有可能在Java中创建泛型类型的实例吗?我在想,根据我所看到的,答案是否定的(由于类型擦除),但如果有人能看到我遗漏的东西,我会很感兴趣:
class SomeContainer<E>
{
E createContents()
{
return what???
}
}
编辑:事实证明,超级类型令牌可以用来解决我的问题,但它需要大量基于反射的代码,如下面的一些答案所示。
我将把这个问题放一段时间,看看是否有人提出了与Ian Robertson的Artima文章截然不同的东西。
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).