### 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.👇

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.

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--------------                                  0014  -- even       100&       1--------------                                        000

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Happy Coding !!!!!!!!!!!👍👍👍👍👍

Android Interview Question For Senior Android Developer