从Java中的匿名对象访问非静态对象

Question

我使用 Google Web Toolkit 编写了一个基于模型-视图-控制器概念的应用程序。现在我的 ClientController 类有两个类型为 ClientModel 和 ClientView 的对象，它们被声明为私有且非静态的。对于异步调用的对象，我编写了一些确实有效的远程过程调用，但是从 AsyncCallback 类型附加匿名对象给我带来了问题，因为我无法访问这两个 ClientModel 和 ClientView 对象，而无需在我的函数中编写一些讨厌的 final 变量，如下所示：

package com.foo.bar;

/**
 * Represents the main handler holding delegates and controlling the contents of
 * the {@link ClientModel} and pushes notifications and message to both the
 * {@link ClientModel} and the {@link ClientView}.
 */
public class ClientController {
    /**
     * Exposes asynchronous functions to call the server component via RPC.
     */
    public static final MyServiceAsync mySvc = GWT.create(myService.class);

    /**
     * Represents the model associated with the {@link ClientController} object.
     */
    private ClientModel theModel = null;
    /**
     * Represents the view associated with the {@link ClientController} object.
     */
    private ClientView theView = null;

    /**
     * Creates a new {@link ClientController} object and instantiates both the
     * {@link ClientModel} and the {@link ClientView}.
     */
    public ClientController() {
        this.theModel = new ClientModel();
        this.theView = new ClientView(this.theModel);
    }

    /* some more code */

    /**
     * Tries to login the specified user and updates the {@link ClientView}
     * object to either an error message or the main interface.
     * 
     * @param user
     *            {@link User} object representing the user to login
     */
    public void loginUser(final User user) {
        ///////////////////////////////////////////////
        // THIS IS UGLY AND I DON'T KNOW HOW TO FIX THIS
        ///////////////////////////////////////////////
        final ClientModel currentModel = this.theModel;

        // Execute the login protocol
        ClientController.mySvc.login(user, new AsyncCallback<Boolean>() {
            /**
             * The request was successfully executed. Returns a boolean value
             * indicating whether the user was logged in.
             * 
             * @param result
             *            true, if the user was logged in; otherwise, false.
             */
            @Override
            public void onSuccess(Boolean result) {
                // The user was successfully logged in and we must both store
                // him in the model and then update the view.
                if (result) {
                    // TODO: Update the view to show the chat itself and save
                    // the current User to the ClientModel.
                    System.out.println("The User " + user.getUsername()
                            + " is now logged in!");
                    // Anonymous object can not access the theModel and theView
                    // objects of ClientController directly ...
                    // drunk, fix later!
                    // UGLY FIX FOR NOW
                    currentModel.setCurrentUser(user);
                } else {
                    // TODO: Unhide the error label of the login form and output
                    // some nice error message.
                    System.out.println("Login failed for User "
                            + user.getUsername() + "!");
                }
            }

            /**
             * The request provoked an error.
             */
            @Override
            public void onFailure(Throwable up) {
                try {
                    throw up; // Ha ha
                } catch (Throwable e) {
                    // Who cares?
                }
            }
        });
    }
}

现在我使用指向内部模型的最终指针在loginUser 被高亮显示两次。谁能解释一下是否有更好的解决方案而不将 theModel 和 theView 移动到静态成员？在您访问任一组件的每个函数中编写这样的“最终包装器”是很烦人的......

Answer 1

当您创建非静态内部类的实例或匿名类的实例时，该实例会隐式绑定到创建的外部类的实例它。您可以使用 OuterClassName.this.member 从内部类中访问外部类的成员。

在您的情况下：ClientController.this.theModel

Answer 2

尝试声明您的成员为最终成员，如下所示：

...

private final ClientModel theModel;

private final ClientView theView;

public ClientController() {
    this.theModel = new ClientModel();
    this.theView = new ClientView(this.theModel);
}