8085 program to exchange a block of bytes in memory

Problem – Write an assembly level program in 8085 microprocessor to exchange a block of 4 bytes staring from address 2001 with data starting from address 3001.

Algorithm –

1. Take a count equal to 4
2. Store the starting address of both blocks in 2 different register pairs
3. Now exchange the contents at the addresses in both register pairs
4. Increment the values of both register pairs
5. Decrements count by 1
6. If count is not equal to 0 repeat steps 3 to 5

MEMORY ADDRESS	MNEMONICS	COMMENTS
2500	LXI D 2001	D <= 20, E <= 01
2503	LXI H 3001	H <= 20, L <= 01
2506	MVI C 04	C <= 04
2508	MOV B, M	B <= M[ H-L ]
2509	LDAX D	A <= M[ D-E ]
250A	MOV M, A	M[ H-L ] <= A
250B	MOV A, B	A <= B
250C	STAX D	M[ D-E ] <= A
250D	INX H	[ H-L ] <= [ H-L ] + 1
250E	INX D	[ D-E ] <= [ D-E ] + 1
250F	DCR C	C <= C – 1
2510	JNZ 2508	JUMP TO 2508 IF C NOT EQUAL TO 0
2513	HLT	STOP THE PROGRAM

Explanation –

1. LXI D 2001 – Loads register pair, that is in this case, D=20 and E=01
   LXI H 3001 – H=30 and L=01
2. MVI C 04 – Assigns immediate data, eg.- here C=04
   MVI A 45 – assigns A(accumulator) with 45, A=45
3. MOV B, M – Here M is the data in H – L register pair and it serves as an address. Copies content at address stored in M to register B
4. LDAX D – Here Accumulator is loaded with the data stored at address formed by register pair D – E
5. MOV M, A – Here A’s content is copied to address which is stored in M.
   MOV A, B – Copies content of register B to A
6. STAX D – Stores the content of A (accumulator) in the address formed by register pair D – E.
7. INX H – Increment the content of register pair H – L
8. INX H – Increment the content of register pair D – E
9. DCR C – Decrements the content of register C
10. JNZ 2508 – If value of register C is not equal to 0 then jump to address 2508
11. HLT – Stop execution of program

---

---