8085 program to transfer a block in reverse order

Here we will see how we transfer a block of data in reverse order using 8085.

Problem Statement

Write 8085 program to transfer a block of N-bytes in reverse order. The block is stored at location 8001 onwards, the size of the block is stored at 8000. The block will be moved at location 9000 onwards.

Discussion

To solve this problem, we are taking the size of the block at first. The DE register pair is set to point at destination address 9000H. The HL pair is set to point to the last element of the block. If the block size is 0A, then the last block will be at 800A. At first HL is pointing to 8000, and took the block size from there and store to C. Now add C with the L register to get the last block address. Now take each element from memory pointed by HL, and store it back to the memory pointed by the DE. Then increase the DE, and decrease the HL. Thus the entire block will be moved in reverse direction.

Input

Address	Data
…	…
8000	0A
8001	11
8002	22
8003	33
8004	44
8005	55
8006	66
8007	77
8008	88
8009	99
800A	AA
…	…

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	21, 00, 80		LXI H,8000	Point to 8000 to get block size
F003	4E		MOV C,M	Take the block size into C
F004	11, 00 90		LXI D,9000	Point to the destination address
F007	7D		MOV A,L	Load L into A
F008	81		ADD C	Add C to point to last address of block
F009	6F		MOV L,A	Store A to L again
F00A	7E	LOOP	MOV A,M	Load memory to A
F00B	12		STAX D	Store A into destination pointed by DE
F00C	13		INX D	Point destination to next address
F00D	2B		DCX H	Point source to previous address
F00E	0D		DCR C	Decrease C by 1
F00F	C2, 0A, F0		JNZ LOOP	if Z is not set jump to LOOP
F012	76		HLT	Terminate the program