**Problem –** Write an assembly language program for calculating the factorial of a number using 8085 microprocessor.

**Example –**

Input : 04H Output : 18H as 04*03*02*01 = 24 in decimal => 18H

In 8085 microprocessor, no direct instruction exists to multiply two numbers, so multiplication is done by repeated addition as 4×3 is equivalent to 4+4+4 (i.e., 3 times).

Load 04H in D register -> Add 04H 3 times -> D register now contains 0CH -> Add 0CH 2 times -> D register now contains 18H -> Add 18H 1 time -> D register now contains 18H -> Output is 18H

**Algorithm –**

- Load the data into register B
- To start multiplication set D to 01H
- Jump to step 7
- Decrements B to multiply previous number
- Jump to step 3 till value of B>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

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

2000H | Data | Data Byte | |

2001H | Result | Result of factorial | |

2002H | LXI H, 2000H | Load data from memory | |

2005H | MOV B, M | Load data to B register | |

2006H | MVI D, 01H | Set D register with 1 | |

2008H | FACTORIAL | CALL MULTIPLY | Subroutine call for multiplication |

200BH | DCR B | Decrement B | |

200CH | JNZ FACTORIAL | Call factorial till B becomes 0 | |

200FH | INX H | Increment memory | |

2010H | MOV M, D | Store result in memory | |

2011H | HLT | Halt | |

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

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 –**

- First set register B with data.
- Set register D with data by calling MULTIPLY subroutine one time.
- Decrement B and add D to itself B times by calling MULTIPLY subroutine as 4*3 is equivalent to 4+4+4 (i.e., 3 times).
- Repeat the above step till B reaches 0 and then exit the program.
- The result is obtained in D register which is stored in memory

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