java 泛型的问题

Question

我想写一个方法list->array 想利用泛型，代码如下

public static <T> T[] list2Array(List<T> list){
    T[] array = (T[])list.toArray(new T[list.size()]);
    return array;
}

也就是传过来的String就返回String数组 Integer就返回Integer数组。但是这个new T[list.size]这里有问题，请问怎么解决？谢谢。

Answer 1

這個Effective Java第二版裡有說到, Item25.

Because of these fundamental differences, arrays and generics do not
mix well. For example, it is illegal to create an array of a generic
type, a parameterized type, or a type parameter. None of these array
creation expressions are legal: new List[], new List[], new
E[]. All will result in generic array creation errors at compile time.

Generally speaking, arrays and generics don’t mix well. If you find
yourself mixing them and getting compile-time errors or warnings, your
first impulse should be to replace the arrays with lists.

1. 因為數組和泛型不對付, 所以在不做強制類型轉換的情況下, 我們是沒有辦法得到一個泛型數組的實例的, 編譯器會限制此類情況發生(new E[], list .toArray());
2. 為什麼陣列和泛型不對付, Effective Java裡面有例子. 根本原因在於:

Arrays are covariant. This scary-sounding word means simply that if
Sub is a subtype of Super, then the array type Sub[] is a subtype of
Super[].

給出例子說明我的理解, 類似的程式碼, 在泛型集合下, 會在靜態編譯階段報錯; 而泛型數組在運行階段給出ArrayStoreException:

private <T> void doSomethingToGenArr(T[] tArr){
    Object[] oArr = tArr;
    oArr[0] = new String("aaa");
}

private <T> void doSomethingToGenList(List<T> list){
    List<Object> l1 =  list; /* compile error */
    l1.set(0, new String("aaa"));
}

@Test
public void test111(){
    doSomethingToGenArr(new Integer[2]);
}

@Test
public void test112(){
    doSomethingToGenList(new ArrayList<Integer>());
}

Answer 2

沒有泛型數組，這一說

Answer 3

toArray()方法有兩種,帶參和不帶參.
帶參的情況下,參數就是一個泛型數組,如T[] a,相當於你手動構造一個數組,傳入方法中.
toArray()內部是一個copy的實現.

舉兩個例子.
1:String[][] str_a = (String [][]) arr.toArray(new String[0][0]);

2 :String[][] a = new String[<size>][size];
String [][] str_a = (String [][]) arr.toArray(a);

當然 要保證轉型成功,不然會引發ClassCastException.

--------------------------分割線------------------- ------

以下是原始碼中是實作,其實就是一個copy操作.

 /**
     * Returns an array containing all of the elements in this list
     * in proper sequence (from first to last element).
     *
     * <p>The returned array will be "safe" in that no references to it are
     * maintained by this list.  (In other words, this method must allocate
     * a new array).  The caller is thus free to modify the returned array.
     *
     * <p>This method acts as bridge between array-based and collection-based
     * APIs.
     *
     * @return an array containing all of the elements in this list in
     *         proper sequence
     */
    public Object[] toArray() {
        return Arrays.copyOf(elementData, size);
    }