java - Java 8 中的方法引用

Question

public class Car {

    private int maxSpeed;

    public Car(int maxSpeed) {
        this.maxSpeed = maxSpeed;
    }

    public int getMaxSpeed() {
        return maxSpeed;
    }
}

我们可以对汽车列表进行排序，

    Car carX = new Car(155);
    Car carY = new Car(140);

    List<Car> cars = new ArrayList<>();
    cars.add(carX);
    cars.add(carY);

    cars.sort(Comparator.comparing(Car::getMaxSpeed));

如果我们看到方法的签名 Comparator.comparing ，入参类型为Function<? super T, ? extends U>

在上面的例子中，Car::getMaxSpeed 是怎样的？被转换到Function<? super T, ? extends U>而以下不编译？

  Function<Void, Integer> function = Car::getMaxSpeed;

Answer 1

那是因为 getMaxSpeed方法是 Function<Car, Integer> .

即:

<Car, Integer> Comparator<Car> java.util.Comparator.comparing(
    Function<? super Car, ? extends Integer> keyExtractor
)

注意

为了引用getMaxSpeed来自 Car 的实例与 ::成语，你必须声明一个Car带有签名的方法:getMaxSpeed(Car car) .