# 8085 program to multiply two 8 bit numbers

**Problem –** Multiply two 8 bit numbers stored at address 2050 and 2051. Result is stored at address 3050 and 3051. Starting address of program is taken as 2000.

**Example –**

**Algorithm –**

- We are taking adding the number 43 seven(7) times in this example.
**As the multiplication of two 8 bit numbers can be maximum of 16 bits**so we need register pair to store the result.

**Program –**

Memory Address | Mnemonics | Comment |
---|---|---|

2000 | LHLD 2050 | H←2051, L←2050 |

2003 | XCHG | H↔D, L↔E |

2004 | MOV C, D | C←D |

2005 | MVI D 00 | D←00 |

2007 | LXI H 0000 | H←00, L←00 |

200A | DAD D | HL←HL+DE |

200B | DCR C | C←C-1 |

200C | JNZ 200A | If Zero Flag=0, goto 200A |

200F | SHLD 3050 | H→3051, L→3050 |

2012 | HLT |

**Explanation –** Registers used: **A, H, L, C, D, E**

**LHLD 2050**loads content of 2051 in H and content of 2050 in L**XCHG**exchanges contents of H with D and contents of L with E**MOV C, D**copies content of D in C**MVI D 00**assigns 00 to D**LXI H 0000**assigns 00 to H and 00 to L**DAD D**adds HL and DE and assigns the result to HL**DCR C**decreaments C by 1**JNZ 200A**jumps program counter to 200A if zero flag = 0**SHLD**stores value of H at memory location 3051 and L at 3050**HLT**stops executing the program and halts any further execution

Read next: Assembly language program (8085 microprocessor) to add two 8 bit numbers

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