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.

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.

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

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。

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。