java.util.BitSet — set() doesn't work as expected

Question

Am I missing something painfully obvious? Or does just nobody in the world actually use java.util.BitSet?

The following test fails:

@Test
public voi         


        

Answer 1

// Abhay Dandekar

import java.util.BitSet;

public class TestBitSet {

    public static void main(String[] args) {

        BitSet bitSet = new BitSet();
        System.out.println("State 0 : " + bitSet.size() + " : " + bitSet.length() );

        bitSet.set(0, true);
        bitSet.set(1, true);
        System.out.println("State 1 : " + bitSet.size() + " : " + bitSet.length() );

        bitSet.set(2, false);
        bitSet.set(3, false);
        System.out.println("State 2 : " + bitSet.size() + " : " + bitSet.length() );

        bitSet.set(4, true);
        System.out.println("State 3 : " + bitSet.size() + " : " + bitSet.length() );

    }
}

A simple java program to show what happens inside. Some points to note :

1. BitSet is backed by a long

2. All the default values are false

3. While returning the length, it returns the index+1 of the highest "true" value in the set.

The output below should be able to explain itself :

State 0 : 64 : 0

State 1 : 64 : 2

State 2 : 64 : 2

State 3 : 64 : 5

So points to conclude :

1. Do not use the length to conclude the no of bits modified

2. Can be used in scenarios like bloom filters. More on bloom filters can be googled .. ;)

Hope this helps

Regards,

Abhay Dandekar