I am trying to achieve something like this:
interface IAbstract
{
string A { get; }
object B { get; }
}
interface IAbstract<T> : IAbstract
{
T B { get; }
}
class RealThing<T> : IAbstract<T>
{
public string A { get; private set; }
public T B { get; private set; }
}
So I can do something like this:
RealThing<string> rt = new RealThing<string>();
IAbstract ia = rt;
IAbstract<string> ias = rt;
object o = ia.B;
string s = ias.B;
Is this possible?
Very nearly. Three things:
You should use new
in IAbstract<T>
to indicate that you know you're hiding an existing member:
new T B { get; }
But even without that, you'll still only get a warning.
You need to implement the IAbstract.B
within RealThing
, which you should almost certainly do using explicit interface implementation, delegating to the strongly-typed member:
object IAbstract.B { get { return B; } }
Within your test code, you need to specify a type argument for RealThing
:
RealThing<string> rt = new RealThing<string>();
This is fine, and even a reasonably common pattern for times where you want to be able to get a non-generic form of an interface.
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