Can someone explain to me why
@Override public void fooMethod(Class<?> c)
doesn't override
public void fooMethod(Class c)
and gives me the following errors instead:
- Name clash: The method fooMethod(Class<?>) of type SubClass has the same erasure as fooMethod(Class) of type SuperClass but does not override it - The method fooMethod(Class<?>) of type SubClass must override a superclass method
?
Edit: "java -version
" says Java(TM) 2 Runtime Environment, Standard Edition (build 1.5.0_16-b06-284). As for the code snippet, it's already above, pretty much; the above extends the one below.
Type erasure is a process in which compiler replaces a generic parameter with actual class or bridge method. In type erasure, compiler ensures that no extra classes are created and there is no runtime overhead.
- Erasure is a type of alteration in document. It can be classified as chemical erasure and physical erasure.
Generics were introduced to the Java language to provide tighter type checks at compile time and to support generic programming. To implement generics, the Java compiler applies type erasure to: Replace all type parameters in generic types with their bounds or Object if the type parameters are unbounded.
Type-erasure simply means "erasing" a specific type to a more abstract type in order to do something with the abstract type (like having an array of that abstract type).
The signature of fooMethod(Class<?>)
is the same as the signature of fooMethod(Class)
after erasure, since the erasure of Class<?>
is simply Class
(JLS 4.6). Hence, fooMethod(Class)
is a subsignature of the fooMethod(Class<?>)
but not the opposite (JLS 8.4.2).
For overriding with instance methods you need the overriding method to be a subsignature of the overridden method (JLS 8.4.8.1). This is clearly not the case here.
Now that we have established the fact that your subclass method doesn't override the superclass method according to the JLS, let's look at the runtime implications when type erasure has occured. We now have two methods that look exactly the 'same' (same name, same parameter types) but do not override each other. If they don't override, they must be both available on the subtype as separate methods, but they have identical runtime signatures: conflict. So Java has to disallow it.
Overriding generic parameter types using raw parameter types is allowed because raw types exist just for this reason: they are a convenient mechanism with specific unsound type rules to accommodate interaction with legacy code. So the type system here will decide that the subclass method does override the superclass one, they are identical after type erasure and we can never have a conflict. As a consequence of this libraries can be generified independently of existing non-generic code.
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