8085 Program to Exchange 10 bytes

In this program we will see how to exchange a block of 10-byte data using 8085.

Problem Statement

Write 8085 Assembly language program to exchange a block of data, where block size is 10.

Discussion

The data are stored at location 8010H to 8019H and 9010H to 9019H. The location 8000H is holding the number of bytes to exchange. In this case number of bytes are 10D so it will be 0AH.

The logic is very simple, The HL and DE register pair is pointing the first and second data block respectively. By taking the data we are just swapping the values of each memory locations. Then repeating this process to swap two blocks completely.

Input

Address	Data
. . .	. . .
8000	0A
. . .	. . .
8010	00
8011	11
8012	22
8013	33
8014	44
8015	55
8016	66
8017	77
8018	88
8019	99
. . .	. . .
9010	84
9011	63
9012	12
9013	47
9014	48
9015	AD
9016	BC
9017	2A
9018	8F
9019	99
. . .	. . .

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	21, 10, 80		LXI H, 8000H	Point 8000Hto get byte count
F003	4E		MOV C, M	Load Count from memory
F004	21, 10, 80		LXI H,8010H	Point first block address
F007	11, 10, 90		LXI D,9010H	Point second block address
F00A	46	LOOP	MOV B, M	Take element from first block to B
F00B	1A		LDAX D	Take element from second block to Acc
F00C	77		MOV M, A	Store Acc content to second block
F00D	78		MOV A, B	Load B to A
F00E	12		STAX D	Store into second block
F00F	23		INX H	Point to next address of first block
F010	13		INX D	Point to next address of second block
F011	0D		DCR C	Decrease the count variable
F012	C2, 0A, F0		JNZ LOOP	When block is not completed, jump to LOOP
F015	76		HLT	Terminate the program

Output

Address	Data
. . .	. . .
8010	84
8011	63
8012	12
8013	47
8014	48
8015	AD
8016	BC
8017	2A
8018	8F
8019	99
. . .	. . .
9010	00
9011	11
9012	22
9013	33
9014	44
9015	55
9016	66
9017	77
9018	88
9019	99
. . .	. . .

Anvi Jain

Updated on: 30-Jul-2019

3K+ Views

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