java比较器比较类型推断

Question

让我们假设更改Comparator.comparing源代码

public static <T, U extends Comparable<? super U>> Comparator<T> comparing(
    Function<? super T, ? extends U> keyExtractor)
{
  Objects.requireNonNull(keyExtractor);
  return (Comparator<T> & Serializable) (c1, c2) -> keyExtractor.apply(c1).compareTo(keyExtractor.apply(c2));
}

到

public static <T, U extends Comparable<? super U>> Comparator<T> comparing(
    Function<T, U> keyExtractor)
{
  Objects.requireNonNull(keyExtractor);
  return (Comparator<T> & Serializable) (c1, c2) -> keyExtractor.apply(c1).compareTo(keyExtractor.apply(c2));
}

我们有以下课程

class PhysicalObject {
  double weight;
  public Double getWeight(){
    return weight;  
  }
}
class Car extends PhysicalObject {}

以下statement未编译

Function<PhysicalObject, Double> weight = p->p.getWeight();

Comparator<Car> c = HypotheticComparators.comparing(weight);

而这个编译Comparator<Car> c3_1 = HypotheticComparators.comparing(PhysicalObject::getWeight);

我知道第一条语句无法编译是因为修改后的比较函数没有有界通配符（? super T），但是为什么第二条语句可以编译而没有任何问题呢

Answer 1

将comparing定义为：

Comparator<T> comparing(Function<T, U> keyExtractor) // abbreviated

声明：

Comparator<Car> c = comparing(weight);

要求参数是Function<Car, ?>，但weight是Function<PhysicalObject, Double>，因此会出现编译错误

然而，当

Comparator<Car> c3_1 = comparing(PhysicalObject::getWeight);

超类PhysicalObject的Double getWeight()充分实现了Function<Car, ?>方法? apply(Car t)，因为t->getWeight()是对该方法的调用

PhysicalObject::getWeight方法引用类似于以下lambda：

Comparator<Car> c3_1 = comparing((Car t) -> {
    PhysicalObject p = t;
    return p.getWeight(); // call PhysicalObject::getWeight
});

或以下匿名类：

Comparator<Car> c3_1 = comparing(new Function<Car, Double>() {
    @Override
    public Double apply(Car t) {
        PhysicalObject p = t;
        return p.getWeight();
    }
});

方法引用中允许从Car到PhysicalObject的加宽转换