Cole Johnson Cole Johnson - 9 months ago 34
C# Question

"out T" vs. "T" in Generics

What is the difference between

<out T>
? For example:

public interface IExample<out T>


public interface IExample<T>

The only info I have gotten from MSDN was that

You can use the out keyword in generic interfaces and delegates.


The out keyword in generics is used to denote that the type T in the interface is covariant. See Covariance and contravariance for details.

The classic example is IEnumerable<out T>. Since IEnumerable<out T> is covariant, you're allowed to do the following:

IEnumerable<string> strings = new List<string>();
IEnumerable<object> objects = strings;

The second line above would fail if this wasn't covariant, even though logically it should work, since string derives from object. Before variance in generic interfaces was added to C# and VB.NET (in .NET 4 with VS 2010), this was a compile time error.

After .NET 4, IEnumerable<T> was marked covariant, and became IEnumerable<out T>. Since IEnumerable<out T> only uses the elements within it, and never adds/changes them, it's safe for it to treat an enumerable collection of strings as an enumerable collection of objects, which means it's covariant.

This wouldn't work with a type like IList<T>, since IList<T> has an Add method. Suppose this would be allowed:

IList<string> strings = new List<string>();
IList<object> objects = strings;  // NOTE: Fails at compile time

You could then call:

objects.Add(new Image()); // This should work, since IList<object> should let us add **any** object

This would, of course, fail - so IList<T> can't be marked covariant.

There is also, btw, an option for in - which is used by things like comparison interfaces. IComparer<in T>, for example, works the opposite way. You can use a concrete IComparer<Foo> directly as an IComparer<Bar> if Bar is a subclass of Foo, because the IComparer<in T> interface is contravariant.