Serialization

[Henry Zhi Lin]

Hi guys,

the following code is from Devaka's exam lab,

 import java.io.*;
     public class A implements Serializable{
     
          int a=1;  static int b=2;  transient int c=2;
          A(int a, int b, int c){this.a=a; this.b=b; this.c=c;}
     
          public static void main(String args[])throws Exception{
               A ob1=new A(4,5,6);
               new ObjectOutputStream(new FileOutputStream("A.ser")).writeObject(ob1);
               A ob2=(A)new ObjectInputStream(new FileInputStream("A.ser")).readObject();
               System.out.print(ob2.a+" "+ob2.b+" "+ob2.c);
          }
     }   

The result of the code is 4 5 0. I do understand why c = 0, but have no clue why b = 5 not 0 can someone help me here?

[Henry Zhi Lin]

I guess I have to answer this myself.

There is a compiler warning "The static field A.b should be accessed by static way"

Which means the compiler treat ob2.b as A.b and at this point A.b = 5 that is why it returns 5.

[Rekha Srinath]

Yes you are right Henry.

Include the code A.b = 10; just after writeObject() call. You will get the value as 4 10 0

[Fabio Nascimento]

You can access static members of a given class by its reference, this is kinda bizarre but the compiler just check the reference's type of the variable and switch it for you on the bytecode, for example:

package br.com.kym.testes;
 
public class A {

static int a = 10;

static int getA() { return A.a; }

public static void main(String[] args) {

A a = new A();

                // All prints 10
System.out.println(a.a);
System.out.println(a.getA());
System.out.println(A.a);
System.out.println(A.getA());

}
 
}

What the compiler do is to check what type "a" is and then switch it for you. All the lower a's becomes upper A's in the bytecode. But be carefull, the compiler does it based on the reference type, not the objec. For instance:

package br.com.kym.testes;
 
public class A {

static int a = 10;

static int getA() { return A.a; }

public static void main(String[] args) {

A a = new B();

System.out.println(a.a);
System.out.println(a.getA());
System.out.println(A.a);
System.out.println(A.getA());

}
 
}
 
class B extends A {
 
static int a = 5;
static int getA() { return B.a; }
 
}

Even your object is of B type, you are accessing the A members!

[]