Why cannot IEnumerable<struct> be cast as IEnumerable<object>?

Question

Why is the last line not allowed?

IEnumerable<double> doubleenumerable = new List<double> { 1, 2 }; IEnumerable<string> stringenumerable = new List<string> { "a", "b" }; IEnumerable<object> objects1 = stringenumerable; // OK IEnumerable<object> objects2 = doubleenumerable; // Not allowed 

Is this because double is a value type that doesn't derive from object, hence the covariance doesn't work?

Does that mean that there is no way to make this work:

public interface IMyInterface<out T> {     string Method();  }  public class MyClass<U> : IMyInterface<U> {     public string Method()     {         return "test";     } }  public class Test {     public static object test2()      {         IMyInterface<double> a = new MyClass<double>();         IMyInterface<object> b = a; // Invalid cast!         return b.Method();     } } 

And that I need to write my very own IMyInterface<T>.Cast<U>() to do that?

Answer 1

Why is the last line not allowed?

Because double is a value type and object is a reference type; covariance only works when both types are reference types.

Is this because double is a value type that doesn't derive from object, hence the covariance doesn't work?

No. Double does derive from object. All value types derive from object.

Now the question you should have asked:

Why does covariance not work to convert IEnumerable<double> to IEnumerable<object>?

Because who does the boxing? A conversion from double to object must box the double. Suppose you have a call to IEnumerator<object>.Current that is "really" a call to an implementation of IEnumerator<double>.Current. The caller expects an object to be returned. The callee returns a double. Where is the code that does the boxing instruction that turns the double returned by IEnumerator<double>.Current into a boxed double?

It is nowhere, that's where, and that's why this conversion is illegal. The call to Current is going to put an eight-byte double on the evaluation stack, and the consumer is going to expect a four-byte reference to a boxed double on the evaluation stack, and so the consumer is going to crash and die horribly with an misaligned stack and a reference to invalid memory.

If you want the code that boxes to execute then it has to be written at some point, and you're the person who gets to write it. The easiest way is to use the Cast<T> extension method:

IEnumerable<object> objects2 = doubleenumerable.Cast<object>(); 

Now you call a helper method that contains the boxing instruction that converts the double from an eight-byte double to a reference.

UPDATE: A commenter notes that I have begged the question -- that is, I have answered a question by presupposing the existence of a mechanism which solves a problem every bit as hard as a solution to the original question requires. How does the implementation of Cast<T> manage to solve the problem of knowing whether to box or not?

It works like this sketch. Note that the parameter types are not generic:

public static IEnumerable<T> Cast<T>(this IEnumerable sequence)  {     if (sequence == null) throw ...     if (sequence is IEnumerable<T>)          return sequence as IEnumerable<T>;     return ReallyCast<T>(sequence); }  private static IEnumerable<T> ReallyCast<T>(IEnumerable sequence) {     foreach(object item in sequence)         yield return (T)item; } 

The responsibility for determining whether the cast from object to T is an unboxing conversion or a reference conversion is deferred to the runtime. The jitter knows whether T is a reference type or a value type. 99% of the time it will of course be a reference type.