为什么不可能重写静态方法?
如果可能,请举例说明。
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
通过重写,我们可以根据对象类型创建一个多态性质。静态方法与对象无关。因此java不支持静态方法重写。
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.
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)。
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.
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
;)
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。