement

A BCD number is stored at location 802BH. Convert the number into its binary equivalent andstore it to the memory location 802CH.

Discussion

In this problem we are taking a BCD number from the memory and converting it to its binaryequivalent. At first we are cutting each nibble of the input. So ifthe input is 52 (0101 0010) then we can simply cut it by masking the number by 0FH and F0H. When the Higher order nibble is cut, thenrotate it to the left four times to transfer it to lower nibble.

Now simply multiply the numbers by using decimal adjust method to get final decimal result.

Input

Address Data
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 .
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802B 52
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Flow Diagram

Program

Address HEX Codes Labels Mnemonics Comments
8000 31, FF, 80
LXI SP,80FFH Initialize stack pointer
8003 21, 2B, 80
LXI H, 802BH  Pointer to the IN-BUFFER
8006 01, 2C, 80
LXI B, 802CH  Pointer to the OUT-BUFFER
8009 7E 
MOV A, M  Move the contents of 802BH to A
800A  CD, 0F, 80
CALL BCDBIN Subroutine to convert a BCD number to HEX
800D  02
STAX B  Store Acc to memory location pointed by BC
800E 76
HLT  Terminate the program
800F  C5  BCDBIN PUSH B  Saving B
8010 47
MOV B, A  Copy A to B
8011 E6, 0F
ANI 0FH  Mask of the most significant four bits
8013 4F 
MOV C, A  Copy A to C
8014 78
MOV A, B  Copy B to A
8015 E6, F0
ANI F0H  Mask of the least significant four bits
8017 0F 
RRC  Rotate accumulator right 4 times
8018 0F
RRC 
8019 0F
RRC 
801A  0F
RRC 
801B  57
MOV D, A  Load the count value to the Reg. D
801C  AF 
XRA A  Clear the contents of the accumulator
801D  1E, 0A
MVI E, 0AH  Initialize Reg. E with 0AH
801F  83 SUM ADD E  Add the contents of Reg. E to A
8020 15
DCR D  Decrement the count by 1 until 0 is reached
8021 C2, 1F, 80
JNZ SUM 
8024 81
ADD C  Add the contents of Reg. C to A
8025 C1 
POP B  Restoring B
8026 C9 
RET  Returning control to the calling program


Output

Address Data
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 .
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802C 34
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 .
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Updated on: 2019-07-30T22:30:24+05:30

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