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# 8085 Program to Divide two 8 Bit numbers

In this program, we will see how to divide two 8-bit numbers using 8085 microprocessor.

## Problem Statement

Write 8085 Assembly language program to divide two 8-bit numbers and store the result at locations **8020H** and **8021H**.

## Discussion

The 8085 has no division operation. To get the result of the division, we should use the repetitive subtraction method.

By using this program, we will get the quotient and the remainder. 8020H will hold the quotient, and 8021H will hold the remainder.

We are saving the data at location 8000H and 8001H. The result is storing at location 8050H and 8051H.

## Input

The Dividend: 0EH

The Divisor 04H

The Quotient will be 3, and the remainder will be 2

## Flow Diagram

## Program

Address | HEX Codes | Labels | Mnemonics | Comments |
---|---|---|---|---|

F000 | 21,0E, 00 | START | LXIH,0CH | Load 8-bit dividend in HL register pair |

F003 | 06,04 | MVIB,04H | Load divisor in B to perform num1 / num2 | |

F005 | 0E,08 | MVIC, 08 | Initialize the counter | |

F007 | 29 | UP | DADH | Shifting left by 1 bit HL = HL + HL |

F008 | 7C | MOVA, H | Load H in A | |

F009 | 90 | SUB B | perform A = A – B | |

F00A | DA,0F, F0 | JC DOWN | If MSB < divisor then shift to left | |

F00D | 67 | MOVH, A | If MSB > divisor, store the current value of A in H | |

F00E | 2C | INR L | Tracking quotient | |

F00F | 0D | DOWN | DCRC | Decrement the counter |

F010 | C2,07, F0 | JNZ UP | If not exhausted then go again | |

F013 | 22,20, 80 | SHLD 8020 | Store the result at 8020 H | |

F016 | 76 | HLT | Stop |

## Output

Address | Data |
---|---|

. . . | . . . |

8020 | 03 |

8021 | 02 |

. . . | . . . |

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