C# variance problem: Assigning List<Derived> as List<Base>

AndiDog picture AndiDog · Jan 9, 2010 · Viewed 17.2k times · Source

Look at the following example (partially taken from MSDN Blog):

class Animal { }
class Giraffe : Animal { }

static void Main(string[] args)
{
    // Array assignment works, but...
    Animal[] animals = new Giraffe[10]; 

    // implicit...
    List<Animal> animalsList = new List<Giraffe>();

    // ...and explicit casting fails
    List<Animal> animalsList2 = (List<Animal>) new List<Giraffe>();
}

Is this a covariance problem? Will this be supported in the future C# release and are there any clever workarounds (using only .NET 2.0)?

Answer

Jon Skeet picture Jon Skeet · Jan 9, 2010

Well this certainly won't be supported in C# 4. There's a fundamental problem:

List<Giraffe> giraffes = new List<Giraffe>();
giraffes.Add(new Giraffe());
List<Animal> animals = giraffes;
animals.Add(new Lion()); // Aargh!

Keep giraffes safe: just say no to unsafe variance.

The array version works because arrays do support reference type variance, with execution time checking. The point of generics is to provide compile-time type safety.

In C# 4 there will be support for safe generic variance, but only for interfaces and delegates. So you'll be able to do:

Func<string> stringFactory = () => "always return this string";
Func<object> objectFactory = stringFactory; // Safe, allowed in C# 4

Func<out T> is covariant in T because T is only used in an output position. Compare that with Action<in T> which is contravariant in T because T is only used in an input position there, making this safe:

Action<object> objectAction = x => Console.WriteLine(x.GetHashCode());
Action<string> stringAction = objectAction; // Safe, allowed in C# 4

IEnumerable<out T> is covariant as well, making this correct in C# 4, as pointed out by others:

IEnumerable<Animal> animals = new List<Giraffe>();
// Can't add a Lion to animals, as `IEnumerable<out T>` is a read-only interface.

In terms of working around this in your situation in C# 2, do you need to maintain one list, or would you be happy creating a new list? If that's acceptable, List<T>.ConvertAll is your friend.