为什么收益率与常规可枚举的行为不同？

Question

虽然我正在制作一个逻辑来迭代两个不同类型的枚举,通过比较,我发现了这个:

class Program
{
    public static IEnumerable<mm> YieldlyGet()
    {
        yield return new mm { Int = 0 };
        yield return new mm { Int = 1 };
        yield return new mm { Int = 2 };
        yield return new mm { Int = 3 };
        yield return new mm { Int = 4 };
        yield return new mm { Int = 5 };
    }

    public static IEnumerable<int> YieldlyGetInt()
    {
        yield return 0;
        yield return 1;
        yield return 2;
        yield return 3;
        yield return 4;
        yield return 5;
    }

    public static IEnumerable<int> Get() 
    {
        return new List<int> { 0, 1,2,3,4,5 };
    }

    static void Main(string[] args) 
    {
        var yieldr = YieldlyGet().GetEnumerator();
        var yieldv = YieldlyGetInt().GetEnumerator();

        var list = Get().GetEnumerator();

        int i = -1;
        Console.WriteLine("For the current index: {0}", ++i);
        Console.WriteLine("y-r: Should I move next? {0}, if yes, value: {1}", yieldr.MoveNext(), yieldr.Current != null ? yieldr.Current.Int : 0);
        Console.WriteLine("y-v: Should I move next? {0}, if yes, value: {1}", yieldv.MoveNext(), yieldv.Current);
        Console.WriteLine("l: Should I move next? {0}, if yes, value: {1}", list.MoveNext(), list.Current);

        Console.WriteLine("For the current index: {0}", ++i);
        Console.WriteLine("y-r: Should I move next? {0}, if yes, value: {1}", yieldr.MoveNext(), yieldr.Current != null ? yieldr.Current.Int : 0);
        Console.WriteLine("y-v: Should I move next? {0}, if yes, value: {1}", yieldv.MoveNext(), yieldv.Current);
        Console.WriteLine("l: Should I move next? {0}, if yes, value: {1}", list.MoveNext(), list.Current);

        Console.WriteLine("For the current index: {0}", ++i);
        Console.WriteLine("y-r: Should I move next? {0}, if yes, value: {1}", yieldr.MoveNext(), yieldr.Current != null ? yieldr.Current.Int : 0);
        Console.WriteLine("y-v: Should I move next? {0}, if yes, value: {1}", yieldv.MoveNext(), yieldv.Current);
        Console.WriteLine("l: Should I move next? {0}, if yes, value: {1}", list.MoveNext(), list.Current);

        Console.WriteLine("For the current index: {0}", ++i);
        Console.WriteLine("y-r: Should I move next? {0}, if yes, value: {1}", yieldr.MoveNext(), yieldr.Current != null ? yieldr.Current.Int : 0);
        Console.WriteLine("y-v: Should I move next? {0}, if yes, value: {1}", yieldv.MoveNext(), yieldv.Current);
        Console.WriteLine("l: Should I move next? {0}, if yes, value: {1}", list.MoveNext(), list.Current);

        Console.WriteLine("For the current index: {0}", ++i);
        Console.WriteLine("y-r: Should I move next? {0}, if yes, value: {1}", yieldr.MoveNext(), yieldr.Current != null ? yieldr.Current.Int : 0);
        Console.WriteLine("y-v: Should I move next? {0}, if yes, value: {1}", yieldv.MoveNext(), yieldv.Current);
        Console.WriteLine("l: Should I move next? {0}, if yes, value: {1}", list.MoveNext(), list.Current);

        Console.WriteLine("For the current index: {0}", ++i);
        Console.WriteLine("y-r: Should I move next? {0}, if yes, value: {1}", yieldr.MoveNext(), yieldr.Current != null ? yieldr.Current.Int : 0);
        Console.WriteLine("y-v: Should I move next? {0}, if yes, value: {1}", yieldv.MoveNext(), yieldv.Current);
        Console.WriteLine("l: Should I move next? {0}, if yes, value: {1}", list.MoveNext(), list.Current);

        Console.WriteLine("For the current index: {0}", ++i);
        Console.WriteLine("y-r: Should I move next? {0}, if yes, value: {1}", yieldr.MoveNext(), yieldr.Current != null ? yieldr.Current.Int : 0);
        Console.WriteLine("y-v: Should I move next? {0}, if yes, value: {1}", yieldv.MoveNext(), yieldv.Current);
        Console.WriteLine("l: Should I move next? {0}, if yes, value: {1}", list.MoveNext(), list.Current);


        Console.ReadLine();
}

问题在于,当我在最后一个位置之后,列表向我显示默认值,而yield创建的Iterator,继续显示最后一个值.

对于当前指数:6
年:我应该下一步吗？假的,如果是,值:5
YV:我应该搬到下一个？假的,如果是,值:5
L:我应该搬到下一个？如果是,则为假,值:0

为什么？

Answer 1

根据该IEnumerator<T>.Current属性的MSDN文档:

Current未定义[when]:最后一次MoveNext返回的调用false,表示集合的结束.

这意味着枚举器的底层实现可以自由地从Current一次MoveNext返回返回任意值false.它可以是0,6,-1,2147483647,或者只是随机选取的值; 无论如何,你不应该使用它.