8085 Program for subtraction of multi-Byte BCD numbers

Microcontroller Microprocessor 8085

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to subtract multi-Byte BCD numbers.

Problem Statement

Write 8085 Assembly language program to subtract two multi-Byte BCD numbers.

Discussion

The numbers are stored into memory, and one additional information is stored. It will show us the Byte count of the multi-Byte BCD number. Here we are choosing 3-Byte BCD numbers. They are stored at location 8001H to 8003H, and another number is stored at location 8004H to 8006H. The location 8000H is holding the Byte count. In this case the Byte count is 03H.

For the subtraction we are using the 10’s complement method for subtraction.

In this case the numbers are: 672173 – 275188 = 376985

Input −

Address	Data
. . .	. . .
8000	03
8001	73
8002	21
8003	67
8004	88
8005	51
8006	27
. . .	. . .

Flow Diagram −

Program −

Address	HEX Codes	Labels	Mnemonics	Comments
F000	21, 00, 80		LXI H,8000H	Point to get the count
F003	4E		MOV C,M	Get the count to C
F004	11, 01, 80		LXI D,8001H	Point to first number
F007	21, 04, 80		LXI H,8004H	Point to second number
F00A	37		STC	Set the carry flag
F00B	3E, 99	LOOP	MVI A,99H	Load 99H into A
F00D	CE,00		ACI 00H	Add 00H and Carry with A
F00F	96		SUB M	Subtract M from A
F010	EB		XCHG	Exchange DE and HL
F011	86		ADD M	Add M to A
F012	27		DAA	Decimal adjust
F013	77		MOV M,A	Store A to memory
F014	EB		XCHG	Exchange DE and HL
F015	23		INX H	Point to next location by HL
F016	13		INX D	Point to next location by DE
F017	0D		DCR C	Decrease C by 1
F018	C2, 0B, F0		JNZ LOOP	Jump to LOOP if Z = 0
F01B	76		HLT	Terminate the program

Output −

Address	Data
. . .	. . .
8001	85
8002	69
8003	37
. . .	. . .

Ankith Reddy

Updated on: 30-Jun-2020

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