在C＃中创建类型列表

Question

I'd like to create a list of types, each of which must implement a particular interface. Like:

interface IBase { }
interface IDerived1 : IBase { }
interface IDerived2 : IBase { }

class HasATypeList
{
    List<typeof(IBase)> items;
    HasATypeList()
    {
        items.Add(typeof(IDerived1));
    }

}

So I know I can do

List<Type> items;

But that won't limit the allowable types in the list to ones that implement IBase. Do I have to write my own list class? Not that it's a big deal, but if I don't have to...

Answer 1

typeof(IBase) , typeof(object) , typeof(Foo) , all return an instance of Type , with the same members and so on.

I don't see what you're trying to achieve and why you want to make a distinction between those ?

In fact, the code you're writing here:

List<typeof(IBase)> items;

(i don't even know if this compiles ? ) Is exactly the same as this:

List<Type> items;

So in fact, what you're trying to achieve is imho useless.

If you really want to achieve this -but I do not see why ... -, you can always create your own collection-type like Olivier Jacot-Descombes is suggesting, but in that case, I'd rather create a type that inherits from Collection<T> instead:

public class MyTypeList<T> : Collection<Type>
{
    protected override InsertItem( int index, Type item )
    {
        if( !typeof(T).IsAssignableFrom(item) )
        {
            throw new ArgumentException("the Type does not derive from ... ");
        }

        base.InsertItem(index, item);
    }
}

Answer 2

Yes. You have to implement a List that throws exceptions if type is not a subclass from IBase.

There is no built in way to do what you want.

Answer 3

The only way to do that is to create your own type collection

public class MyTypeList
{
    List<Type> _innerList;

    public void Add(Type type)
    {
        if (typeof(IBase).IsAssignableFrom(type)) {
             _innerList.Add(type);
        } else {
            throw new ArgumentException(
                "Type must be IBase, implement or derive from it.");
        }
    }

    ...
}