使用Java 8中的供应商填充具有通用列表的数组会抛出类型擦除的ClassCastEx b/c

Question

我想使用Supplier和Stream.generate将带有通用列表的数组填充为元素.

看起来像这样:

    Supplier<List<Object>> supplier = () -> new ArrayList<Object>();
    List<Object>[] test = (List<Object>[]) Stream.generate(supplier).limit(m).toArray();

错误输出为:

Exception in thread "main" java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to [Ljava.util.List;

现在,如何使用Java 8提供的技术填充具有泛型类型的数组？或者这是不可能的(还),我必须以"经典"的方式做到这一点？

此致,Claas M.

编辑

根据@ Water的要求,我使用stream.collect(使用Cast测试数组)和传统的迭代方法对填充数组/列表进行了一些性能测试.

首先使用列表进行性能测试:

private static int m = 100000;

/**
 * Tests which way is faster for LISTS.
 * Results:
 * 1k Elements: about the same time (~5ms)
 * 10k Elements: about the same time (~8ms)
 * 100k Elements: new way about 1.5x as fast (~18ms vs ~27ms)
 * 1M Elements: new way about 2x as fast (~30ms vs ~60ms)
 * NOW THIS IS INTERESTING:
 * 10M Elements: new way about .1x as fast (~5000ms vs ~500ms)
 * (100M OutOfMemory after ~40Sec)
 * @param args
 */

public static void main(String[] args) {

    Supplier<String> supplier = () -> new String();
    long startTime,endTime;

    //The "new" way
    startTime = System.currentTimeMillis();
    List<String> test1 =  Stream.generate(supplier).limit(m ).collect(Collectors.toList());
    endTime = System.currentTimeMillis();
    System.out.println(endTime - startTime);


    //The "old" way
    startTime = System.currentTimeMillis();
    List<String> test2 = new ArrayList();
    Iterator<String> i = Stream.generate(supplier).limit(m).iterator();
    while (i.hasNext()) {
        test2.add(i.next());
    }
    endTime = System.currentTimeMillis();
    System.out.println(endTime - startTime);


}

第二个使用数组的性能测试:

    private static int m = 100000000;

    /**
     * Tests which way is faster for ARRAYS.
     * Results:
     * 1k Elements: old way much faster (~1ms vs ~6ms)
     * 10k Elements: old way much faster (~2ms vs ~7ms)
     * 100k Elements: old way about 2x as fast (~7ms vs ~14ms)
     * 1M Elements: old way a bit faster (~50ms vs ~60ms)
     * 10M Elements: old way a bit faster (~5s vs ~6s)
     * 100M Elements: Aborted after about 5 Minutes of 100% CPU Utilisation on an i7-2600k
     * @param args
     */

    public static void main(String[] args) {

        Supplier<String> supplier = () -> new String();
        long startTime,endTime;

        //The "new" way
        startTime = System.currentTimeMillis();
        String[] test1 =  (String[]) Stream.generate(supplier).limit(m ).collect(Collectors.toList()).toArray(new String[m]);
        endTime = System.currentTimeMillis();
        System.out.println(endTime - startTime);


        //The "old" way
        startTime = System.currentTimeMillis();
        String[] test2 = new String[m];
        Iterator<String> it = Stream.generate(supplier).iterator();
        for(int i = 0; i < m; i++){
            test2[i] = it.next();
        }
        endTime = System.currentTimeMillis();
        System.out.println(endTime - startTime);


    }

}

正如你所看到的,Water确实是对的 - Cast让它变慢了.但对于Lists,新方法更快; 至少从100k-1M元素.我还是不知道为什么它对于10M Elements这么慢,我真的很想听到一些评论.

Answer 1

Stream 生成器仍然生成您想要的对象，唯一的问题是调用 toArray() 会给您返回一个对象数组，并且您不能从对象数组向下转换为子对象数组（因为您已经得到了类似的东西：对象[] { ArrayList, ArrayList })。

以下是正在发生的情况的示例：

你认为你有这个：

    String[] hi = { "hi" };
    Object[] test = (Object[]) hi; // It's still a String[]
    String[] out = (String[]) test;
    System.out.println(out[0]); // Prints 'hi'

但你实际上有：

    String[] hi = { "hi" };
    Object[] test = new Object[1]; // This is not a String[]
    test[0] = hi[0];
    String[] out = (String[]) test; // Cannot downcast, throws an exception.
    System.out.println(out[0]);

您将返回上面的直接块，这就是您收到转换错误的原因。

有几种方法可以解决这个问题。如果你想查看你的列表，你可以轻松地用它们创建一个数组。

    Supplier<List<Integer>> supplier = () -> { 
        ArrayList<Integer> a = new ArrayList<Integer>();
        a.add(5);
        a.add(8);
        return a;
    };

    Iterator<List<Integer>> i = Stream.generate(supplier).limit(3).iterator();

    // This shows there are elements you can do stuff with.
    while (i.hasNext()) {
        List<Integer> list = i.next();
        // You could add them to your list here.
        System.out.println(list.size() + " elements, [0] = " + list.get(0));
    }

如果您决定处理该函数，您可以执行以下操作：

    Supplier<List<Integer>> supplier = () -> { 
        ArrayList<Integer> a = new ArrayList<Integer>();
        a.add(5);
        a.add(8);
        return a;
    };

    Object[] objArr = Stream.generate(supplier).limit(3).toArray();
    for (Object o : objArr) {
        ArrayList<Integer> arrList = (ArrayList<Integer>) o; // This is not safe to do, compiler can't know this is safe.
        System.out.println(arrList.get(0)); 
    }

根据Stream Javadocs，如果您想将其转换为数组，可以使用其他 toArray() 方法，但我还没有探索过这个函数，所以我不想讨论我不知道的东西。