考虑一种更函数化的方法:与其凭空创建一些E(这显然是一种代码味道)，不如传递一个知道如何创建E的函数。

E createContents(Callable<E> makeone) {
     return makeone.call(); // most simple case clearly not that useful
}

这是我想到的一个选择，可能会有帮助:

public static class Container<E> {
    private Class<E> clazz;

    public Container(Class<E> clazz) {
        this.clazz = clazz;
    }

    public E createContents() throws Exception {
        return clazz.newInstance();
    }
}

EDIT:你也可以使用这个构造函数(但它需要一个E的实例):

@SuppressWarnings("unchecked")
public Container(E instance) {
    this.clazz = (Class<E>) instance.getClass();
}

在Java 8中，你可以使用Supplier函数接口很容易地实现这一点:

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

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

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

你可以这样构造这个类:

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

这一行上的String::new语法是一个构造函数引用。

如果你的构造函数接受参数，你可以使用lambda表达式:

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

如果你在泛型类中需要一个类型参数的新实例，那么让你的构造函数要求它的类…

public final class Foo<T> {

    private Class<T> typeArgumentClass;

    public Foo(Class<T> typeArgumentClass) {

        this.typeArgumentClass = typeArgumentClass;
    }

    public void doSomethingThatRequiresNewT() throws Exception {

        T myNewT = typeArgumentClass.newInstance();
        ...
    }
}

用法:

Foo<Bar> barFoo = new Foo<Bar>(Bar.class);
Foo<Etc> etcFoo = new Foo<Etc>(Etc.class);

优点:

比Robertson的超级类型令牌(STT)方法简单得多(而且问题更少)。 比STT方法更有效(STT方法会把你的手机当早餐吃)。

缺点:

Can't pass Class to a default constructor (which is why Foo is final). If you really do need a default constructor you can always add a setter method but then you must remember to give her a call later. Robertson's objection... More Bars than a black sheep (although specifying the type argument class one more time won't exactly kill you). And contrary to Robertson's claims this does not violate the DRY principal anyway because the compiler will ensure type correctness. Not entirely Foo<L>proof. For starters... newInstance() will throw a wobbler if the type argument class does not have a default constructor. This does apply to all known solutions though anyway. Lacks the total encapsulation of the STT approach. Not a big deal though (considering the outrageous performance overhead of STT).

摘自Java教程-泛型的限制:

不能创建类型参数的实例

不能创建类型参数的实例。例如，下面的代码会导致编译时错误:

public static <E> void append(List<E> list) {
    E elem = new E();  // compile-time error
    list.add(elem);
}

作为一种变通方法，你可以通过反射创建一个类型参数的对象:

public static <E> void append(List<E> list, Class<E> cls) throws Exception {
    E elem = cls.getDeclaredConstructor().newInstance();   // OK
    list.add(elem);
}

你可以像下面这样调用append方法:

List<String> ls = new ArrayList<>();
append(ls, String.class);

下面是基于ParameterizedType的改进解决方案。getActualTypeArguments， @noah、@Lars Bohl和其他一些人已经提到过。

首先是实施上的小改进。Factory不应该返回实例，而是返回类型。一旦使用Class.newInstance()返回实例，就减少了使用范围。因为只有无参数构造函数可以像这样调用。更好的方法是返回一个类型，并允许客户端选择他想调用的构造函数:

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);
    }

  }
}

下面是一个用法示例。@Lars Bohl只展示了一种通过扩展具体化通用的方式。@noah只能通过使用{}创建实例。下面是演示这两种情况的测试:

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);
  }
}

注意:你可以强制TypeReference的客户端在创建实例时总是使用{}，方法是让这个类成为抽象类:我没有这样做，只是为了显示被擦除的测试用例。