Convert 16-Bit Binary Number to BCD in 8085

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to convert 16-bit binary data to BCD data.

Problem Statement:

Write 8085 Assembly language program to convert 16-bit binary data to BCD data. The binary data is stored at location 8000H and 8001H.

Discussion:

This problem is solved by implementing 16-bit counter. We are storing the 16-bit number at first, then decreasing the numbers one by one, and increasing the decimal value by adjusting the decimal value. To increase the value, we can use the INR instruction, but INR instruction does not affect the carry flag. So here we are using ADI 01H for increasing it by 1.

The binary number is taken from location 8000H and 8001H, and the final result is stored at location 8050H to 8052H.

Input:

Address	Data
. . .	. . .
8000	FF
8001	FF
. . .	. . .

Flow Diagram:

Program:

Address	HEX Codes	Labels	Mnemonics	Comments
F000	2A, 00, 80		LHLD 8000H	Initialize HL with 16-bit data
F003	11, 00, 00		LXI D,0000H	Clear DE register
F006	AF		XRA A	Clear A register
F007	C6, 01	LOOP	ADI 01H	Add 01 with A
F009	21		DAA	Adjust decimal
F00A	47		MOV B,A	Store A to B
F00B	D2, 1B, F0		JNC SKIP	If CY = 0, go to skip
F00E	7B		MOV A,E	Load E to A
F00F	C6, 01		ADI 01H	Add 01H with A
F011	27		DAA	Decimal adjust
F012	5F		MOV E,A	Get E from A
F013	D2, 1B,F0		JNC SKIP	If CY = 0, go to skip
F016	7A		MOV A,D	Take D to A
F017	C6, 01		ADI 01H	Add 01H with A
F019	27		DAA	Adjust decimal
F01A	57		MOV D,A	Load A to D
F01B	2B	SKIP	DCX H	Decrease DE
F01C	7C		MOV A,H	Load H to A
F01D	85		ORA L	OR A and L
F01E	78		MOV A,B	Load B to A
F01F	C2, 07, F0		JNZ LOOP	Jump to Loop
F022	EB		XCHG	Exchange DE and HL
F023	22, 51, 80		SHLD 8051H	Store HL content into memory
F026	32, 50, 80		STA 8050H	Store A to memory
F029	76		HLT	Terminate the program

Output:

Address	Data
. . .	. . .
8050	35
8051	55
8052	06
. . .	. . .

Arjun Thakur

Updated on: 2019-07-30T22:30:25+05:30

1K+ Views

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