How does java serialization deserialize final fields when no default constructor specified?

Question

I have an class defining an immutable value type that I now need to serialize. The immutability comes from the final fields which are set in the constructor. I've tried serializing, and it works (surprisingly?) - but I've no idea how.

Here's an example of the class

public class MyValueType implements Serializable {     private final int value;      private transient int derivedValue;      public MyValueType(int value)     {         this.value = value;         this.derivedValue = derivedValue(value);     }      // getters etc... } 

Given that the class doesn't have a no arg constructor, how can it be instantiated and the final field set?

(An aside - I noticed this class particularly because IDEA wasn't generating a "no serialVersionUID" inspection warning for this class, yet successfully generated warnings for other classes that I've just made serializable.)

Answer 1

Deserialization is implemented by the JVM on a level below the basic language constructs. Specifically, it does not call any constructor.