重写依赖于类的实例。多态性的意义在于，您可以子类化一个类，而实现这些子类的对象对于父类中定义的相同方法将具有不同的行为(并且在子类中被重写)。静态方法不与类的任何实例相关联，因此这个概念不适用。

There were two considerations driving Java's design that impacted this. One was a concern with performance: there had been a lot of criticism of Smalltalk about it being too slow (garbage collection and polymorphic calls being part of that) and Java's creators were determined to avoid that. Another was the decision that the target audience for Java was C++ developers. Making static methods work the way they do had the benefit of familiarity for C++ programmers and was also very fast, because there's no need to wait until runtime to figure out which method to call.

静态方法被JVM视为全局方法，根本不绑定到对象实例。

如果可以从类对象中调用静态方法(就像在Smalltalk等语言中那样)，那么在概念上是可能的，但在Java中却不是这样。

EDIT

你可以重载静态方法，没关系。但是你不能重写静态方法，因为类不是一级对象。您可以使用反射在运行时获取对象的类，但所获得的对象并不与类层次结构并行。

class MyClass { ... }
class MySubClass extends MyClass { ... }

MyClass obj1 = new MyClass();
MySubClass obj2 = new MySubClass();

ob2 instanceof MyClass --> true

Class clazz1 = obj1.getClass();
Class clazz2 = obj2.getClass();

clazz2 instanceof clazz1 --> false

你可以对类进行反射，但它仅限于此。使用clazz1.staticMethod()不会调用静态方法，而是使用MyClass.staticMethod()。静态方法不绑定到对象，因此在静态方法中没有this或super的概念。静态方法是一个全局函数;因此，也没有多态性的概念，因此，方法重写没有意义。

但是，如果MyClass在运行时是一个调用方法的对象，这是可能的，就像在Smalltalk(或者可能是一个评论建议的JRuby，但我对JRuby一无所知)。

哦是的…还有一件事。您可以通过对象obj1.staticMethod()调用静态方法，但这实际上是MyClass.staticMethod()的语法糖，应该避免。在现代IDE中，它通常会引发一个警告。我不知道他们为什么允许走这条捷径。

重写是为实例成员保留的，以支持多态行为。静态类成员不属于特定实例。相反，静态成员属于类，因此不支持重写，因为子类只继承受保护和公共实例成员，而不继承静态成员。您可能希望定义一个接口，并研究工厂和/或策略设计模式，以评估替代方法。

重写静态方法有什么好处呢?不能通过实例调用静态方法。

MyClass.static1()
MySubClass.static1()   // If you overrode, you have to call it through MySubClass anyway.

编辑:似乎由于语言设计中的一个不幸疏忽，您可以通过实例调用静态方法。一般没人会这么做。我的坏。

我个人认为这是Java设计中的一个缺陷。是的，是的，我理解非静态方法附加到实例，而静态方法附加到类，等等。不过，考虑下面的代码:

public class RegularEmployee {
    private BigDecimal salary;

    public void setSalary(BigDecimal salary) {
        this.salary = salary;
    }

    public static BigDecimal getBonusMultiplier() {
        return new BigDecimal(".02");
    }

    public BigDecimal calculateBonus() {
        return salary.multiply(getBonusMultiplier());
    }

    /* ... presumably lots of other code ... */
}

public class SpecialEmployee extends RegularEmployee {
    public static BigDecimal getBonusMultiplier() {
        return new BigDecimal(".03");
    }
}

这段代码不能像您期望的那样工作。也就是说，特殊员工和普通员工一样有2%的奖金。但如果你去掉了“静态”，那么SpecialEmployee就能得到3%的奖金。

(不可否认，这个例子的编码风格很差，因为在现实生活中，你可能希望奖励乘数在数据库的某个地方，而不是硬编码。但这只是因为我不想用大量与主题无关的代码使示例陷入困境。)

在我看来，您可能想要使getBonusMultiplier成为静态的，这似乎很合理。也许您希望能够显示所有员工类别的奖金乘数，而不需要在每个类别中有一个员工实例。搜索这样的例子有什么意义呢?如果我们正在创建一个新的员工类别，并且还没有任何员工分配给它，该怎么办?这在逻辑上是一个静态函数。

但这并不奏效。

是的，是的，我可以想出很多方法来重写上面的代码，让它工作。我的观点不是它产生了一个无法解决的问题，而是它为粗心的程序员制造了一个陷阱，因为这种语言的行为不像我认为一个理性的人所期望的那样。

也许如果我试着为OOP语言编写一个编译器，我很快就会明白为什么实现它以覆盖静态函数是困难的或不可能的。

或许有一些很好的理由来解释为什么Java会这样做。有人能指出这种行为的好处吗，这种行为能让一些问题变得更简单吗?我的意思是，不要只是把我指给Java语言规范，然后说“看，这是它如何行为的文档”。我知道。但是，它为什么会有这样的表现，有一个很好的理由吗?(除了明显的“让它正常工作太难了”……)

更新

@VicKirk: If you mean that this is "bad design" because it doesn't fit how Java handles statics, my reply is, "Well, duh, of course." As I said in my original post, it doesn't work. But if you mean that it is bad design in the sense that there would be something fundamentally wrong with a language where this worked, i.e. where statics could be overridden just like virtual functions, that this would somehow introduce an ambiguity or it would be impossible to implement efficiently or some such, I reply, "Why? What's wrong with the concept?"

I think the example I give is a very natural thing to want to do. I have a class that has a function that does not depend on any instance data, and which I might very reasonably want to call independent of an instance, as well as wanting to call from within an instance method. Why should this not work? I've run into this situation a fair number of times over the years. In practice I get around it by making the function virtual, and then creating a static method whose only purpose in life is to be a static method that passes the call on to the virtual method with a dummy instance. That seems like a very roundabout way to get there.

一般来说，允许“重写”静态方法是没有意义的，因为没有好的方法来确定在运行时调用哪个方法。以Employee为例，如果我们调用regularemploee . getbonusmultiplier()——应该执行哪个方法?

以Java为例，人们可以想象这样一种语言定义:只要静态方法是通过对象实例调用的，就可以“覆盖”它们。然而，这样做只是重新实现常规的类方法，在没有真正带来任何好处的情况下为语言增加冗余。

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)。

下面的代码表明这是可能的:

class OverridenStaticMeth {   

static void printValue() {   
System.out.println("Overriden Meth");   
}   

}   

public class OverrideStaticMeth extends OverridenStaticMeth {   

static void printValue() {   
System.out.println("Overriding Meth");   
}   

public static void main(String[] args) {   
OverridenStaticMeth osm = new OverrideStaticMeth();   
osm.printValue();   

System.out.println("now, from main");
printValue();

}   

} 

方法重写可以通过动态调度实现，这意味着对象的声明类型不决定其行为，而是决定其运行时类型:

Animal lassie = new Dog();
lassie.speak(); // outputs "woof!"
Animal kermit = new Frog();
kermit.speak(); // outputs "ribbit!"

尽管lassie和kermit都声明为Animal类型的对象，但它们的行为(method .speak())会有所不同，因为动态调度只会在运行时将方法调用.speak()绑定到实现，而不是在编译时。

现在，这里是静态关键字开始有意义的地方:单词“静态”是“动态”的反义词。所以你不能重写静态方法的原因是因为静态成员上没有动态分派——因为静态字面上的意思是“非动态的”。如果它们是动态分派的(因此可以被重写)，静态关键字就没有意义了。

简短的回答是:这是完全可能的，但Java没有做到。

下面是一些代码，说明了Java中的当前状态:

文件Base.java:

package sp.trial;
public class Base {
  static void printValue() {
    System.out.println("  Called static Base method.");
  }
  void nonStatPrintValue() {
    System.out.println("  Called non-static Base method.");
  }
  void nonLocalIndirectStatMethod() {
    System.out.println("  Non-static calls overridden(?) static:");
    System.out.print("  ");
    this.printValue();
  }
}

文件Child.java:

package sp.trial;
public class Child extends Base {
  static void printValue() {
    System.out.println("  Called static Child method.");
  }
  void nonStatPrintValue() {
    System.out.println("  Called non-static Child method.");
  }
  void localIndirectStatMethod() {
    System.out.println("  Non-static calls own static:");
    System.out.print("  ");
    printValue();
  }
  public static void main(String[] args) {
    System.out.println("Object: static type Base; runtime type Child:");
    Base base = new Child();
    base.printValue();
    base.nonStatPrintValue();
    System.out.println("Object: static type Child; runtime type Child:");
    Child child = new Child();
    child.printValue();
    child.nonStatPrintValue();
    System.out.println("Class: Child static call:");
    Child.printValue();
    System.out.println("Class: Base static call:");
    Base.printValue();
    System.out.println("Object: static/runtime type Child -- call static from non-static method of Child:");
    child.localIndirectStatMethod();
    System.out.println("Object: static/runtime type Child -- call static from non-static method of Base:");
    child.nonLocalIndirectStatMethod();
  }
}

如果你运行这个(我在Mac上做的，从Eclipse，使用Java 1.6)，你会得到:

Object: static type Base; runtime type Child.
  Called static Base method.
  Called non-static Child method.
Object: static type Child; runtime type Child.
  Called static Child method.
  Called non-static Child method.
Class: Child static call.
  Called static Child method.
Class: Base static call.
  Called static Base method.
Object: static/runtime type Child -- call static from non-static method of Child.
  Non-static calls own static.
    Called static Child method.
Object: static/runtime type Child -- call static from non-static method of Base.
  Non-static calls overridden(?) static.
    Called static Base method.

在这里，唯一可能令人惊讶的情况(也是这个问题所涉及的)似乎是第一种情况:

运行时类型并不用于确定调用哪些静态方法，即使是在使用对象实例(obj.staticMethod())调用时也是如此。

最后一种情况:

当从类的对象方法中调用静态方法时，选择的静态方法是类本身可访问的方法，而不是定义对象运行时类型的类可访问的方法。

使用对象实例调用

静态调用在编译时解析，而非静态方法调用在运行时解析。注意，尽管静态方法是继承的(从父方法)，但它们不会被(子方法)覆盖。如果你另有期待，这可能会是一个惊喜。

从对象方法中调用

对象方法调用使用运行时类型解析，但静态(类)方法调用使用编译时(已声明)类型解析。

改变规则

要更改这些规则，以便示例中的最后一个调用Child.printValue()，静态调用必须在运行时提供类型，而不是编译器在编译时使用对象(或上下文)声明的类解析调用。然后，静态调用可以使用(动态)类型层次结构来解析调用，就像现在的对象方法调用一样。

这是很容易做到的(如果我们改变Java:-O)，并不是完全不合理的，然而，它有一些有趣的考虑。

主要需要考虑的是，我们需要决定哪个静态方法调用应该执行此操作。

目前，Java语言中有这种“怪癖”，即obj.staticMethod()调用被ObjectClass.staticMethod()调用取代(通常带有警告)。[注意:ObjectClass是obj的编译时类型。]这些将是很好的候选人，以这种方式重写，采用obj的运行时类型。

如果我们这样做了，就会使方法体更难阅读:父类中的静态调用可能会被动态地“重新路由”。为了避免这种情况，我们必须使用类名调用静态方法——这使得调用更明显地用编译时类型层次结构来解析(就像现在一样)。

调用静态方法的其他方法更加棘手:this. staticmethod()的意思应该与obj.staticMethod()相同，接受this的运行时类型。然而，这可能会给现有程序带来一些麻烦，这些程序调用(显然是本地的)没有修饰的静态方法(可以说相当于this.method())。

那么无修饰调用staticMethod()会怎样呢?我建议他们像今天一样，使用本地类上下文来决定要做什么。否则就会产生巨大的混乱。当然，如果method是非静态方法，则method()意味着this.method()，如果method是静态方法，则意味着ThisClass.method()。这是另一个困惑的来源。

其他的考虑

如果我们改变了这种行为(并使静态调用具有潜在的动态非本地性)，我们可能会希望重新审视final、private和protected作为类静态方法的限定符的意义。然后，我们都必须习惯这样一个事实:私有静态方法和公共final方法不会被覆盖，因此可以在编译时安全地解析，并且可以“安全”地作为本地引用读取。

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。

I like and double Jay's comment (https://stackoverflow.com/a/2223803/1517187). I agree that this is the bad design of Java. Many other languages support overriding static methods, as we see in previous comments. I feel Jay has also come to Java from Delphi like me. Delphi (Object Pascal) was one of the languages implementing OOP before Java and one of the first languages used for commercial application development. It is obvious that many people had experience with that language since it was in the past the only language to write commercial GUI products. And - yes, we could in Delphi override static methods. Actually, static methods in Delphi are called "class methods", while Delphi had the different concept of "Delphi static methods" which were methods with early binding. To override methods you had to use late binding, declare "virtual" directive. So it was very convenient and intuitive and I would expect this in Java.

其实我们错了。 尽管Java默认情况下不允许重写静态方法，但如果你彻底查看Java中Class和Method类的文档，你仍然可以通过以下工作方法来模拟静态方法重写:

import java.lang.reflect.InvocationTargetException;
import java.math.BigDecimal;

class RegularEmployee {

    private BigDecimal salary = BigDecimal.ONE;

    public void setSalary(BigDecimal salary) {
        this.salary = salary;
    }
    public static BigDecimal getBonusMultiplier() {
        return new BigDecimal(".02");
    }
    public BigDecimal calculateBonus() {
        return salary.multiply(this.getBonusMultiplier());
    }
    public BigDecimal calculateOverridenBonus() {
        try {
            // System.out.println(this.getClass().getDeclaredMethod(
            // "getBonusMultiplier").toString());
            try {
                return salary.multiply((BigDecimal) this.getClass()
                    .getDeclaredMethod("getBonusMultiplier").invoke(this));
            } catch (IllegalAccessException e) {
                e.printStackTrace();
            } catch (IllegalArgumentException e) {
                e.printStackTrace();
            } catch (InvocationTargetException e) {
                e.printStackTrace();
            }
        } catch (NoSuchMethodException e) {
            e.printStackTrace();
        } catch (SecurityException e) {
            e.printStackTrace();
        }
        return null;
    }
    // ... presumably lots of other code ...
}

final class SpecialEmployee extends RegularEmployee {

    public static BigDecimal getBonusMultiplier() {
        return new BigDecimal(".03");
    }
}

public class StaticTestCoolMain {

    static public void main(String[] args) {
        RegularEmployee Alan = new RegularEmployee();
        System.out.println(Alan.calculateBonus());
        System.out.println(Alan.calculateOverridenBonus());
        SpecialEmployee Bob = new SpecialEmployee();
        System.out.println(Bob.calculateBonus());
        System.out.println(Bob.calculateOverridenBonus());
    }
}

输出结果:

0.02
0.02
0.02
0.03

我们想要达到的目标:)

即使我们将第三个变量Carl声明为regulareemployee并给它分配了SpecialEmployee实例，我们仍然会在第一种情况下调用regulareemployee方法，在第二种情况下调用SpecialEmployee方法

RegularEmployee Carl = new SpecialEmployee();

System.out.println(Carl.calculateBonus());
System.out.println(Carl.calculateOverridenBonus());

看看输出控制台:

0.02
0.03

;)

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。

通过重写，可以实现动态多态。 当您说覆盖静态方法时，您试图使用的词语是矛盾的。

静态表示-编译时，重写用于动态多态性。 两者在性质上是相反的，因此不能同时使用。

动态多态行为发生在程序员使用对象并访问实例方法时。JRE将根据您使用的对象类型映射不同类的不同实例方法。

当你说覆盖静态方法时，我们将使用类名访问静态方法，它将在编译时被链接，因此没有在运行时将方法与静态方法链接的概念。因此，术语“重写”静态方法本身没有任何意义。

注意:即使你用一个对象访问一个类方法，java编译器仍然有足够的智能来发现它，并会做静态链接。

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

这个问题的答案很简单，标记为静态的方法或变量只属于类，因此静态方法不能在子类中继承，因为它们只属于超类。

Here is a simple explanation. A static method is associated with a class while an instance method is associated with a particular object. Overrides allow calling the different implementation of the overridden methods associated with the particular object. So it is counter-intuitive to override static method which is not even associated with objects but the class itself in the first place. So static methods cannot be overridden based on what object is calling it, it will always be associated with the class where it was created.

静态方法、变量、块或嵌套类属于整个类而不是对象。

Java中的方法用于公开对象/类的行为。在这里，由于方法是静态的(即静态方法仅用于表示类的行为)，改变/覆盖整个类的行为将违反面向对象编程的基本支柱之一，即高内聚。(记住构造函数在Java中是一种特殊的方法。)

高内聚性——一个类应该只有一个角色。例如:car类应该只生成汽车对象，而不生成自行车、卡车、飞机等。但是Car类可能有一些只属于它自己的特性(行为)。

因此，在设计java编程语言时。语言设计者认为，只有通过使方法本质上是静态的，才能允许开发人员保留类的某些行为。

下面的代码尝试覆盖静态方法，但不会遇到任何编译错误。

public class Vehicle {
static int VIN;

public static int getVehileNumber() {
    return VIN;
}}

class Car extends Vehicle {
static int carNumber;

public static int getVehileNumber() {
    return carNumber;
}}

这是因为，在这里我们没有重写一个方法，而只是重新声明它。Java允许重新声明一个方法(静态/非静态)。

从Car类的getVehileNumber()方法中删除静态关键字将导致编译错误，因为，我们正在尝试改变只属于Vehicle类的静态方法的功能。

此外，如果getVehileNumber()被声明为final，那么代码将无法编译，因为final关键字限制程序员重新声明方法。

public static final int getVehileNumber() {
return VIN;     }

总的来说，这取决于软件设计人员在哪里使用静态方法。 我个人更喜欢使用静态方法来执行某些操作，而不需要创建类的任何实例。第二，对外界隐藏一个类的行为。

现在看到上面的答案，每个人都知道我们不能重写静态方法，但不应该误解从子类访问静态方法的概念。

我们可以通过子类引用访问超类的静态方法，如果这个静态方法没有被子类中定义的新静态方法所隐藏。

例如，请参见下面的代码:-

public class StaticMethodsHiding {
    public static void main(String[] args) {
        SubClass.hello();
    }
}


class SuperClass {
    static void hello(){
        System.out.println("SuperClass saying Hello");
    }
}


class SubClass extends SuperClass {
    // static void hello() {
    // System.out.println("SubClass Hello");
    // }
}

输出:

SuperClass saying Hello

关于在子类中隐藏静态方法的细节，请参阅Java oracle文档并搜索你可以在子类中做什么。

谢谢

Yes. Practically Java allows overriding static method, and No theoretically if you Override a static method in Java then it will compile and run smoothly but it will lose Polymorphism which is the basic property of Java. You will Read Everywhere that it is not possible to try yourself compiling and running. you will get your answer. e.g. If you Have Class Animal and a static method eat() and you Override that static method in its Subclass lets called it Dog. Then when wherever you Assign a Dog object to an Animal Reference and call eat() according to Java Dog's eat() should have been called but in static Overriding Animals' eat() will Be Called.

class Animal {
    public static void eat() {
        System.out.println("Animal Eating");
    }
}

class Dog extends Animal{
    public static void eat() {
        System.out.println("Dog Eating");
    }
}

class Test {
    public static void main(String args[]) {
       Animal obj= new Dog();//Dog object in animal
       obj.eat(); //should call dog's eat but it didn't
    }
}


Output Animal Eating

According to Polymorphism Principle of Java, the Output Should be Dog Eating. But the result was different because to support Polymorphism Java uses Late Binding that means methods are called only at the run-time but not in the case of static methods. In static methods compiler calls methods at the compile time rather than the run-time, so we get methods according to the reference and not according to the object a reference a containing that's why You can say Practically it supports static overring but theoretically, it doesn't.