8085 Program To Subtract Two 16 Bit Numbers

Let us suppose that the first 16 bit number is stored in memory location 9000H and the second number is stored in memory location 9002H. These two 16 numbers are to be subtracted and the result is to be stored in memory location 9004H.

Algorithm:

1. Start.
2. Load first data from 9000H directly into register pair HL.
3. Exchange contents of register pair DE and HL.
4. Load second data from 9002H directly into register pair HL.
5. Move contents of register L into accumulator A.
6. Subtract content of register E from A.
7. Move contents of accumulator A into register L.
8. Move contents of register H into accumulator A.
9. Subtract with borrow contents of register D from accumulator A.
10. Move contents of accumulator A into register H.
11. Store data contained in HL register pair into 9004H.
12. Terminate the program.

Program Code:

LHLD 9000H

XCHG

LHLD 9002H

MOV A, E

SUB L

MOV L, A

MOV A, D

SBB H

MOV H, A

SHLD 9004H

HLT

The first 16 bit number is loaded into HL register pair. The 16 bit number is then moved into DE register pair. The second 16 bit number is then loaded into HL register pair. Since, in 8085 there is no instruction for subtracting two 16 bit numbers (DAD is instruction for 16 bit addition) so we subtract two 8 bit numbers. The content of register E is moved into accumulator A and is the content of accumulator A is subtracted with contents of register L. The difference obtained is again moved from accumulator to L register. The content of D register is now moved into accumulator A and then the contents of register H is subtracted with borrow from the contents of accumulator. The difference thus obtained is moved into the H register. Now HL register pair contains the difference of 16 bit numbers. And the difference is stored in memory location 9004H.

Output:

Example: 1

Here, the first 16 bit number is stored in 9000H and 9001H and the second 16 but number is stored in 9002H and 9003H. Now, since 8085 does not have direct subtraction instruction of 16 bit numbers so we subtract two 8 bit numbers.i.e.

At first,

  34

-34

-----

 00

This result is stored in memory location 9004H

Then,

  12

-12

-----

 00

This result is stored in memory location 9005H

So, we obtain the subtraction of 16 bit numbers.

Example: 2

Here, At first,

 

  08

-05

-----

 03

This result is stored in 9004H.

Then,

  06

-04

-----

 02

This result is then stored in 9005H.

Thus, we obtain the result of subtraction of two 16 bit numbers.