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

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。

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

;)

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

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.

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!";
  }

}