8085 Program to do an operation on two BCD numbers based on the contents of X

Microcontroller Microprocessor 8085

Now let us see a program of Intel 8085 Microprocessor. In this program we will see how to do different operations on BCD numbers based on choice.

Problem Statement:

Write 8085 Assembly language program to perform some operations on two 8-bit BCD numbers base on choice.

Discussion:

In this program we are taking a choice. The choice value is stored at memory location 8000H (named as X). And the BCD numbers are stored at location 8001H and 8002H. We are storing the result at location 8050H and 8051H.

Here if the choice is 00H, then it will perform addition, for 01H, it will perform subtraction, and for 02H, it will do the multiplication operation.

Input:

First input

Address	Data
. . .	. . .
8000	00
8001	97
8002	88
. . .	. . .

Second input

Address	Data
. . .	. . .
8000	01
8001	97
8002	88
. . .	. . .

Third input

Address	Data
. . .	. . .
8000	02
8001	05
8002	04
. . .	. . .

Flow Diagram:

Program:

Address	HEX Codes	Labels	Mnemonics	Comments
F000	21, 00, 80		LXI H,8000H	Point to get the choice
F003	7E		MOV A,M	Load choice into A
F004	FE, 00		CPI 00H	Compare for ADD
F006	CA,14, F0		JZ ADD	Jump to do Addition
F009	FE, 01		CPI 01H	Compare for SUB
F00B	CA,27, F0		JZ SUB	Jump to do Subtraction
F00E	FE, 02		CPI 02H	Compare for MUL
F010	CA, 3F, F0		JZ MUL	Jump to do Multiplication
F013	76		HLT	Terminate the program
F014	23	ADD	INX H	Point to first operand
F015	7E		MOV A,M	Load operand to A
F016	23		INX H	Point to next operand
F017	86		ADD M	Add M with A
F018	27		DAA	Decimal adjust
F019	6F		MOV L,A	Store A to L
F01A	D2, 22, F0		JNC SKP1	If CY = 0, jump to SKP1
F01D	26, 01		MVI H,01H	Load H with 01H
F01F	C3, 62, F0		JMP STORE	Store result
F022	26, 00	SKP1	MVI H,00H	Clear HL
F024	C3, 62, F0		JMP STORE	Store HL as result
F027	23	SUB	INX H	Point to first operand
F028	46		MOV B,M	Load operand to B
F029	3E, 99		MVI A,99H	Load A with 99H
F02B	23		INX H	Point to next operand
F02C	96		SUB M	Subtract M from A
F02D	C6, 01		ADI 01H	Add 01H to get 10's complement
F02F	80		ADD B	Add B with A
F030	27		DAA	Adjust decimal
F031	6F		MOV L,A	Store A to L
F032	DA, 3A, F0		JC SKP2	If CY = 1, jump to SKP2
F035	26, FF		MVI H,FFH	Load H with FFH
F037	C3, 62, F0		JMP STORE	Store result
F03A	26, 00	SKP2	MVI H,00H	Clear HL
F03C	C3, 62, F0		JMP STORE	Store HL as result
F03F	23	MUL	INX H	Point to first operand
F040	46		MOV B,M	Load operand to B
F041	23		INX H	Point to next operand
F042	4E		MOV C,M	Load the second operand
F043	26, 00		MVI H,00H	Clear H register
F045	AF		XRA A	Clear A register
F046	B9		CMP C	Compare C with A
F047	CA, 5E, F0		JZ DONE	Jump to Done if Z = 1
F04A	80	LOOP	ADD B	Add B with A
F04B	27		DAA	Adjust decimal
F04C	57		MOV D,A	Move A to D
F04D	D2, 55, F0		JNC NINC	if CY = 0, not increment H
F050	7C		MOV A,H	Load H to A
F051	C6, 01		ADI 01H	Increase A
F053	27		DAA	Adjust decimal
F054	67		MOV H,A	Get back to H
F055	79	NINC	MOV A,C	Load A to C
F056	C6, 99		ADI 99H	Add A and 99H
F058	27		DAA	Adjust Decimal
F059	4F		MOV C,A	Load A to C again
F05A	7A		MOV A,D	Get back the data from D to A
F05B	C2, 4A, F0		JNZ LOOP	Jump to Loop if Z = 0
F05E	6F	DONE	MOV L,A	Take A to L
F05F	C3, 62, F0		JMP STORE	Store HL as result
F062	22, 50, 80	STORE	SHLD 8050H	Store result from HL
F065	76		HLT	Terminate the program

Output:

First output

Address	Data
. . .	. . .
8050	85
8051	01
. . .	. . .

Second output

Address	Data
. . .	. . .
8050	09
8051	00
. . .	. . .

Third output

Address	Data
. . .	. . .
8050	20
8051	00
. . .	. . .

George John

Updated on: 30-Jul-2019

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