Java 8构造函数方法引用

Question

我正在阅读perform message它附带的样本我重现..

@FunctionalInterface
public interface Action {
    public void perform();
}

实施者

public final class ActionImpl implements Action {
    public ActionImpl() {
        System.out.println("constructor[ActionIMPL]");
    }

    @Override
    public void perform() {
        System.out.println("perform method is called..");
    }    
}

来电者.

public final class MethodReferences {

    private final Action action;

    public MethodReferences(Action action) {
        this.action = action;
    }

    public void execute() {
        System.out.println("execute->called");
        action.perform();
        System.out.println("execute->exist");
    }

    public static void main(String[] args) {
        MethodReferences clazz = new MethodReferences(new ActionImpl());
        clazz.execute();
    }
}

如果调用此方法则将以下内容打印到输出中

constructor[ActionIMPL]
execute->called
perform method is called..
execute->exist

一切都很好,但如果我使用方法参考不MethodReference打印方法!为什么我错过了什么？

如果我使用此代码

MethodReferences clazz = new MethodReferences(() -> new ActionImpl());
clazz.execute();

或者这个代码

final MethodReferences clazz = new MethodReferences(ActionImpl::new);

打印出来

execute->called
constructor[ActionIMPL]
execute->exist

perform message

任何帮助都非常感谢我使用

public class MethodReferenceCall {
    public MethodReferenceCall() {
        System.out.println("MethodReferenceCall class constructor called");
    }

    public static void main(String[] args) {
        MethodReferenceCall clazz = new MethodReferenceCall();
        MethodReferences constructorCaller = new MethodReferences(MethodReferenceCall::new);
        constructorCaller.execute();
    }
}

更新

对于像我这样学习的人来说,这是正确运行的代码.

我创建了一个调用者类.

因为我需要实现一个空方法MethodReference,我需要作为参数传递给类构造函数perform message我引用的构造函数MethodReference

@FunctionalInterface
public interface Action {
    public void perform();
}

我希望能帮助委内瑞拉人提出最好的问候.

Answer 1

这个

MethodReferences clazz = new MethodReferences(() -> new ActionImpl());

不使用方法引用,它使用lambda表达式.所述功能接口Action的

public void perform();

所以

() -> new ActionImpl()

被翻译成类似的东西

new Action() {
    public void perform() {
        new ActionImpl();
    }
}

同样地,在

MethodReferences clazz = new MethodReferences(ActionImpl::new);

该

new Action() {
    public void perform() {
        new ActionImpl();
    }
}

它使用构造函数引用,被翻译成类似的东西

MethodReferences clazz = new MethodReferences(() -> new ActionImpl());

这ActionImpl::new不会调用ActionImpl::new.它解析为期望类型的实例,其功能接口方法实现为调用该构造函数.