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If you implement an interface compiler asks you to provide implementation of those methods. But in case of calling overridden clone() method how compiler come to know that a particular interface is not mentioned (here in this case Cloneable) in class declaration. How compiler does this mapping to tell user that CloneNotSupported ? Is it has something to do with late-binding?

I think JVM has information about each class in method area like what are classes it is extending and what are interfaces it is implementing. So calling clone() method will check into those stored metadata about the class and finds out that Cloneable is not in that list of interfaces ? Is this somewhere close to correct answer ?

Amaresh
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1 Answers1

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You got it. Basically, the Object.clone() method does the following:

if (!(this instanceof Cloneable)) {
    throw new CloneNotSupportedException(); 
}

The javadoc explains it:

First, if the class of this object does not implement the interface Cloneable, then a CloneNotSupportedException is thrown.

Note that the compiler doesn't have anything to do with that. It all happens at runtime.

JB Nizet
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  • clone is native though :) +1 for the code snippet – mprabhat May 22 '12 at 07:34
  • Where can I find this piece of checking? It is coming from native implementation of clone()? Please note that I have overridden clone() method and not calling super.clone(). And my enclosing class doesn't implements clonable. – Amaresh May 22 '12 at 07:37
  • Yes, it's done in the native code of the method. The javadoc says: *By convention, the returned object should be obtained by calling super.clone*. Follow this convention, or choose another method name, or implement a copy constructor. – JB Nizet May 22 '12 at 07:40