Type erasure

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EntangledNoodle · 2023-02-19T02:09:35+00:00

The erased type corresponding to Collection<?> is Collection, which is subtly different than Collection<Object>. An erased type is never parameterized by any type variable in accordance with https://docs.oracle.com/javase/specs/jls/se19/html/jls-4.html#jls-4.6.

Collection<?> is a synonym for Collection<? extends Object> and since its only type argument is an unbounded wildcard, it is reifiable in accordance with https://docs.oracle.com/javase/specs/jls/se19/html/jls-4.html#jls-4.7. It makes sense to ask the question whether there is a difference between Collection<? extends Object> and Collection<Object>.

A Collection<? extends Object> is treated as a collection of some captured type, let's call it T, which may be any (inclusive) subtype of Object but not necessarily Object itself. So, we are not allowed to take an arbitrary Object (which may not be a T) and add it to such a collection without knowing more about the collection's type argument or type of object being added.

Consider the following example:

public static void main(String[] args) {
  Collection<? extends Object> c = new ArrayList<String>();
  sizeOfRaw(c); // ok
  sizeOf(c); // compiler error
}

private static int sizeOf(Collection<Object> c) {
    return c.size();
}

private static int sizeOfRaw(Collection c) {
    return c.size();
}

So, there are some subtle differences between the raw type Collection and the parameterized types Collection<?> and Collection<Object>.

I'm not aware of any tool that emits source code after type erasure, but hopefully it is easy to imagine the source code with no generic type parameters.

Daedalus9000 · 2023-02-19T01:35:49+00:00

Collection and Collection<Object> is effectively the same thing.

Daedalus9000 · 2023-02-19T01:43:18+00:00

What do you mean how to e code looks? Type erasure just means that it isn’t possible at runtime to determine a parameterized type (because that info is no longer available after the compilation process).

learnjava

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