java继承概念一例子打印不解解决方法
日期：2014-05-20 浏览次数：20746 次
java继承概念一例子打印不解
class Base {
public void foo(Base x)
{
System.out.println("Base.Base");
}
public void foo(Derived x)
{
System.out.println("Base.Derived");
}
}
class Derived extends Base
{
public void foo(Base x)
{
System.out.println("Derived.Base");
}
public void foo(Derived x)
{
System.out.println("Derived.Derived");
}
}
public class StaticParamsDemo
{
public static void whichFoo(Base arg1, Base arg2)
{
//It is guaranteed that we will call foo(Base)
//Only issue is which class's version of foo(Base)
//is called; the dynamic type of arg1 is used
//to decide
arg1.foo(arg2);
}
public static void main(String []args)
{
Base b = new Base();
Derived d = new Derived();
whichFoo(b,b);
whichFoo(b,d);
whichFoo(d,b);
whichFoo(d,d);
}
}
打印结果为：
Base.Base
Base.Base
Derived.Base
Derived.Base
想请问大虾，结果为什么不是：
Base.Base
Base.Derived
Derived.Base
Derived.Derived.
------解决方案--------------------
这个代码中继承的体现就是父类中的方法子类也有，但是会被子类Derived 覆盖
whichFoo的一个参数是什么类型，就调用对应类型的foo方法，例如，arg1是Base类型就调用Base类的foo方法，
第二个参数是什么类型，就调用该类中对应的foo方法
------解决方案--------------------
其实这段代码的突破口可以从
class Base {
public void foo(Base x)
{
System.out.println("Base.Base");
}
public void foo(Derived x)
{
System.out.println("Base.Derived");
}
}
class Derived extends Base
{
public void foo(Base x)
{
System.out.println("Derived.Base");
}
public void foo(Derived x)
{
System.out.println("Derived.Derived");
}
}
public class StaticParamsDemo
{
public static void whichFoo(Base arg1, Base arg2)
{
//It is guaranteed that we will call foo(Base)
//Only issue is which class's version of foo(Base)
//is called; the dynamic type of arg1 is used
//to decide
arg1.foo(arg2);
}
public static void main(String []args)
{
Base b = new Base();
Derived d = new Derived();
whichFoo(b,b);
whichFoo(b,d);
whichFoo(d,b);
whichFoo(d,d);
}
}
打印结果为：
Base.Base
Base.Base
Derived.Base
Derived.Base
想请问大虾，结果为什么不是：
Base.Base
Base.Derived
Derived.Base
Derived.Derived.
------解决方案--------------------
这个代码中继承的体现就是父类中的方法子类也有，但是会被子类Derived 覆盖
whichFoo的一个参数是什么类型，就调用对应类型的foo方法，例如，arg1是Base类型就调用Base类的foo方法，
第二个参数是什么类型，就调用该类中对应的foo方法
------解决方案--------------------
其实这段代码的突破口可以从
- Java code
public class StaticParamsDemo
{
public static void whichFoo(Base arg1, Base arg2)
{
arg1.foo(arg2);
}
public static void main(String []args)
{
Base b = new Base();
Derived d = new Derived();
whichFoo(b,b); //第一个参数是[color=#FF0000]b[/color]那就是进类Base 然后调用whichFoo，那第一个参数是类Base 自然就打印出Base 类中的第一个System.out.println("Base.Base");
whichFoo(b,d);//第一个参数是[color=#FF0000]b[/color]那就是进类Base 然后调用whichFoo，那第一个参数是类Base 自然就打印出Base 类中的第一个System.out.println("Base.Base");
whichFoo(d,b);//第一个参数是[color=#FF0000]d[/color]那就是进类Derived然后调用whichFoo，那第一个参数是类Derived 自然就打印出Derived类中的第一个System.out.println("Derived.Base");
whichFoo(d,d);//第一个参数是[color=#FF0000]d[/color]那就是进类Derived然后调用whichFoo，那第一个参数是类Derived 自然就打印出Derived类中的第一个System.out.println("Derived.Base");
}
}
------解决方案--------------------
- Java code
public class StaticParamsDemo
{
public static void whichFoo(Base arg1, Base arg2)
{
arg1.foo(arg2);
}
public static void whichtwo(Base arg1,Derived arg2){
arg1.foo(arg2);
}
public static void main(String []args)
{
Base b = new Base();
Derived d = new Derived();
whichFoo(b,b);
whichtwo(b,d);
whichFoo(d,b);
whichtwo(d,d);
}
}
------解决方案--------------------
结果是：
Base.Base
Base.Derived
Derived.Base
Derived.Derived
------解决方案--------------------
------解决方案--------------------
摸索下创建对象的流程吧。。。
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