Interface 8255 with 8085 microprocessor for 1’s and 2’s complement of a number

**Problem –** Interface 8255 with 8085 microprocessor and write an assembly language program to display 99 in Port A, 1’s complement of 99 in Port B, and 2’s complement of 99 in Port C. If Port addresses are 30H, 32H, and 33H resp.

**Example –**

D7 |
D6 |
D5 |
D4 |
D3 |
D2 |
D1 |
D0 |

1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |

**Algorithm –**

- Construct the control word register.
- Input value of accumulator A.
- Display value of A in Port A.
- Now 1’s complement of A is calculated and result is displayed in Port B.
- Now 2’s complement of A is calculated by adding 1 to 1’s complement of A. The result is displayed in Port C.

**Program –**

Mnemonics |
Comments |
---|---|

MVI A, 80 | A<–80 |

OUT 33 | Control Register<–A |

MVI A, 99 | A<–99 |

OUT 30 | Port A<–A |

CMA | 1’s complement of A |

OUT 31 | Port B<–A |

INR A | A<–A+1 |

OUT 32 | Port C<–A |

RET | Return |

**Explanation –**

**.MVI A, 80:**Value of control register is 80.**OUT 33:**Putting the value of A in 33H which is port number of port control register.**.MVI A, 99:**Value of A is equals to 99.**OUT 30:**Displaying the value of A in 30H which is port number of Port A.**CMA:**Calculates 1’s complement of A.**OUT 31:**Displaying 1’s complement of A in 31H which is port number of Port B.**INR A:**1’s complement of A is incremented by 1 i.e. 2’s complement of A is calculated.**OUT 32**: Displaying 2’s complement of A in 32 H which is port number of Port C.**RET:**Return.

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