Open Generic as a type parameter

[TonyValenti]

Let's say I have the following code:

public abstract class ChildBase {
  public DateTimeOffset Date {get; set;}
}

public class Parent<TChild> where TChild : ChildBase, new() {
  public TChild Child {get; set;} = new();
}

Given the above, Parent.Child.Date is always valid to get or set.

How would I write the following code?

object Input = SomeMethod();
if(Input is Parent<> {} V1) {
  V1.Child.Date = DateTimeOffset.UtcNow;
}

Thanks!

[svick]

Effectively, what you're asking for is covariance. As you probably know, you can use covariance if you change Parent to:

public interface IParent<out TChild> where TChild : ChildBase {
  TChild Child { get; }
}

Then your condition would be:

object Input = SomeMethod();
if (Input is IParent<ChildBase> V1) {
  V1.Child.Date = DateTimeOffset.UtcNow;
}

Full code.

If you can't change Parent like this, then I think your best option is reflection.

[Richiban]

Actually I think what's closer to what the OP wants is existential types:

Given

public interface IParent
{
	type TChild : ChildBase;

	TChild Child { get; }
}

You could then perform a regular type check (because the generic type doesn't form part of the parent type directly) and the following simply becomes legal:

if (SomeMethod() is IParent v1)
{
	v1.Child.Date = DateTimeOffset.UtcNow;
}

[333fred]

I don't think this is likely to be a solution, as I don't think IParent will be directly usable there if it has an existential type.