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。

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

;)

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();

}   

} 

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。

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。