Java OCA OCP Practice Question 1912

Question

What is the output of the following program? Assume the file paths referenced in the class exist and are able to be written to and read from. 

package mypkg; //from   w  w w  . j av a  2 s .c om
import java.io.*; 
public class MyClass implements Serializable { 
   private int chambers = -1; 
   private transient Double size = null; 
   private static String color; 
   public MyClass() { 
      this.chambers = 3; 
      color = "BLUE"; 
   } 
  
   public static void main(String[] love) throws Throwable { 
      try (ObjectOutputStream o = new ObjectOutputStream( 
            new FileOutputStream("scan.txt"))) { 
         final MyClass v = new MyClass(); 
         v.chambers = 2; 
         v.size = 10.0; 
         v.color = "RED"; 
         o.writeObject(v); 
      } 
      new MyClass(); 
      try (ObjectInputStream o = new ObjectInputStream( 
            new FileInputStream("scan.txt"))) { 
         MyClass v = (MyClass)o.readObject(); 
         System.out.print(v.chambers+","+v.size+","+v.color); 
      } 
   } 
   { chambers = 4; } 
}
  • A. 2,null,RED
  • B. 2,null,BLUE
  • C. 3,10.0,RED
  • D. The code does not compile.


B.

Note

The class compiles and runs without issue, so Option D is incorrect.

The class defines three variables, only one of which is serializable.

The first variable, chambers, is serializable, with the value 2 being written to disk and then read from disk.

Note that constructors and instance initializers are not executed when a class is deserialized.

The next variable, size, is transient.

It is discarded when it is written to disk, so it has the default object value of null when read from disk.

Finally, the variable color is static, which means it is shared by all instances of the class.

Even though the value was RED when the instance was serialized, this value was not written to disk, since it was not part of the instance.

The constructor call new Value() between the two try-with-resources blocks sets this value to BLUE, which is the value printed later in the application.

For these reasons, the class prints 2,null,BLUE, making Option B the correct answer.




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