将 Enumerable.Empty<>() 转换为另一个实现 IEnumerable 的 class 会返回 null

Question

I have a class that implements IEnumerable:

public class A { }
public class B : IEnumerable<A> { } 

How can i use class B as Enumerable.Empty<A>() in this case? I mean that casting like this Enumerable.Empty<A>() as B returns null. Why it does? Should i implement any specific constructor or a method? Or is it an forbidden operation and I should to do it in the other way?

Answer 1

Enumerable.Empty<T>() is implemented as :

internal class EmptyEnumerable<T>
{
    public static readonly T[] Instance = new T[0];
}

public static IEnumerable<T> Empty<T>()
{
    return EmptyEnumerable<T>.Instance;
}

If you remove the optimization to cache the empty array, you could rewrite this as:

public static IEnumerable<T> Empty<T>()
{
    return new T[0];
}

So Enumerable.Empty<T>() just returns an empty array of type T .

You wouldn't write:

B b = new A[0];

This doesn't make sense: a B isn't an array of instances of A .

For the same reason, you can't write:

B b = Enumerable.Empty<A>();