我有2个类serial1和serial 2. serial1实现Serializable而serial2没有。根据理论,我应该得到以下代码的Exception,但它工作正常。为什么会这样?
import java.io.*;
public class SerialTest {
public static void main(String args[]){
FileOutputStream fos=null;
ObjectOutputStream oos =null;
serial1 se = new serial1();
serial1 sd = null;
se.mets();
try {
fos= new FileOutputStream("serialtest");
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
oos =new ObjectOutputStream(fos);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
oos.writeObject(se);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
FileInputStream fis=null;
ObjectInputStream ois = null;
try {
fis = new FileInputStream("serialtest");
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
ois = new ObjectInputStream(fis);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
sd = (serial1) ois.readObject();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ClassNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
sd.mets();
}
}
import java.io.Serializable;
public class serial1 implements Serializable{
/* public static void main(String []args){
serial1 ss = new serial1();
ss.mets();
}*/
public void mets(){
serial2 s2 = new serial2();
System.out.println( "serial 1 + mets");
s2.met1();
}
}
public class serial2 {
public void met1(){
System.out.println("serial2 + met1");
}
}
--------------------------- * 输出是
序列1 + mets
serial2 + met1
序列1 + mets
serial2 + met1
答案 0 :(得分:3)
我没有看到你序列化serial2
答案 1 :(得分:3)
您实际上并未对serial2
进行序列化。您的mets
方法创建了一个局部变量,但是一旦该方法返回,它就会超出范围并且有资格进行垃圾收集。
如果serial2
中有serial1
类型的实例变量,那么在尝试序列化时会看到异常(假设它是一个非空值),但是局部变量不会有问题。