Create instance of generic type in Java?

Question

Is it possible to create an instance of a generic type in Java? I'm thinking based on what I've seen that the answer is no (due to type erasure), but I'd be interested if anyone can see something I'm missing:

class SomeContainer<E>
{
    E createContents()
    {
        return what???
    }
}

EDIT: It turns out that Super Type Tokens could be used to resolve my issue, but it requires a lot of reflection-based code, as some of the answers below have indicated.

I'll leave this open for a little while to see if anyone comes up with anything dramatically different than Ian Robertson's Artima Article.

Answer 1

In Java 8 you can use the Supplier functional interface to achieve this pretty easily:

class SomeContainer<E> {
  private Supplier<E> supplier;

  SomeContainer(Supplier<E> supplier) {
    this.supplier = supplier;
  }

  E createContents() {
    return supplier.get();
  }
}

You would construct this class like this:

SomeContainer<String> stringContainer = new SomeContainer<>(String::new);

The syntax String::new on that line is a constructor reference.

If your constructor takes arguments you can use a lambda expression instead:

SomeContainer<BigInteger> bigIntegerContainer
    = new SomeContainer<>(() -> new BigInteger(1));

Answer 2

As you said, you can't really do it because of type erasure. You can sort of do it using reflection, but it requires a lot of code and lot of error handling.

Answer 3

If you mean new E() then it is impossible. And I would add that it is not always correct - how do you know if E has public no-args constructor? But you can always delegate creation to some other class that knows how to create an instance - it can be Class<E> or your custom code like this

interface Factory<E>{
    E create();
}    

class IntegerFactory implements Factory<Integer>{    
  private static int i = 0; 
  Integer create() {        
    return i++;    
  }
}

Answer 4

return   (E)((Class)((ParameterizedType)this.getClass().getGenericSuperclass()).getActualTypeArguments()[0]).newInstance();

Answer 5

Here is an improved solution, based on ParameterizedType.getActualTypeArguments, already mentioned by @noah, @Lars Bohl, and some others.

First small improvement in the implementation. Factory should not return instance, but a type. As soon as you return instance using Class.newInstance() you reduce a scope of usage. Because only no-arguments constructors can be invoke like this. A better way is to return a type, and allow a client to choose, which constructor he wants to invoke:

public class TypeReference<T> {
  public Class<T> type(){
    try {
      ParameterizedType pt = (ParameterizedType) this.getClass().getGenericSuperclass();
      if (pt.getActualTypeArguments() == null || pt.getActualTypeArguments().length == 0){
        throw new IllegalStateException("Could not define type");
      }
      if (pt.getActualTypeArguments().length != 1){
        throw new IllegalStateException("More than one type has been found");
      }
      Type type = pt.getActualTypeArguments()[0];
      String typeAsString = type.getTypeName();
      return (Class<T>) Class.forName(typeAsString);

    } catch (Exception e){
      throw new IllegalStateException("Could not identify type", e);
    }

  }
}

Here is a usage examples. @Lars Bohl has shown only a signe way to get reified geneneric via extension. @noah only via creating an instance with {}. Here are tests to demonstrate both cases:

import java.lang.reflect.Constructor;

public class TypeReferenceTest {

  private static final String NAME = "Peter";

  private static class Person{
    final String name;

    Person(String name) {
      this.name = name;
    }
  }

  @Test
  public void erased() {
    TypeReference<Person> p = new TypeReference<>();
    Assert.assertNotNull(p);
    try {
      p.type();
      Assert.fail();
    } catch (Exception e){
      Assert.assertEquals("Could not identify type", e.getMessage());
    }
  }

  @Test
  public void reified() throws Exception {
    TypeReference<Person> p = new TypeReference<Person>(){};
    Assert.assertNotNull(p);
    Assert.assertEquals(Person.class.getName(), p.type().getName());
    Constructor ctor = p.type().getDeclaredConstructor(NAME.getClass());
    Assert.assertNotNull(ctor);
    Person person = (Person) ctor.newInstance(NAME);
    Assert.assertEquals(NAME, person.name);
  }

  static class TypeReferencePerson extends TypeReference<Person>{}

  @Test
  public void reifiedExtenension() throws Exception {
    TypeReference<Person> p = new TypeReferencePerson();
    Assert.assertNotNull(p);
    Assert.assertEquals(Person.class.getName(), p.type().getName());
    Constructor ctor = p.type().getDeclaredConstructor(NAME.getClass());
    Assert.assertNotNull(ctor);
    Person person = (Person) ctor.newInstance(NAME);
    Assert.assertEquals(NAME, person.name);
  }
}

Note: you can force the clients of TypeReference always use {} when instance is created by making this class abstract: public abstract class TypeReference<T>. I've not done it, only to show erased test case.

Answer 6

You can with a classloader and the class name, eventually some parameters.

final ClassLoader classLoader = ...
final Class<?> aClass = classLoader.loadClass("java.lang.Integer");
final Constructor<?> constructor = aClass.getConstructor(int.class);
final Object o = constructor.newInstance(123);
System.out.println("o = " + o);