When Cloning a subclass a nonfinal class that defines a clone() method that fails to call super.clone(), cloning a subclass of this class will produce an object of the wrong class.
The Java API [API 20112013] for the clone() method says:
By convention, the returned object should be obtained by calling
super.clone. If a class and all of its superclasses (exceptObject) obey this convention, it will be the case thatx.clone().getClass() == x.getClass().
Noncompliant Code Example
In this noncompliant code example, the clone() method in the class Base fails to call super.clone():
...
Consequently, the object devClone ends up being of type Base instead of Derived and , and the doLogic() method is incorrectly applied.
Compliant Solution
This compliant solution correctly calls super.clone() in the Base class's clone() method:
| Code Block | ||
|---|---|---|
| ||
class Base implements Cloneable {
public Object clone() throws CloneNotSupportedException {
return super.clone();
}
protected void doLogic() {
System.out.println("Superclass doLogic");
}
}
class Derived extends Base {
public Object clone() throws CloneNotSupportedException {
return super.clone();
}
protected void doLogic() {
System.out.println("Subclass doLogic");
}
public static void main(String[] args) {
Derived dev = new Derived();
try {
// Has type Derived, as expected
Base devClone = (Base)dev.clone(); // Has type Derived, as expected
devClone.doLogic(); // Prints "Subclass doLogic", as expected
} catch (CloneNotSupportedException e) { /* ... */ }
}
}
|
Applicability
Failing to call super.clone() may cause a cloned object to have the wrong type, with unexpected or incorrect results when it is used.
Bibliography
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