C# unbounded generic type as constrain

Question

Is it possible to have a generic constraint which is an unbounded generic type?

For example:

public T DoSomething<T>(T dictionary) where T : IDictionary<,>
{
    ...
}

---

Edit: Just to explain the context, I want to constrain the usage of the method to an IDictionary, but for the method itself it does not matter exactly what TKey and TValue are.

Gadget

public V DoSomething<K,V>(IDictionary<K,V> dictionary) Is this what you meant? – Silvio

Amir Abiri, why do you want the return type to be the same as the parameter type T? Are you going to return the parameter? There's probably no need for that. And you can't easily construct another T. Unless you want to constraint T : new() as well. – Barrows

Answer 1

This isn't possible, you need to specify the type parameters:

public T DoSomething<T, TKey, TValue>(T dictionary) where T : IDictionary<TKey, TValue> { ... }

Hebe

You see the funny thing is that TKey and TValue can be inferred from T and I don't need them anywhere in the body of the method. So I had hoped to make the generic signature cleaner. – Gadget

@AmirAbiri - Could you constrain T to the non-generic IDictionary instead? – Hebe

Well actually I used IDictionary only to simplify the example. My real type is "any subclass of BaseGameState<?>" (i.e T isn't an interface, it's my own base class). – Gadget

Answer 2

No, you can't do that. Maybe you can use

public IDictionary<TKey, TValue> DoSomething<TKey, TValue>(IDictionary<TKey, TValue> dictionary)
{
...
}

Barrows

Answer 3

There is an issue in rosylin github, however, it's not done yet. So it's not possible right now, but I expect it will be relatively soon.

Cultivate