Recursive generics | 易学教程

问题

Is there a way to make this method properly generic and do away with the warnings?

/**
 * <p>Sort a collection by a certain "value" in its entries. This value is retrieved using
 * the given <code>valueFunction</code> which takes an entry as argument and returns
 * its value.</p>
 * 
 * <p>Example:</p>
 * <pre>// sort tiles by number
 *Collects.sortByValue(tileList, true, new Function<Integer,NormalTile>() {
 *  public Integer call(NormalTile t) {
 *      return t.getNumber();
 *  }
 *});</pre>
 *
 * @param list The collection.
 * @param ascending Whether to sort ascending (<code>true</code>) or descending (<code>false</code>).
 * @param valueFunction The function that retrieves the value of an entry.
 */
public static <T> void sortByValue(List<T> list, final boolean ascending, @SuppressWarnings("rawtypes") final Function<? extends Comparable, T> valueFunction) {
    Collections.sort(list, new Comparator<T>() {
        @SuppressWarnings({ "unchecked", "rawtypes" })
        @Override public int compare(T o1, T o2) {
            final Comparable v1 = valueFunction.call(o1);
            final Comparable v2 = valueFunction.call(o2);
            return v1.compareTo(v2) * (ascending ? 1 : -1);
        }
    });
}

I tried Function<? extends Comparable<?>, T> and Function<? extends Comparable<? extends Comparable>, T> but neither compiled, with an error on the call to compareTo. For the former that is:

The method compareTo(capture#9-of ?) in the type Comparable is not applicable for the arguments (capture#10-of ? extends Comparable)

回答1:

Try this:

public static <T, C extends Comparable<? super C>> void sortByValue(List<T> list, final boolean ascending, final Function<C, T> valueFunction) {
    Collections.sort(list, new Comparator<T>() {
        @Override public int compare(T o1, T o2) {
            final C v1 = valueFunction.apply(o1);
            final C v2 = valueFunction.apply(o2);
            return v1.compareTo(v2) * (ascending ? 1 : -1);
        }
    });
}

you also need the super to allow comparators defined for sub types. More explanations here: http://docs.oracle.com/javase/tutorial/extra/generics/morefun.html

UPDATE

Also, looking at your code I see yet another bicycle, there is a good library the Google Collections, which provides very convenient Ordering notion to handle it.

So, your code would look like:

Ordering<NormalTile> myOrdering = Ordering.natural()
  .onResultOf(new Function<Integer,NormalTile>() {
  public Integer call(NormalTile t) {
      return t.getNumber();
  }))
  .nullsLast();
...
Collections.sort(list, myOrdering);
//or
newList = myOrdering.sortedCopy(readonlyList);

回答2:

This works for me (Eclipse compiler)

public static <T, U extends Comparable<U>> void sortByValue(
  List<T> list, final boolean ascending, final Function<U, T> valueFunction) {

  Collections.sort(list, new Comparator<T>() {
    @Override
    public int compare(T o1, T o2) {
      final U v1 = valueFunction.call(o1);
      final U v2 = valueFunction.call(o2);
      return v1.compareTo(v2) * (ascending ? 1 : -1);
    }
  });
}

As others posted, you may even go further and declare U as

U extends Comparable<? super U>

That will come in handy if you have more method arguments / return values depending on U

回答3:

What if you declare two parameters for the function?

public static <T,C extends Comparable<C>> void sortByValue(List<T> list,
    final boolean ascending, final Function<C, T> valueFunction) {
...
final C v1 = ...
final C v2  ...

Haven't sanity checked myself with a compiler (don't have your interfaces and am too hungry to mock them :) ), but give it a shot.

I'm also too groggy to reason over whether it should be C extends Comparable<C> or C extends Comparable<? super C>. I think the former would work and be a tad more general, although in practice, most classes don't implement Comparable except against themselves.

来源：https://stackoverflow.com/questions/8415604/recursive-generics

标签

java

generics

unchecked