Program to convert BCD to HEX in 8085 Microprocessor

Here we will see one 8085 program, that program will convert BCD numbers to HEX equivalent.

Problem Statement −

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

Discussion −

In this problem we are taking a BCD number from the memory and converting it to its binary equivalent. At first we are cutting each nibble of the input. So if the 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, then rotate 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
…	…
802B	52
…	…

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