Question

我想创建用于执行GetInstance的父类。但我总是崩溃。 mInstance = new Instance<>();总是崩溃此行

public class Instance<T> {

    public Instance<T> mInstance;

    public Instance<T> GetInstance() {
        return mInstance;
    }

    public <T> Instance() {
        mInstance = new Instance<>();
    }
}

public class Timer extends Instance<Timer> {

}

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Answer 1

你做错了

在构造函数中，您以递归方式调用自身，因此stackoverflow。

将mInstance的初始化移到课外或使其静止。

试试这个：

public static class Instance<T> {

    private static final Instance<?> mInstance = new Instance<>();

    @SuppressWarnings("unchecked")
    public Instance<T> GetInstance() {
        // Make generic static instance. 
        // Strategy used similar to Collections.emptyList() implementation
        return (Instance<T>) mInstance;
    }

    private Instance() {

    }
}

Answer 2

IntelliJ使用他的向导编写Singleton。

public final class Singleton{

private static final Singleton ourInstance = new Singleton();

public static Singleton getInstance() {
    return ourInstance;
}

/**
 * Private constructor
 */
private Singleton() {}
}

这样，单例只能创建一次。请注意私有构造函数，它不允许您创建更多实例，以及只有一个此类对象的static final实例。至少，public static Singleton getInstance方法会返回实例。请记住，在某些情况下，您会将单例类的其他方法声明为synchronized，但这取决于您的类的业务逻辑。

编辑：正如另一位用户建议的那样，最好将该类标记为最终类，以防止对单例逻辑进行不必要的修改。

Answer 3

您有2个轻微错误，导致您的单身人士课程无效。首先，你的mInstance不是类的静态字段，这意味着类Instance的每个实例（严重选择名称TBH，使用更具体的东西）都会让它自己的mInstance失败整个目的，一个单一的实例。

其次，你有一个泛型类，遗憾的是在单例上下文中不能很好地工作。因为当你总是只有一个实例时，引入泛型是没有意义的。因为那样你就可以用T代替通用变量String或者你正在使用的类型。

要解决这些问题，您需要按以下方式更改课程：

public final class Instance {

     private static Instance mInstance;

     public static Instance getInstance(){
         if(mInstance == null){
             mInstance = new Instance();
         }
         return mInstance;
     }

     private Instance(){
         // your initialization logic
     }
}

我也做了一些正常的决定，例如制作类final，不要让第二个实例弹出，例如通过子类化。然后我也建立了构造函数和mInstnace字段private。要再次防止对类本身进行不必要的更改，并且不允许用户允许创建多个实例。

Java Generics类和构造函数

3 个答案: