协变接口不可能包含类型参数是协变类型的泛型类型吗？

Question

I have a covariant interface and generic class that is unrelated. I'd like the covariant interface to have a property that is an instance of that generic class on the covariant type, like so.

public interface IFoo<out T>
{
    Bar<T> barobj { get; set; }
}

public class Bar<T>
{
}

Unfortunately I'm getting an error

Error CS1961 Invalid variance: The type parameter 'T' must be invariantly valid on 'IFoo<T>.barobj'. 'T' is covariant.

Does this mean that it's impossible to have a covariant interface with a generic type that uses the covariant type as the parameter? Am I doing something wrong here?

Answer 1

An interface can only be covariant if it only allows outputs of the generic type. Your interface is not covariant because you can set the value of barobj . If you make the property read-only, then it can be covariant if barobj is covariant. So that means you need a covariant interface for Bar :

public interface IFoo<out T>
{
    IBar<T> barobj { get; }
}

public class Bar<T> : IBar<T>
{
}

public interface IBar<out T> 
{
}