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))

Answer 2

what happens when you shift an int left

As the name suggests, left-shifting a number by n places, multiplies it by 2^n . The following code will help you understand it better.

public class Main {
    public static void main(String[] args) {
        int x = 5;
        System.out.println("In decimal: " + x + ", In Binary: "
                + "0000000000000000".substring(Integer.toBinaryString(x).length()) + Integer.toBinaryString(x));
        int y = x << 1;
        System.out.println("In decimal: " + y + ", In Binary: "
                + "0000000000000000".substring(Integer.toBinaryString(y).length()) + Integer.toBinaryString(y));
        int z = x << 2;
        System.out.println("In decimal: " + z + ", In Binary: "
                + "0000000000000000".substring(Integer.toBinaryString(z).length()) + Integer.toBinaryString(z));
    }
}

Output:

In decimal: 5, In Binary: 0000000000000101
In decimal: 10, In Binary: 0000000000001010
In decimal: 20, In Binary: 0000000000010100

Answer 3

Notice line int bit = key % INT_SIZE; // which is 5 % 32 according to your code int bit = key % INT_SIZE; // which is 5 % 32 according to your code

So now bit = 5 // because 5 % 32 = 5

Now int bitMask = 1 << bit; // you're saying shit 1 to five bits to left int bitMask = 1 << bit; // you're saying shit 1 to five bits to left

in 32 bits 1 = 000000000000000000000000000000000001

After shifting 5 bits to left it becomes:

000000000000000000000000000000000100000 = 32 //in binary that is why bitmask continas 32

Now let's talk about |= operator

| is a bitwise OR operator you are just using as compound assignment

bitArr[ind] |= bitMask is similar to bitArr[ind] = bitArr[ind] | bitmask

For example say bitArr[ind] equals 5 and bitmask equals 3

now bitArr[ind] | bitmask bitArr[ind] | bitmask means 5 | 3 5 | 3

It will do bitwise OR with binary representation

means 101 | 011 101 | 011 which becomes 111 which is 7