您必须提供 the非通用GetEnumerator（）方法，并包含非通用名称空间：

You must provide an explicit implementation of the non-generic GetEnumerator() method and include the non-generic namespace:

using namespace System::Collections;

....

virtual IEnumerator^ EnumerableGetEnumerator() = IEnumerable::GetEnumerator
{
    return GetEnumerator<MyClass^>();
}

更新：如评论中所述， GetEnumerator的显式版本必须命名为不同名称以避免名称冲突，因此我将其命名为EnumerableGetEnumerator。

Update: As mentioned in the comments, the explicit version of GetEnumerator must be named different to avoid name ***, thus I've named it EnumerableGetEnumerator.

类似地，在C＃中，您必须这样做： / p>

Similarly, in C# you would have to do it like this:

using System.Collections.Generic;

public class MyCollection : IEnumerable<MyClass>
{
    public MyCollection()
    {
    }  

    public IEnumerator<MyClass> GetEnumerator()
    {
        return null;
    }

    IEnumerator IEnumerable.GetEnumerator()
    {
        return GetEnumerator<MyClass>();
    }
}