# 8085 program to find square root of a number

**Problem –** Write an assembly language program in 8085 microprocessor to find square root of a number.

**Example –**

**Assumptions –**

Number, whose square root we need to find is stored at memory location 2050 and store the final result in memory location 3050.

**Algorithm –**

- Assign 01 to register D and E
- Load the value, stored at memory location 2050 in accumulator A
- Subtract value stored at accumulator A from register D
- Check if accumulator holds 0, if true then jump to step 8
- Increment value of register D by 2
- Increment value of register E by 1
- Jump to step 3
- Move value stored at register E in A
- Store the value of A in memory location 3050

**Program –**

MEMORY ADDRESS | MNEMONICS | COMMENT |
---|---|---|

2000 | MVI D, 01 | D <- 01 |

2002 | MVI E, 01 | E <- 01 |

2004 | LDA 2050 | A <- M[2050] |

2007 | SUB D | A <- A – D |

2008 | JZ 2011 | Jump if ZF = 0 to memory location 2011 |

200B | INC D | D <- D + 1 |

200C | INC D | D <- D + 1 |

200D | INC E | E <- E + 1 |

200E | JMP 2007 | Jump to memory location 2007 |

2011 | MOV A, E | A <- E |

2012 | STA 3050 | A -> M[3050] |

2015 | HLT | END |

**Explanation –** Registers used A, D, E:

**MVI D, 01 –**initialize register D with 01**MVI E, 01 –**initialize register E with 01**LDA 2050 –**loads the content of memory location 2050 in accumulator A**SUB D –**subtract value of D from A**JZ 2011 –**make jump to memory location 2011 if zero flag is set**INR D –**increments value of register D by 1. Since it is used two times, therefore value of D is incremented by 2**INR E –**increments value of register E by 1**JMP 2007 –**make jump to memory location 2007**MOV A, E –**moves the value of register E in accumulator A**STA 3050 –**stores value of A in 3050**HLT –**stops executing the program and halts any further execution

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