java - 递归泛型

Question

有没有办法使此方法正确通用并消除警告？

/**
 * <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);
        }
    });
}

我试过了 Function<? extends Comparable<?>, T>和 Function<? extends Comparable<? extends Comparable>, T>但均未编译，调用 compareTo 时出错.对于前者，即:

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

Answer 1

试试这个:

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);
        }
    });
}

您还需要 super 来允许为子类型定义比较器。更多解释在这里:http://docs.oracle.com/javase/tutorial/extra/generics/morefun.html

更新

此外，查看您的代码我又看到了另一辆自行车，Google Collections 有一个很好的库，它提供了非常方便的 Ordering处理它的想法。

因此，您的代码如下所示:

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);