java - 为什么可以用无参方法来代替Function

Question

对于以下代码片段，在 calculateOnShipments 中，一个参数接受以 Shipment 作为输入、以 Double 作为输出的函数。

Function<Shipment, Double> f

1) 为什么可以使用Shipment::calculateWeight调用？ calculateWeight() 不接受任何参数，尽管它返回 Double

public double calculateWeight()

2) 我尝试注释掉 calculateWeight() 但保留 calculateWeight(Shipment s1)，然后出现错误消息

"Cannot make a static reference to the non-static method calculateWeight(Shipment) from the type Shipment"

这条消息是什么意思？为什么说静态引用？在我评论 calculateWeight() 之前，为什么没有错误消息？

class Shipment {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        List<Shipment> l = new ArrayList<Shipment>();
        l.add(new Shipment());
        l.add(new Shipment());
        Shipment shipment1 = new Shipment();
        // Using an anonymous class
        List<Double> lw = shipment1.calculateOnShipments(l, Shipment::calculateWeight);
        System.out.println(lw);//[0.0,0.0]
    }

    public double calculateWeight() {
        double weight = 0;
        // Calculate weight
        return weight;
    }

    public double calculateWeight(Shipment s1) {
        double weight = 1;
        // Calculate weight
        return weight;
    }

    public List<Double> calculateOnShipments(
            List<Shipment> l, Function<Shipment, Double> f) {
        List<Double> results = new ArrayList<>();
        for (Shipment s : l) {
            results.add(f.apply(s));
        }
        return results;
    }
}

Answer 1

why it can be called by Shipment::calculateWeight? calculateWeight() does not accept any parameter, although it return double

可以使用方法引用来调用它，因为方法引用代表调用方法的代码 calculateWeight在shipment上它被接受作为 Function<Shipment, Double> 的输入。阅读这些内容以获得进一步的相互关系——

shipment -> shipment.calculateWeight()

或更进一步

new Function<Shipment, Double>() {
    @Override
    public Double apply(Shipment shipment) {
        return shipment.calculateWeight();
    }
})

<小时/>

I try to comment out the calculateWeight() but keep the calculateWeight(Shipment s1), then there is a error msg

现在，如果您已经理解了上面的 lambda 表示，那么如果没有调用 calculateWeight ，则方法引用是不可能的。方法需要一个装运，其中一个是调用该方法的实例，另一个应该传递给方法以用作参数。这可能是这样的——

shipment -> shipment.calculateWeight(shipment1)

<小时/>

请注意，将方法转换为 static比如

public static double calculateWeight(Shipment s1)

将导致推断方法引用接受实例作为参数本身，并且工作方式如下:

List<Double> lw = shipment1.calculateOnShipments(l, Shipment::calculateWeight)

哪里Shipment::calculateWeight代表shipment -> calculateWeight(shipment) .