Cast List<T> to List<Interface>

Question

You can't cast it (preserving reference identity) - that would be unsafe. For example:

public interface IFruit {}

public class Apple : IFruit {}
public class Banana : IFruit {}

...

List<Apple> apples = new List<Apple>();
List<IFruit> fruit = apples; // Fortunately not allowed
fruit.Add(new Banana());

// Eek - it's a banana!
Apple apple = apples[0];

Now you can convert a List<Apple> to an IEnumerable<IFruit> in .NET 4 / C# 4 due to covariance, but if you want a List<IFruit> you'd have to create a new list. For example:

// In .NET 4, using the covariance of IEnumerable<T>
List<IFruit> fruit = apples.ToList<IFruit>();

// In .NET 3.5
List<IFruit> fruit = apples.Cast<IFruit>().ToList();

But this is not the same as casting the original list - because now there are two separate lists. This is safe, but you need to understand that changes made to one list won't be seen in the other list. (Modifications to the objects that the lists refer to will be seen, of course.)

A Cast iterator and .ToList():

List<IDic> casted = input.Cast<IDic>().ToList() will do the trick.

Originally I said covariance would work - but as Jon has rightly pointed out; no it won't!

And originally I also stupidly left off the ToList() call

I too had this problem and after reading Jon Skeet's answer I modified my code from using List<T> to use IEnumerable<T>. Although this does not answer the OP's original question of How can I cast List<Client> to List<IDic>, it does avoid the need to do so and thus may be helpful to others who encounter this issue. This of course assumes that the code that requires the use of List<IDic> is under your control.

E.g.:

public void ProcessIDic(IEnumerable<IDic> sequence)
{
   // Implementation
}

Instead of:

public void ProcessIDic(List<IDic> list)
{
   // Implementation
}

If you can use LINQ then you can do this...

List<Client> clientList = new List<Client>();
List<IDic> list = clientList.Select(c => (IDic)c).ToList();