# 8085 program to find nth power of a number

**Problem –** Write an assembly language code for calculating the nth power of a number using 8085 microprocessor.

**Example –**

Input : Base=>02H Exponent=>03H Output :08H

In 8085 microprocessor, no direct instruction exists to multiply two numbers, so multiplication is done by repeated addition as 4*4 is equivalent to 4+4+4+4(ie 4 times).

Load 02H(base) to register B and 03H(exponent) to register C -> set D register to 02H -> Add 02H B(ie 2) times -> D register now contains 04H -> Add 04H B(ie 2) times -> D register now contains 08H -> Output is 08H.

**Algorithm –**

- Load the base into register B and exponent into register C.
- To start multiplication set D to 01H.
- Jump to step 7.
- Decrements C.
- Jump to step 3 till value of C>0.
- Take memory pointer to next location and store result.
- Load E with contents of B and clear accumulator.
- Repeatedly add contents of D to accumulator E times.
- Store accumulator content to D.
- Go to step 4.

**Program –**

Address | Label | Mnemonic | Comment |
---|---|---|---|

2000H | Base | Data Byte for base | |

2001H | Exponent | Data Byte for exponent | |

2002H | Result | Result of factorial | |

2003H | LXI H, 2000H | Load data from memory | |

2006H | MOV B, M | Load base to B register | |

2007H | INX H | Increment memory | |

2008H | MOV C, M | Load exponent to C register | |

2009H | MVI D, 01H | Set D register to 1 | |

200BH | POWER_LOOP | CALL MULTIPLY | Subroutine call for multiplication |

200EH | DCR C | Decrement C | |

200FH | JNZ POWER_LOOP | Call power_loop till C becomes 0 | |

2012H | INX H | Increment memory | |

2013H | MOV M, D | Store result in memory | |

2014H | HLT | Halt | |

2100H | MULTIPLY | MOV E, B | Transfer contents of B to E |

2101H | MVI A, 00H | Clear accumulator to store result | |

2103H | MULTIPLYLOOP | ADD D | Add contents of D to A |

2104H | DCR E | Decrement E | |

2105H | JNZ MULTIPLYLOOP | Repeated addition | |

2108H | MOV D, A | Transfer contents of A to D | |

2109H | RET | Return from subroutine |

**Explanation –**

- Set register B with base and register C with exponent.
- Set register D with base by calling MULTIPLY subroutine one time.
- Decrement C and add D to itself B times by calling MULTIPLY subroutine.
- Repeat the above step till C reaches 0 and then exit the program.
- The result is obtained in D register which is stored in memory

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