Bit Manipulation

Given a 32 bit unsigned integer num and an integer i. Perform following operations on the number - 
1. Get ith bit
2. Set ith bit
3. Clear ith bit
Note : For better understanding, we are starting bits from 1 instead 0. (1-based). You have to print space three space 
separated values ( as shown in output without a line break) and do not have to return anything.

Example 1:
Input: 70 3
Output: 1 70 66
Explanation: Bit at the 3rd position from LSB is 1. (1 0 0 0 1 1 0) .The value of the given number after setting 
the 3rd bit is 70. The value of the given number after clearing 3rd bit is 66. (1 0 0 0 0 1 0)

Example 2:
Input: 8 1
Output: 0 9 8
Explanation: Bit at the first position from LSB is 0. (1 0 0 0)  .The value of the given number after setting 
the 1st bit is 9. (1 0 0 1).  The value of the given number after clearing 1st bit is 8. (1 0 0 0)
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// tc=O(1) sc=O(1)
class Solution {
  public:
    void bitManipulation(int num, int i) {
        // your code here
        // int mask= i<<i;
        //get ith bit
        int get;
        if(num & (1<<i-1))
            get=1;
        else
            get=0;
            
        cout<<get<<" " ;
        // set ith bit
        cout<<(num | (1<<i-1))<<" ";
        
        //clear the ith bit
        
        cout<<(num & (~(1<<i-1)))<<" ";
    }
};