java - 如何在 Java 8 中将方法作为参数传递？

Question

我不明白如何使用 lambda 将方法作为参数传递。

考虑以下(未编译)代码，我怎样才能完成它以使其工作？

public class DumbTest {
    public class Stuff {
        public String getA() {
            return "a";
        }
        public String getB() {
            return "b";
        }
    }

    public String methodToPassA(Stuff stuff) {
        return stuff.getA();
    }

    public String methodToPassB(Stuff stuff) {
        return stuff.getB();
    }

    //MethodParameter is purely used to be comprehensive, nothing else...
    public void operateListWith(List<Stuff> listStuff, MethodParameter method) {
        for (Stuff stuff : listStuff) {
            System.out.println(method(stuff));
        }
    }

    public DumbTest() {
        List<Stuff> listStuff = new ArrayList<>();
        listStuff.add(new Stuff());
        listStuff.add(new Stuff());

        operateListWith(listStuff, methodToPassA);
        operateListWith(listStuff, methodToPassB);
    }

    public static void main(String[] args) {
        DumbTest l = new DumbTest();

    }
}

Answer 1

声明您的方法以接受与您的方法签名匹配的现有功能接口(interface)类型的参数:

public void operateListWith(List<Stuff> listStuff, Function<Stuff, String> method) {
    for (Stuff stuff : listStuff) {
        System.out.println(method.apply(stuff));
    }
}

并这样调用它:

operateListWith(listStuff, this::methodToPassA);

作为进一步的见解，您不需要 methodToPassA 的间接访问:

operateListWith(listStuff, Stuff::getA);