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How to use java.Set

I'm trying to make it working for quite some time,but just can't seem to get it. I have object Tower built of Block's. I've already made it working using arrays, but I wanted to learn Set's. I'd like to get similar functionality to this:

public class Tower {


public Tower(){
}

public Tower add(Block k1){

    //(...)
    //if block already in tower, return "Block already in tower"
}

public Tower delete(Block k1){

    //(...)
    //if block already dleted, show "No such block in tower"
}

}

Someone gave me some code, but I constantly get errors when trying to use it :

Set<Block> tower = new HashSet<Block>();

boolean added = tower.add( k1 );
if( added ) {
System.out.println("Added 1 block.");
} else {
System.out.println("Tower already contains this block.");
}

How to implement it ?

like image 454
sasklacz Avatar asked Mar 22 '10 05:03

sasklacz


1 Answers

The first thing you need to study is the java.util.Set API.

Here's a small example of how to use its methods:

    Set<Integer> numbers = new TreeSet<Integer>();

    numbers.add(2);
    numbers.add(5);

    System.out.println(numbers); // "[2, 5]"
    System.out.println(numbers.contains(7)); // "false"

    System.out.println(numbers.add(5)); // "false"
    System.out.println(numbers.size()); // "2"

    int sum = 0;
    for (int n : numbers) {
        sum += n;
    }
    System.out.println("Sum = " + sum); // "Sum = 7"

    numbers.addAll(Arrays.asList(1,2,3,4,5));
    System.out.println(numbers); // "[1, 2, 3, 4, 5]"

    numbers.removeAll(Arrays.asList(4,5,6,7));
    System.out.println(numbers); // "[1, 2, 3]"

    numbers.retainAll(Arrays.asList(2,3,4,5));
    System.out.println(numbers); // "[2, 3]"

Once you're familiar with the API, you can use it to contain more interesting objects. If you haven't familiarized yourself with the equals and hashCode contract, already, now is a good time to start.

In a nutshell:

  • @Override both or none; never just one. (very important, because it must satisfied property: a.equals(b) == true --> a.hashCode() == b.hashCode()
    • Be careful with writing boolean equals(Thing other) instead; this is not a proper @Override.
  • For non-null references x, y, z, equals must be:
    • reflexive: x.equals(x).
    • symmetric: x.equals(y) if and only if y.equals(x)
    • transitive: if x.equals(y) && y.equals(z), then x.equals(z)
    • consistent: x.equals(y) must not change unless the objects have mutated
    • x.equals(null) == false
  • The general contract for hashCode is:
    • consistent: return the same number unless mutation happened
    • consistent with equals: if x.equals(y), then x.hashCode() == y.hashCode()
      • strictly speaking, object inequality does not require hash code inequality
      • but hash code inequality necessarily requires object inequality
  • What counts as mutation should be consistent between equals and hashCode.

Next, you may want to impose an ordering of your objects. You can do this by making your type implements Comparable, or by providing a separate Comparator.

Having either makes it easy to sort your objects (Arrays.sort, Collections.sort(List)). It also allows you to use SortedSet, such as TreeSet.


Further readings on stackoverflow:

  • Overriding equals and hashCode in Java
  • When to use Comparable vs Comparator
like image 98
polygenelubricants Avatar answered Oct 04 '22 13:10

polygenelubricants