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java 泛型的问题

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。

天蓬老师天蓬老师2788 天前582

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  • PHP中文网

    PHP中文网2017-04-17 11:24:00

    这个Effective Java第二版里有说到, 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. 因为数组和泛型不对付, 所以在不做强制类型转换的情况下, 我们是没有办法得到一个泛型数组的实例的, 编译器会限制此类情况发生(new E[], list.toArray());
    2. 为什么数组和泛型不对付, Effective Java里有例子. 根本原因在于:

    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[].

    给出例子说明我的理解, 类似的代码, 在泛型集合下, 会在静态编译阶段报错; 而泛型数组在运行阶段给出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>());
    }
    

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  • 伊谢尔伦

    伊谢尔伦2017-04-17 11:24:00

    没有泛型数组,这一说

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  • 天蓬老师

    天蓬老师2017-04-17 11:24:00

    toArray()方法有两种,带参和不带参.
    带参的情况下,参数就是一个一个泛型数组,如T[] a,相当于你手动构造一个数组,传入方法中.
    toArray()内部是一个copy的实现.

    举两个例子.
    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);

    当然 要保证转型成功,不然会引发ClassCastException.

    --------------------------分割线-------------------------

    以下是源码中是实现,实际上就是一个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);
        }
    

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