How can I get a list of interfaces of only inheritors of a certain base interface from a Type instance?

Question

C++ / C#

Let's say there is a basic interface IBaseIntarface and several interfaces inheriting it IMyIntarface1, IMyIntarface2 and so on.
There is a class that implements these heirs (maybe several of them), plus there may be some more

//IBaseIntarface base_intarface;
Type type = base_intarface.GetType();
 List<Type> list_types = type.GetInterfaces().ToList();

In this case, all interfaces get into list_types (plus IBaseIntarface for some reason (I think if it still had a parent, it would also get there))
If I do this:

List<Type> list_types = type
                .GetInterfaces()
                .Where(it => it != typeof(IBaseIntarface))
                .ToList();

Then the base interface is no longer included in the list. but there is still everything that the object has.
If you write like this:

List<Type> list_types = type
                .GetInterfaces()
                .Where(it => it != typeof(IBaseIntarface) && it is IBaseIntarface)
                .ToList();

That won't make it to the list.
How to write a condition so that the list contains only child interfaces from a certain base?

Answer 1

https://stackoverflow.com/questions/26733/getting-...
The approach is to go through all the types in the assembly and see if they implement the given interface.