为什么不可能重写静态方法?

如果可能,请举例说明。


当前回答

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。

其他回答

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!";
  }

}

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。

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.

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。