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Find odd even using bitwise operator

 As we have already seen in the previous post to swap two numbers without using the third variable and using the bitwise operator.πŸ‘‡

https://www.javaoneworld.com/2020/03/swap-two-numbers-using-bitwise-operator.html

 As I already mentioned that bitwise operators are always computational fast than any mathematical operators.



Why bitwise operators are computational fast than Mathematical operators πŸ‘ˆ

so using a bitwise operator instead of a mathematical operator is always a good approach.

Here we are using & operator to find that number is Odd or Even.

public class JavaOneWorld {
    public static void main(String args[]) {
    int x= 7;

if((x&1)==0){
    System.out.println("even");
}else{
    System.out.println("odd");
}
    }
}



The idea behind using the & operator !!!!!

As we already know binary works only on 1 and 0.
and a number is either odd or even,
so if a number is even then the last bit of that number must be 0 and if odd then the last bit must be 1.

i.e.-

5     odd
101 last bit is 1.

4 even

100 last bit is 0.

so here we have to use this property to find out the odd/even.

if we & 1 in any binary number it will result ---

1    --- if the given number's last bit will 1 (means number is odd)

0    --- if the given number's last bit will 0 (means number is even)
   

7  -- odd

       111
&       1
--------------                            
      001

4  -- even

       100
&       1
--------------                                  
      000

   
***********************
***********************
***********************

Happy Coding !!!!!!!!!!!πŸ‘πŸ‘πŸ‘πŸ‘πŸ‘

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