How to set all the even positioned bits of a number

Problem statement

Given a number n, set all the even positioned bits of n.

Example 1

Input: n=10
Output: 10

Example 2

Input: n=8
Output: 10

Solution

The steps in the algorithm are as follows:

Generate a bit mask where the even positioned bits are set.
Perform a bitwise OR operation and mask the given number.

How to generate the bit mask

The steps to generate the bit mask are as follows:

We have two variables, such as count and mask.
We use a temporary variable temp in a for loop, and perform the following steps until temp becomes zero. The temp is right-shifted until it becomes zero.
1. If the count is odd, we then left shift count by 1 and perform bitwise OR with the mask.
2. Then, we increment the count by 1.

Code

class Main{
    static int generateMask(int n){
        int count = 0, mask = 0;
        for (int temp = n; temp > 0; temp >>= 1) {
            if ((count & 1) > 0) mask |= (1 << count);
            count++;
        }
        return mask;
    }
    static int setEvenPositionedBits(int n) {
        int mask = generateMask(n);
        return (n | mask);
    }
    public static void main(String[] args) {
        int n = 8;
        System.out.println("Setting even positioned bits of " + n + " we get " + setEvenPositionedBits(n));
    }
}

How to set all the even positioned bits of a number

Problem statement

Example 1

Example 2

Solution

How to generate the bit mask

Code

Explanation