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下面这段代码中,为什么

//对引用的对象也进行复制
o.p=(Professor)p.clone(); 

就能够实现深拷贝呢?

class Professor implements Cloneable 
{ 
  String name; 
  int age; 
  Professor(String name,int age) 
  { 
  this.name=name; 
  this.age=age; 
  } 
 public Object clone() 
  { 
   Object o=null; 
  try 
   { 
    o=super.clone(); 
   } 
  catch(CloneNotSupportedException e) 
   { 
    System.out.println(e.toString()); 
   } 
  return o; 
  } 
} 
public class Student implements Cloneable 
{ 
  String name; 
  int age; 
  Professor p; 
  Student(String name,int age,Professor p) 
  { 
  this.name=name; 
  this.age=age; 
  this.p=p; 
  } 
 public Object clone() 
  { 
   Student o=null; 
  try 
   { 
    o=(Student)super.clone(); 
   } 
  catch(CloneNotSupportedException e) 
   { 
    System.out.println(e.toString()); 
   } 
   //对引用的对象也进行复制
   o.p=(Professor)p.clone(); 
  return o; 
  }  
 public static void main(String[] args) 
  { 
  Professor p=new Professor("wangwu",50); 
  Student s1=new Student("zhangsan",18,p); 
  Student s2=(Student)s1.clone(); 
  s2.p.name="lisi"; 
  s2.p.age=30; 
  //学生1的教授不 改变。
  System.out.println("name="+s1.p.name+","+"age="+s1.p.age); 
  System.out.println("name="+s2.p.name+","+"age="+s2.p.age); 
 } 
} 

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1 Answer

这只是表面上看起来是“深拷贝”, 实际上Student、Professor都没有实现深拷贝。

你在main方法增加几个输出:

Professor p=new Professor("wangwu",50);
        Student s1=new Student("zhangsan",18,p);
        Student s2=(Student)s1.clone();

        System.out.println(s1.p == s2.p); //false

        System.out.println(s1.name == s2.name); // true

        System.out.println(s1.p.name == s2.p.name); //true

        s2.p.name="lisi";
        s2.p.age=30;
        //学生1的教授不 改变。
        System.out.println("name="+s1.p.name+","+"age="+s1.p.age);
        System.out.println("name="+s2.p.name+","+"age="+s2.p.age);

        System.out.println(s1.name == s2.name); //true

        System.out.println(s1.p.name == s2.p.name); //false

即可看到,s1,s2的name仍然是"==", 在未set p.name时其name也为"==", 所以说都没有实现深拷贝。
当你set s2.p.name时,s2.p.name指向了另一个字符串常量的地址,所以(s1.p.name == s2.p.name); //false


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