How to set all odd position bits of a number

Problem statement

Given a number n, set all the odd positioned bits of n. The position of LSB is considered to be 1.

Example 1:

Input: n=8
Output: 13

Example 2:

Input: n=7
Output: 7

Solution

We’ll generate a bit mask that has all the odd positioned bits set and use the bitwise OR operation.

How do we generate a bit mask?

First, we’ll make a copy of n and call it temp. Two more variables are created as follows:

count: This is used to generate the power of 2, which is used to create the mask.
mask: This stores the bit mask.

The steps of the bit mask generation are as follows:

Loop until the value of temp is greater than zero.
1. If the count is even, then we set the count-th bit in the mask.
2. We increment the count by 1.
3. We right shift the temp by 1.

Code

class Main{
    static int generateMask(int n){
        int count = 0;
        int mask = 0;
        int temp = n;
        while(temp > 0){
            if((count & 1) != 1)
                mask = mask | (1 << count);
            count++;
            temp >>=1;
        }
        return mask;
    }
    static int setOddBits(int n) {
        int mask = generateMask(n);
        return (n | mask);
    }
    public static void main(String[] args) {
        int n = 10;
        System.out.println("Setting odd positioned bits of " + n + " we get " + setOddBits(n));
    }
}

How to set all odd position bits of a number

Problem statement

Solution

Code

Explanation