java - Why HashSet order always same for my program?

Question

For some tutorial, they said:

HashSet doesn’t maintain any order, the elements would be returned in any random order.

But I write a test program, the result is always same.

import java.util.*;

public class HashSetDemo {

    public static void main(String[] args) {
        HashSet<String> hs1 = new HashSet<String>();
        hs1.add("a");
        hs1.add("b");
        hs1.add("c");
        hs1.add("d");
        hs1.add(null);
        hs1.add(null);
        System.out.println(hs1);
        System.out.println(hs1);
    }
}

Output:

[null, a, b, c, d]
[null, a, b, c, d]

I tried many times, but the order is always same. Why? Hope someone can help me, Thanks in advance!

Answer 1

The reason for this behaviour is that a HashSet is backed by a HashMap, which in turn is backed by an array of Entry-objects. where the hash is used to find the index of the array. So there is always an order of elements in a HashSet (the order of the array), you just don't have any guarantees as to what this order is.

As far as I can tell from the code, the order of the HashSet is determined (or at least affected) by the order of the computed hashes of its elements. Then, with relatively simple inputs (like your single character string), one might assume that there is a strict ordering of the hashes, which will give you what seems to be a natural ordering. With more complex objects, and thus more complex hash computations, the hashes will be more spread, and the ordering "more random".

Also, like it's been pointed out, "no guarantee of ordering" does not imply "guaranteed random ordering".

The hashcode-method of the String class also comes into play here, for single character Strings the hashcode will just be the int value of the one char in the String. And since char's int values are ordered alphabetically, so will the computed hashes of single char Strings.