Java Generics Question

Question

I have a class

public abstract class FakeClass<T extends MyClass> {
    protected HashMap<Character, T> myMap;

    private void myMethod(){
        myMap.put('c', /*???? I need to instatiate something of type T here.*/)
    }
}

As you can see, I can't figure out how to instantiate something of type T. Is this possible? If so can someone point me in the right direction to do it?

Thanks!

@Mr. MacGyver That probably isn't going to help because `MyClass` isn't always going to match `T extends MyClass`. What would be helpful is some context as to what that map is going to be used for. — trutheality, Jul 25 '11 at 00:26

Ted Hopp · Accepted Answer · 2017-09-18T01:27:44.240

7

This can only be done by passing some information about T to myMethod(). One approach, described in this thread, is to pass a factory object for T. A special case of this, described in this thread, is to pass a Class<T> type argument to myMethod(), which can then use type.newInstance() to create a new T object using the default constructor. (This will fail if T does not have a default constructor.) The Class object serves as the factory object for T.

The reason we need all this is due to type erasure in Java.

edited Sep 18 '17 at 01:27

answered Jul 25 '11 at 00:25

Ted Hopp

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It is so unfortunate that Java generics were designed/implemented like this. – luiscubal Jul 25 '11 at 00:30
How does type erasure play into this? I'd say the reason we need all this is that constructors don't get inherited in the same way that other methods are. – trutheality Jul 25 '11 at 00:32
@trutheality If there was no type erasure, the JVM(rather than just the Java compiler) would know what T really is, so "new T()" would be meaningful. – luiscubal Jul 25 '11 at 00:34
@luiscubal: Generics have very little to do with this. You'd have the same problem without generics (and the same solution) if you wanted to create an "instance" of an interface. – trutheality Jul 25 '11 at 00:34
@luiscubal: Calling `new T()` is a compiler error if `T` is a class with a private default constructor. So `new T()` should be illegal no matter how you twist it. Unless you resolve all the types at compile-time a-la C++ templates. – trutheality Jul 25 '11 at 00:36
@trutheality - In C#, there is no type erasure, so (with some extra tricks such as explicitly indicating that T must have a public default constructor) "new T()" is valid. So, if it wasn't for type erasure, it would be trivial to add support for this in Java. – luiscubal Jul 25 '11 at 00:38
@luiscubal The problem is the "explicitly indicating that T must have a public default constructor" part. Not type erasure. – trutheality Jul 25 '11 at 00:43
1

@trutheality - I agree that no type erasure is a necessary but not sufficient condition for "new T()" to be valid, if that's what you meant. It is "necessary" because even if T had a default constructor, it's meaningless if the JVM doesn't know what T is to begin with. – luiscubal Jul 25 '11 at 00:44
@luiscubal let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/1788/discussion-between-trutheality-and-luiscubal) – trutheality Jul 25 '11 at 00:50

score 0 · Answer 2 · answered Jul 25 '11 at 00:19

Usually, the point of generics is to be able to accept an unknown-at-compile-time type as input.

You can't instantiate an unknown class (because it might not have a visible constructor). If you just need a placeholder to put into the map, you can put a null value.

Without knowing more context, there isn't much more that anyone can do.

Java Generics Question

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