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

这是因为Java的数组(与泛型不同)在运行时包含关于其组件类型的信息。因此，在创建数组时必须知道组件类型。因为您不知道T在运行时是什么，所以您不能创建数组。

类可以声明类型为T[]的数组，但它不能直接实例化这样的数组。相反，一种常见的方法是实例化一个Object[]类型的数组，然后对类型T[]进行窄化转换，如下所示:

  public class Portfolio<T> {
  T[] data;
 public Portfolio(int capacity) {
   data = new T[capacity];                 // illegal; compiler error
   data = (T[]) new Object[capacity];      // legal, but compiler warning
 }
 public T get(int index) { return data[index]; }
 public void set(int index, T element) { data[index] = element; }
}

试试这个:

List<?>[] arrayOfLists = new List<?>[4];

T vals [];/ /好吧

但是，你不能实例化T的数组 // vals = new T[10];//不能创建T的数组

你不能创建T的数组的原因是没有办法 编译器来了解实际创建的数组类型。

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