What happens when you shift an int left

Question

I'm trying to understand part of this code that was created here to create a my_hashset: https://leetcode.com/problems/design-hashset/discuss/548792/Java-solution-faster-than-99

   class HashCode_from_leet {
    int[] bitArr;
    private static final int MAX = 100000;
    private static final int INT_SIZE = Integer.SIZE;

    public HashCode_from_leet() {
        bitArr = new int[MAX / INT_SIZE + 1];
    }
    
    public void add(int key) {  //pass in 5
        int ind = key / INT_SIZE;  
        int bit = key % INT_SIZE;
        int bitMask = 1 << bit;  //QUESTION: why does bitMask = 32?
        bitArr[ind] |= bitMask;  // QUESTION: what does '|=' mean?
    }
    
    // NOTE I REMOVED SOME METHODS 
    
    public boolean contains(int key) {
        int ind = key / INT_SIZE;
        int bit = key % INT_SIZE;
        int bitMask = 1 << bit;
        return (bitArr[ind] & bitMask) != 0;
    }
    
    
    public static void main(String[] args) {
        HashCode_from_leet hfl = new HashCode_from_leet();
        hfl.add(5);
        System.out.println(hfl.contains(5));
    }
}

When I pass in 5 into the add(int key) method bitMask = 32

I'm not sure why -- I understand << means shift left, so we take 5 and shift it left, but would think that equals 5*10^2?

Answer 1

When you shift left an integer by X eg: Z =(Y << X) it will acutally do this operation : Z = (Y * (2^X))

When you shift right an integer by X eg: Z =(Y >> X) it will acutally do this operation : Z = (Y / (2^X))