Making a generic type constraint on Func<T>

Question

I wanted to know if this is possible:

public class Foo<T> where T : Func<T>

or

public class Foo<T> where T : Func<>

It seems like the compiler is telling me it not possible. I suppose I can throw a runtime exception in the constructor, but was hoping to have it a compiler error.

Any ways about doing this?

Delitescent

Probably the proper solution here is to just accept T, then use Func<T> in your class. – Renteria

I thought about that, but then I'd have to go like the .NET folks create 17 overloads to support the base Func<T .... T16> case. :) I know it's silly, just wanting to know. – Delitescent

Not saying this is a good idea but you could follow this roslyn.codeplex.com/discussions/543871 thread and make a change in Roslyn and compile your code using that compiler. – Inhere

Answer 1

Unfortunately, it looks like you are out of luck. Func<> and Action<> are both delegate types, which cannot be used as a generic type constraint.

This answer sums it up pretty well C# Generics won't allow Delegate Type Constraints

Goodnatured

Answer 2

As the other answer informs, you can't achieve a generic type constraint for Func<>, however C# 7.3 allows you to do a constraint like this

public class Foo<T> where T : Delegate

or

public class Foo<T> where T : MulticastDelegate

that's the closest you can get.

Milord