**Problem –** Write an assembly language program in 8085 microprocessor to check whether the 8 bit number which is stored at memory location 2050 is even or odd. If even, store 22 at memory location 3050 otherwise store 11 at memory location 3050.

**Example –**

A number is said to be odd if its lower bit is 1 otherwise even. Therefore to identify whether the number is even or odd, we perform AND operation with 01 by the help of **ANI** instruction. If number is odd then we will get 01 otherwise 00 in accumulator. **ANI** instruction also affect the flags of 8085. Therefore if accumulator contains 00 then zero flag becomes set otherwise reset.

**Algorithm –**

- Load the content of memory location 2050 in accumulator A.
- Perform AND operation with 01 in value of accumulator A by the help of
**ANI**instruction. - Check if zero flag is set, i.e if ZF = 1 then store 22 in accumulator A otherwise store 11 in A.
- Store the value of A in memory location 3050

**Program –**

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

2000 | LDA 2050 | A <- M[2050] |

2003 | ANI 01 | A <- A (AND) 01 |

2005 | JZ 200D | Jump if ZF = 1 |

2008 | MVI A 11 | A <- 11 |

200A | JMP 200F | Jump to memory location |

200D | MVI A 22 | A <- 22 |

200F | STA 3050 | M[3050] <- A |

2012 | HLT | END |

**Explanation –** Registers used A:

**LDA 2050 –**loads the content of memory location 2050 in accumulator A**ANI 01 –**performs AND operation between accumulator A and 01 and store the result in A**JZ 200D –**jump to memory location 200D if ZF = 1**MVI A 11 –**assign 11 to accumulator**JMP 200F –**jump to memory location 200F**MVI A 22 –**assign 22 to accumulator**STA 3050 –**stores value of A in 3050**HLT –**stops executing the program and halts any further execution

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