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

正如注释中提到的,其中一些在Java 8中变得无关紧要,在Java 8中final可以是隐式的。但是,只有有效的final变量才能用于匿名内部类或lambda表达式。


这主要是由于Java管理闭包的方式。

When you create an instance of an anonymous inner class, any variables which are used within that class have their values copied in via the autogenerated constructor. This avoids the compiler having to autogenerate various extra types to hold the logical state of the "local variables", as for example the C# compiler does... (When C# captures a variable in an anonymous function, it really captures the variable - the closure can update the variable in a way which is seen by the main body of the method, and vice versa.)

As the value has been copied into the instance of the anonymous inner class, it would look odd if the variable could be modified by the rest of the method - you could have code which appeared to be working with an out-of-date variable (because that's effectively what would be happening... you'd be working with a copy taken at a different time). Likewise if you could make changes within the anonymous inner class, developers might expect those changes to be visible within the body of the enclosing method.

Making the variable final removes all these possibilities - as the value can't be changed at all, you don't need to worry about whether such changes will be visible. The only ways to allow the method and the anonymous inner class see each other's changes is to use a mutable type of some description. This could be the enclosing class itself, an array, a mutable wrapper type... anything like that. Basically it's a bit like communicating between one method and another: changes made to the parameters of one method aren't seen by its caller, but changes made to the objects referred to by the parameters are seen.

如果您对Java和c#闭包之间更详细的比较感兴趣,我有一篇文章对此进行了进一步的讨论。在这个答案中,我想专注于Java方面:)

其他回答

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.

在Java中,变量不仅可以作为参数,还可以作为类级字段,例如

public class Test
{
 public final int a = 3;

或者作为一个局部变量

public static void main(String[] args)
{
 final int a = 3;

如果希望访问和修改匿名类中的变量,则可能希望将该变量设置为外围类中的类级变量。

public class Test
{
 public int a;
 public void doSomething()
 {
  Runnable runnable =
   new Runnable()
   {
    public void run()
    {
     System.out.println(a);
     a = a+1;
    }
   };
 }
}

你不能把一个变量作为final,然后给它一个新值。Final的意思就是:值是不可改变的和Final的。

由于它是最终的,Java可以安全地将其复制到本地匿名类。你没有得到对int的引用(特别是因为你在Java中不能有对int这样的原语的引用,只能引用对象)。

它只是将a的值复制到一个隐式int,在匿名类中称为a。

由于Jon有实现细节的答案,另一个可能的答案是JVM不想处理已经结束他的激活的写入记录。

考虑这样一个用例,在这个用例中,你的lambdas不是被应用的,而是被存储在某个地方并稍后运行。

我记得在Smalltalk中,当你做这样的修改时,你会得到一个非法的商店。

Java内部类中的final变量[关于]

内部类只能使用

外部类引用 final局部变量来自范围外的引用类型(例如Object…) 值(原语)(例如int…)类型可以由最终引用类型包装。IntelliJ IDEA可以帮助您将其转换为一个元素数组

当编译器生成一个非静态嵌套(内部类)时-一个新类- <OuterClass>$<InnerClass>.class创建并将有界参数传递给构造函数[堆栈上的局部变量],它类似于闭包[Swift about]

Final变量是不能重新赋值的变量。最终引用变量仍然可以通过修改状态来更改

如果这是可能的,那就奇怪了,因为作为一个程序员,你可以做出这样的东西

//Not possible 
private void foo() {

    MyClass myClass = new MyClass(); //Case 1: myClass address is 1
    int a = 5;                       //Case 2: a = 5

    //just as an example
    new Button().addClickHandler(new ClickHandler() {
        
        @Override
        public void onClick(ClickEvent event) {

            /*
            myClass.something(); //<- what is the address - 1 or 2?
            int b = a; //<- what is the value - 5 or 10 ?

            //illusion that next changes are visible for Outer class
            myClass = new MyClass();
            a = 15;
            */
        }
    });

    myClass = new MyClass(); //Case 1: myClass address is 2
    int a = 10; //Case 2: a = 10
}

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);