C++PKJava的动态绑定与静态绑定

在我前面的一篇文章中,有写到C++的静态绑定和动态绑定.期间遇到了很多问题.
而用Java的话,可以很轻松的实现所要的需求.
但是...

/**
 * @time 上午10:23:34
 * @fileName TestStaticDynamicBinding.java
 * @package  me.banxi.oo
 */
package me.banxi.oo;

class Figure{
	public static final String name = "Figure";

	public Figure(){
		
	}
	public double area(){
		return 0.0;
	}

}
class Circle extends Figure{
	public static final String name = "Circle";
	private double _radius;
	public Circle() {
		
	}
	public Circle(double radius){
		_radius = radius;
	}
	public double area(){
		return Math.PI*_radius*_radius;
	}
}

class Rectangle extends Figure{
	public static final String name = "Rectangle";
	private double _length;
	private double _width;
	
	public Rectangle(){
		
	}
	public Rectangle(double length,double width){
		_length = length;
		_width = width;
	}
	
	public double area(){
		return _length*_width;
	}
}
/**
 * @author banxi1988
 */
public class TestStaticDynamicBinding {

	static void printArea(Figure figure){
		System.out.println(figure.name+"的面积是:"+figure.area());
	}
	
	public static void main(String[] args) {
		Figure figure = new Figure();
		printArea(figure);
		
		Circle circle = new Circle(2.0);
		printArea(circle);
		
		Rectangle rect = new Rectangle(3.0,4.0);
		printArea(rect);

	}

}
/**
 *
banxi1988@banxi:~/cppworkspace/C++_pk_Java/bin$ java me.banxi.oo.TestStaticDynamicBinding
Figure的面积是:0.0
Figure的面积是:12.566370614359172
Figure的面积是:12.0
banxi1988@banxi:~/cppworkspace/C++_pk_Java/bin$ 

 ***/

由此可见Java中类成员方法都是动态绑定的.
但是类中静态成员还是静态绑定的.
经过测试发现,对于类的中字段属性,也是静态绑定的..
所以就算我把name声明改成下面的,最后结果输出的还是Figure.

public  final String name = "Circle";

修改后如下,完成,静态绑定:

/**
 * @time 上午10:23:34
 * @fileName TestStaticDynamicBinding.java
 * @package  me.banxi.oo
 */
package me.banxi.oo;

class Figure{
	private static final String name = "Figure";

	public Figure(){
		
	}
	public String getName(){
		return name;
	}
	
	public double area(){
		return 0.0;
	}

}
class Circle extends Figure{
	private static  final String name = "Circle";
	private double _radius;
	public Circle() {
		
	}
	public Circle(double radius){
		_radius = radius;
	}
	public String getName(){
		return name;
	}
	
	public double area(){
		return Math.PI*_radius*_radius;
	}
}

class Rectangle extends Figure{
	private static final String name = "Rectangle";
	private double _length;
	private double _width;
	
	public Rectangle(){
		
	}
	public Rectangle(double length,double width){
		_length = length;
		_width = width;
	}
	public String getName(){
		return name;
	}
	
	public double area(){
		return _length*_width;
	}
}
/**
 * @author banxi1988
 */
public class TestStaticDynamicBinding {

	static void printArea(Figure figure){
		System.out.println(figure.getName()+"的面积是:"+figure.area());
	}
	
	public static void main(String[] args) {
		Figure figure = new Figure();
		printArea(figure);
		
		Circle circle = new Circle(2.0);
		printArea(circle);
		
		Rectangle rect = new Rectangle(3.0,4.0);
		printArea(rect);

	}

}
/**
 *
banxi1988@banxi:~/cppworkspace/C++_pk_Java/bin$ java me.banxi.oo.TestStaticDynamicBinding
Figure的面积是:0.0
Circle的面积是:12.566370614359172
Rectangle的面积是:12.0
banxi1988@banxi:~/cppworkspace/C++_pk_Java/bin$ 
 ***/

下面测试一下Java中同类中构造函数的调用.
如下,将Rectangle类的name属性.写成下面的形式:

	private static final String name;//error
	private double _length;
	private double _width;
	
	public Rectangle(){
		name = "Rectangle";
	}
	public Rectangle(double length,double width){
		this();
		_length = length;
		_width = width;
	}
	public String getName(){
		return name;
	}
	

上面有一句语句是有错的,就是static final同时修改的字段.name没有直接初始化.
就违反了Java中上述两个修饰的属性了.
支持任意一个都可以.
修改后如下:

class Rectangle extends Figure{
	private  static String name;
	private double _length;
	private double _width;
	
	public Rectangle(){
		name = "Rectangle";
	}
	public Rectangle(double length,double width){
		this();
		_length = length;
		_width = width;
	}
	public String getName(){
		return name;
	}
	
	public double area(){
		return _length*_width;
	}
}

总结Java中的方法是动态绑定的.而字段则是静态绑定.
在Java中可以直接对字段,进行初始化.
而C++中则静态属性字段要在类外部初始化.实例属性字段要在构造函数中初始化.
而且在C++中调用同一类中构造函数好像没有java中的this()机制.
如果在C++中直接调用另一个构造函数的话,生成了另一个隐藏的对象.