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
– trutheality, Commented Jul 25, 2011 at 0:26

Ted Hopp · Accepted Answer · 2017-09-18 01:27:44Z

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, 2017 at 1:27

answered Jul 25, 2011 at 0:25

Ted Hopp

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9 Comments

luiscubal Over a year ago

It is so unfortunate that Java generics were designed/implemented like this.

trutheality Over a year ago

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.

luiscubal Over a year ago

@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.

trutheality Over a year ago

@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.

luiscubal Over a year ago

@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.

|

trutheality · Accepted Answer · 2011-07-25 00:19:23Z

0

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.

answered Jul 25, 2011 at 0:19

trutheality

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