答案已经给出了，但如果你已经有一个T的实例，那么你可以这样做:

T t; //Assuming you already have this object instantiated or given by parameter.
int length;
T[] ts = (T[]) Array.newInstance(t.getClass(), length);

希望，我能帮忙， Ferdi265

Oracle教程:

You cannot create arrays of parameterized types. For example, the following code does not compile: List<Integer>[] arrayOfLists = new List<Integer>[2]; // compile-time error The following code illustrates what happens when different types are inserted into an array: Object[] strings = new String[2]; strings[0] = "hi"; // OK strings[1] = 100; // An ArrayStoreException is thrown. If you try the same thing with a generic list, there would be a problem: Object[] stringLists = new List<String>[]; // compiler error, but pretend it's allowed stringLists[0] = new ArrayList<String>(); // OK stringLists[1] = new ArrayList<Integer>(); // An ArrayStoreException should be thrown, // but the runtime can't detect it. If arrays of parameterized lists were allowed, the previous code would fail to throw the desired ArrayStoreException.

对我来说，这听起来很软弱。我认为任何对泛型有充分理解的人，都完全可以理解，甚至期望，在这种情况下ArrayStoredException不会被抛出。

引用:

Arrays of generic types are not allowed because they're not sound. The problem is due to the interaction of Java arrays, which are not statically sound but are dynamically checked, with generics, which are statically sound and not dynamically checked. Here is how you could exploit the loophole: class Box<T> { final T x; Box(T x) { this.x = x; } } class Loophole { public static void main(String[] args) { Box<String>[] bsa = new Box<String>[3]; Object[] oa = bsa; oa[0] = new Box<Integer>(3); // error not caught by array store check String s = bsa[0].x; // BOOM! } } We had proposed to resolve this problem using statically safe arrays (aka Variance) bute that was rejected for Tiger. -- gafter

(我相信是尼尔·盖特，但不确定)

在这里查看上下文:http://forums.sun.com/thread.jspa?threadID=457033&forumID=316

这是因为泛型是在他们创建后才添加到java的，所以它有点笨拙，因为java的原始创造者认为在创建数组时，类型应该在创建时指定。这对泛型不起作用，所以你必须这么做 E[] array=(E[]) new Object[15]; 这将编译，但会给出警告。

主要原因是Java中的数组是协变的。

这里有一个很好的概述。

在我的例子中，我只是想要一个堆栈数组，就像这样:

Stack<SomeType>[] stacks = new Stack<SomeType>[2];

由于这是不可能的，我使用了以下作为解决方案:

创建一个非泛型包装器类围绕堆栈(例如MyStack) MyStack[] stacks = new MyStack[2]工作得很好

丑陋，但Java是快乐的。

注意:正如BrainSlugs83在问题的评论中提到的，在.NET中完全可以有泛型数组