java为什么没有输入参数的方法可以用来替换函数<T，R>

Question

对于以下代码段，在calculateOnShipments中，一个参数接受一个函数，其中Shipment作为输入Double作为输出

Function<Shipment, Double> f

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

public double calculateWeight()

2）我试图注释掉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"

这是什么意思？为什么说是静态引用？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

可以使用方法引用调用它，因为方法引用表示调用shipment上的方法calculateWeight的代码，该方法被接受为Function<Shipment, Double>的输入。阅读以下内容，了解更多相关信息

shipment -> shipment.calculateWeight()

或者更进一步

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

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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calculateWeight方法调用，方法引用是不可能的，因为调用calculateWeight方法需要一个装运，一个是实例，在该实例上调用该方法，另一个是传递给该方法作为参数。那可能是

shipment -> shipment.calculateWeight(shipment1)

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请注意，将方法转换为static，例如

public static double calculateWeight(Shipment s1)

将导致推断方法引用以接受实例作为参数本身，并将作为：

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

其中Shipment::calculateWeight表示shipment -> calculateWeight(shipment)

Answer 2

1) By why it can be called by Shipment::calculateWeight? calculateWeight() does not accept any parameter, […]

calculateWeight()是类Shipment上的一个非静态方法，因此在某种意义上，它采用类型为Shipment的隐式this参数。所以Shipment::calculateWeight是一个接受这个参数的函数。（如果calculateWeight()是一个非静态方法，或者如果您编写了shipment1::calculateWeight而不是Shipment::calculateWeight以便将实例烘焙到方法引用中，那么这将是一个没有参数的函数。）

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

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

这是什么意思？为什么说是静态引用？[…]

这是个好问题。潜在的问题是，就方法引用而言，calculateWeight(Shipment s1)需要两个参数：类型Shipment的隐式this参数（因为它是类Shipment上的实例方法），以及类型Shipment的显式s1参数

错误信息会很奇怪；本质上，编译器是在说“这不是一个静态方法，所以你需要给它一个实例”，意思是你可以写shipment1::calculateWeight而不是Shipment::calculateWeight来指定应该调用它的实例