Question

I think what I'm trying to do is clear, but I'm no Generics expert.

import java.util.ArrayList;
public class MinHeap<E extends Comparable> extends ArrayList<E>  {
    /* A simple wrapper around a List to make it a binary min-heap. */
    public MinHeap() {
        super();
    }

    @Override
    public boolean add(E e) {
        super.add(e);
        int i = this.size() - 1;
        int parent;

        while (i > 0) {
            parent = this.getParentIndex(i);

            if (this.get(i).compareTo(this.get(parent)) < 0) {
                this.swap(i, parent);
            } else {
                break;
            }
        }
        return true;
    }

    public int getParentIndex(int i) {
        if (i % 2 == 1) {
            return (i - 1) / 2;
        } else {
            return (i - 2) / 2;
        }
    }

    private void swap(int i, int j) {
        E temp = this.get(i);
        this.set(i, this.get(j));
        this.set(j, temp);
    }
}

I get a warning at compile-time:

MinHeap.java:21: warning: [unchecked] unchecked call to compareTo(T) as a member of the raw type Comparable
            if (this.get(i).compareTo(this.get(parent)) < 0) {
                                 ^
  where T is a type-variable:
    T extends Object declared in interface Comparable
1 warning

which I don't understand. What am I missing?

I initially thought that it had to do with needing to check the that this.get(i) and this.get(parent) were instances of Comparable ... so I added a check:

if (!(this.get(i) instanceof Comparable) ||
    !(this.get(parent) instanceof Comparable)) {
    return false;
}

But that gives the same warning.

Was it helpful?

Solution

public class MinHeap<E extends Comparable> extends ArrayList<E> 

should be

public class MinHeap<E extends Comparable<E>> extends ArrayList<E> 

since Comparable is a generic interface itself.

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