**Q. Write an 8085 program and draw a flowchart to Subtract two 16-bit numbers without considering the borrow.**(8085 Microprocessor Program)

**Flowchart/Algorithm**

**Program**

Address |
Mnemonics |
Operand |
Opcode |
Comments |

2000 | LHLD | 3000H | 2A | Load H-L pair with 1^{st} operand from 3000H. |

2001 | 00 | |||

2002 | 30 | |||

2003 | XCHG | EB | Exchange H-L pair with D-E pair. | |

2004 | LHLD | 3002H | 2A | Load H-L pair with 2^{nd} operand from 3002H. |

2005 | 02 | |||

2006 | 30 | |||

2007 | MOV | A, E | 7B | Move the lower-order of 1^{st} number from reg. E to reg. A. |

2008 | SUB | L | 95 | Subtract the lower-order of 2^{nd} number from lower-order of 1^{st} number. |

2009 | MOV | L, A | 6F | Move the result from reg. A to register L. |

200A | MOV | A, D | 7A | Move the higher-order of 1^{st} number from reg. D to reg. A. |

200B | SBB | H | 9C | Subtract the higher-order of 2^{nd} number from higher-order of 1^{st} number with borrow from the previous subtraction. |

200C | MOV | H, A | 67 | Move the result from reg. A to reg. H. |

200D | SHLD | 3004H | 22 | Store the 16-bit result from H-L pair to memory. |

200E | 04 | |||

200F | 30 | |||

2010 | HLT | 76 | Halt |

**Output**

*Before Execution:*

3000H: 08H

3001H: 06H

3002H: 05H

3003H: 04H

*After Execution:*

3004H: 03H

3005H: 02H

**Program Explanation**

- This program subtracts two 16-bit operands stored in memory locations 3000H-3001H and 3002H-3003H, without considering the borrow taken (if any).
- Let us assume that the operands stored at memory locations 3000H-3001H is 08H-06H and 3002H-3003H is 05H-04H.
- The H-L pair is loaded with the first 16-bit operand 0806H from memory locations 3000H- 3001H.
- Then, the first 16-bit operand is moved to D-E pair.
- The second 16-bit operand 0504H is loaded to H-L pair from memory locations 3002H- 3003H.
- The lower-order of the first number is moved from register E to accumulator.
- The lower-order of 2
^{nd}number in register L is subtracted from lower-order of 1^{st}number in the accumulator. - The result of the subtraction is moved from the accumulator to register L.
- Then, the higher-order of 1st number is moved from register D to accumulator.
- The higher-order of 2
^{nd}number in register H is subtracted from higher-order of the first number in the accumulator, along with the borrow from the previous subtraction. - The result of the subtraction is moved from the accumulator to register H.
- Now, the final result is in H-L pair.
- The result is stored from H-L pair to memory locations 3004H-3005H.

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