8085 program to subtract two 16-bit numbers with or without borrow

Problem – Write an assembly language program in 8085 microprocessor to subtract two 16 bit numbers.

Assumption –

• Starting address of program: 2000
• Input memory location: 2050, 2051, 2052, 2053
• Output memory location: 2054, 2055

Example –

INPUT:
       (2050H) = 19H
       (2051H) = 6AH 
       (2052H) = 15H 
       (2053H) = 5CH

OUTPUT:
        (2054H) = 04H
        (2055H) = OEH

RESULT:
Hence we have subtracted two 16 bit numbers.

Algorithm –

1. Get the LSB in L register and MSB in H register of 16 Bit number.
2. Exchange the content of HL register with DE register.
3. Again Get the LSB in L register and MSB in H register of 16 Bit number.
4. Subtract the content of L register from the content of E register.
5. Subtract the content of H register from the content of D register and borrow from previous step.
6. Store the result in memory location.

Program –

MEMORY ADDRESS	MNEMONICS	COMMENTS
2000	LHLD 2050	Load H-L pair with address 2050
2003	XCHG	EXCHANGE H-L PAIR WITH D-E PAIR
2004	LHLD 2052	Load H-L pair with address 2052
2007	MVI C, 00	C<-00H
2009	MOV A, E	A<-E
200A	SUB L	A<-A-L
200B	STA 2054	2054<-A
200E	MOV A, D	A<-D
200F	SBB H	SUBTRACT WITH BORROW
2010	STA 2055	2055<-A
2013	HLT	TERMINATES THE PROGRAM

Explanation –

1. LHLD 2050: load HL pair with address 2050.
2. XCHG: exchange the content of HL pair with DE.
3. LHLD 2052: load HL pair with address 2050.
4. MOV A, E: move the content of register E to A.
5. SUB L: subtract the content of A with the content of register L.
6. STA 2054: store the result from accumulator to memory address 2054.
7. MOV A, D: move the content of register D to A.
8. SBB H: subtract the content of A with the content of register H with borrow.
9. STA 2055: store the result from accumulator to memory address 2055.
10. HLT: stops executing the program and halts any further execution.

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