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

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)。

其他回答

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

;)

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.

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、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。

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。