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
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
;)
其他回答
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.
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();
}
}
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
;)
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.
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.