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.

匿名类是一个内部类，严格的规则适用于内部类(JLS 8.1.3):

任何在内部类中使用但未声明的局部变量、形式方法参数或异常处理程序参数必须声明为final。在内部类中使用但未声明的任何局部变量必须明确地在内部类的主体之前赋值。

我还没有在jls或jvm上找到一个原因或解释，但我们知道，编译器为每个内部类创建了一个单独的类文件，它必须确保，在这个类文件上声明的方法(在字节码级别上)至少可以访问局部变量的值。

(Jon有完整的答案-我保留这一个未删除，因为有人可能对JLS规则感兴趣)

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if the variable could be modified by the rest of the method - you could have code which appeared to be working with an out-of-date variable (because that's effectively what would be happening...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:)

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;
}

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
}

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。