a can only be final here. Why? How can I reassign a in onClick() method without keeping it as private member? private void f(Button b, final int a){ b.addClickHandler(new ClickHandler() { @Override public void onClick(ClickEvent event) { int b = a*5; } }); } How can I return the 5 * a when it clicked? I mean, private void f(Button b, final int a){ b.addClickHandler(new ClickHandler() { @Override public void onClick(ClickEvent event) { int b = a*5; return b; // but return type is void } }); }


当前回答

试试这段代码,

创建数组列表并将值放入其中并返回:

private ArrayList f(Button b, final int a)
{
    final ArrayList al = new ArrayList();
    b.addClickHandler(new ClickHandler() {

         @Override
        public void onClick(ClickEvent event) {
             int b = a*5;
             al.add(b);
        }
    });
    return al;
}

其他回答

Java匿名类与Javascript闭包非常相似,但Java以不同的方式实现。(请看安徒生的答案)

所以为了不让Java开发人员对那些有Javascript背景的人的奇怪行为感到困惑。我想这就是为什么他们强迫我们使用final,这不是JVM的限制。

让我们看看下面的Javascript例子:

var add = (function () {
  var counter = 0;

  var func = function () {
    console.log("counter now = " + counter);
    counter += 1; 
  };

  counter = 100; // line 1, this one need to be final in Java

  return func;

})();


add(); // this will print out 100 in Javascript but 0 in Java

在Javascript中,计数器的值是100,因为从头到尾只有一个计数器变量。

但在Java中,如果没有final,它将打印出0,因为在创建内部对象时,0值被复制到内部类对象的隐藏属性中。(这里有两个整数变量,一个在局部方法中,另一个在内部类隐藏属性中)

因此,内部对象创建后的任何更改(如第1行)都不会影响内部对象。所以它会混淆两种不同的结果和行为(Java和Javascript之间)。

我相信这就是为什么,Java决定强制它是最终的,所以数据从开始到结束都是“一致的”。

private void f(Button b, final int a[]) {

    b.addClickHandler(new ClickHandler() {

        @Override
        public void onClick(ClickEvent event) {
            a[0] = a[0] * 5;

        }
    });
}

要理解这种限制的基本原理,请考虑以下程序:

public class Program {

    interface Interface {
        public void printInteger();
    }
    static Interface interfaceInstance = null;

    static void initialize(int val) {
        class Impl implements Interface {
            @Override
            public void printInteger() {
                System.out.println(val);
            }
        }
        interfaceInstance = new Impl();
    }

    public static void main(String[] args) {
        initialize(12345);
        interfaceInstance.printInteger();
    }
}

The interfaceInstance remains in memory after the initialize method returns, but the parameter val does not. The JVM can’t access a local variable outside its scope, so Java makes the subsequent call to printInteger work by copying the value of val to an implicit field of the same name within interfaceInstance. The interfaceInstance is said to have captured the value of the local parameter. If the parameter weren’t final (or effectively final) its value could change, becoming out of sync with the captured value, potentially causing unintuitive behavior.

试试这段代码,

创建数组列表并将值放入其中并返回:

private ArrayList f(Button b, final int a)
{
    final ArrayList al = new ArrayList();
    b.addClickHandler(new ClickHandler() {

         @Override
        public void onClick(ClickEvent event) {
             int b = a*5;
             al.add(b);
        }
    });
    return al;
}

也许这个把戏能给你启发

Boolean var= new anonymousClass(){
    private String myVar; //String for example
    @Overriden public Boolean method(int i){
          //use myVar and i
    }
    public String setVar(String var){myVar=var; return this;} //Returns self instane
}.setVar("Hello").method(3);