Overriding in Java simply means that the particular method would be called based on the runtime type of the object and not on the compile-time type of it (which is the case with overridden static methods). As static methods are class methods they are not instance methods so they have nothing to do with the fact which reference is pointing to which Object or instance, because due to the nature of static method it belongs to a specific class. You can redeclare it in the subclass but that subclass won't know anything about the parent class' static methods because, as I said, it is specific to only that class in which it has been declared. Accessing them using object references is just an extra liberty given by the designers of Java and we should certainly not think of stopping that practice only when they restrict it more details and example http://faisalbhagat.blogspot.com/2014/09/method-overriding-and-method-hiding.html

Well... the answer is NO if you think from the perspective of how an overriden method should behave in Java. But, you don't get any compiler error if you try to override a static method. That means, if you try to override, Java doesn't stop you doing that; but you certainly don't get the same effect as you get for non-static methods. Overriding in Java simply means that the particular method would be called based on the run time type of the object and not on the compile time type of it (which is the case with overriden static methods). Okay... any guesses for the reason why do they behave strangely? Because they are class methods and hence access to them is always resolved during compile time only using the compile time type information. Accessing them using object references is just an extra liberty given by the designers of Java and we should certainly not think of stopping that practice only when they restrict it :-)

示例:让我们试着看看如果我们尝试重写一个静态方法会发生什么:-

class SuperClass {
// ......
public static void staticMethod() {
    System.out.println("SuperClass: inside staticMethod");
}
// ......
}

public class SubClass extends SuperClass {
// ......
// overriding the static method
public static void staticMethod() {
    System.out.println("SubClass: inside staticMethod");
}

// ......
public static void main(String[] args) {
    // ......
    SuperClass superClassWithSuperCons = new SuperClass();
    SuperClass superClassWithSubCons = new SubClass();
    SubClass subClassWithSubCons = new SubClass();

    superClassWithSuperCons.staticMethod();
    superClassWithSubCons.staticMethod();
    subClassWithSubCons.staticMethod();
    // ...
}
}

输出: SuperClass:在staticMethod内部 SuperClass:在staticMethod内部 子类:staticMethod内部

注意输出的第二行。如果staticMethod被重写，这一行应该与第三行相同，因为我们在运行时类型的对象上调用'staticMethod()'作为'子类'而不是'超类'。这证实了静态方法总是只使用它们的编译时类型信息进行解析。

简单的解决方案:使用单例实例。它将允许重写和继承。

在我的系统中，我有SingletonsRegistry类，它为传递的class返回实例。如果没有找到instance，则创建它。

Haxe语言类:

package rflib.common.utils;
import haxe.ds.ObjectMap;



class SingletonsRegistry
{
  public static var instances:Map<Class<Dynamic>, Dynamic>;

  static function __init__()
  {
    StaticsInitializer.addCallback(SingletonsRegistry, function()
    {
      instances = null;
    });

  } 

  public static function getInstance(cls:Class<Dynamic>, ?args:Array<Dynamic>)
  {
    if (instances == null) {
      instances = untyped new ObjectMap<Dynamic, Dynamic>();      
    }

    if (!instances.exists(cls)) 
    {
      if (args == null) args = [];
      instances.set(cls, Type.createInstance(cls, args));
    }

    return instances.get(cls);
  }


  public static function validate(inst:Dynamic, cls:Class<Dynamic>)
  {
    if (instances == null) return;

    var inst2 = instances[cls];
    if (inst2 != null && inst != inst2) throw "Can\'t create multiple instances of " + Type.getClassName(cls) + " - it's singleton!";
  }

}

Overriding in Java simply means that the particular method would be called based on the runtime type of the object and not on the compile-time type of it (which is the case with overridden static methods). As static methods are class methods they are not instance methods so they have nothing to do with the fact which reference is pointing to which Object or instance, because due to the nature of static method it belongs to a specific class. You can redeclare it in the subclass but that subclass won't know anything about the parent class' static methods because, as I said, it is specific to only that class in which it has been declared. Accessing them using object references is just an extra liberty given by the designers of Java and we should certainly not think of stopping that practice only when they restrict it more details and example http://faisalbhagat.blogspot.com/2014/09/method-overriding-and-method-hiding.html

通过重写，我们可以根据对象类型创建一个多态性质。静态方法与对象无关。因此java不支持静态方法重写。

在Java(和许多面向对象语言，但我不能说所有;所有的方法都有一个固定的签名——参数和类型。在虚方法中，第一个参数是隐含的:对对象本身的引用，当从对象内部调用时，编译器会自动添加这个参数。

静态方法没有区别——它们仍然有固定的签名。然而，通过将方法声明为静态，您已经显式地声明了编译器不能在该签名的开头包含隐含的对象形参。因此，任何其他调用此方法的代码都不能试图将对象引用放到堆栈上。如果它确实这样做了，那么方法执行将无法工作，因为参数将在堆栈上的错误位置—移位1。

由于两者之间的差异;虚方法总是有一个上下文对象的引用(即this)，这样就可以引用堆中属于该对象实例的任何东西。但是对于静态方法，由于没有传递引用，该方法不能访问任何对象变量和方法，因为上下文是未知的。

如果您希望Java更改定义，以便为每个方法(静态方法或虚拟方法)传递对象上下文，那么实际上您将只有虚拟方法。

就像有人在评论中问的那样——你想要这个功能的原因和目的是什么?

I do not know Ruby much, as this was mentioned by the OP, I did some research. I see that in Ruby classes are really a special kind of object and one can create (even dynamically) new methods. Classes are full class objects in Ruby, they are not in Java. This is just something you will have to accept when working with Java (or C#). These are not dynamic languages, though C# is adding some forms of dynamic. In reality, Ruby does not have "static" methods as far as I could find - in that case these are methods on the singleton class object. You can then override this singleton with a new class and the methods in the previous class object will call those defined in the new class (correct?). So if you called a method in the context of the original class it still would only execute the original statics, but calling a method in the derived class, would call methods either from the parent or sub-class. Interesting and I can see some value in that. It takes a different thought pattern.

由于您正在使用Java工作，您将需要适应这种做事方式。他们为什么这么做?好吧，可能是为了提高当时的性能基于现有的技术和理解。计算机语言在不断发展。回顾过去，并没有OOP这种东西。在未来，还会有其他新的想法。

EDIT: One other comment. Now that I see the differences and as I Java/C# developer myself, I can understand why the answers you get from Java developers may be confusing if you are coming from a language like Ruby. Java static methods are not the same as Ruby class methods. Java developers will have a hard time understanding this, as will conversely those who work mostly with a language like Ruby/Smalltalk. I can see how this would also be greatly confusing by the fact that Java also uses "class method" as another way to talk about static methods but this same term is used differently by Ruby. Java does not have Ruby style class methods (sorry); Ruby does not have Java style static methods which are really just old procedural style functions, as found in C.

顺便说一下，谢谢你的问题!今天我学到了一些关于类方法的新知识(Ruby风格)。