Making a generic type constraint on Func

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?

Probably the proper solution here is to just accept `T`, then use `Func` in your class. — Kendall Frey, Oct 04 '14 at 22:30
I thought about that, but then I'd have to go like the .NET folks create 17 overloads to support the base Func case. :) I know it's silly, just wanting to know. — halivingston, Oct 04 '14 at 22:33
Not saying this is a good idea but you could follow this https://roslyn.codeplex.com/discussions/543871 thread and make a change in Roslyn and compile your code using that compiler. — mjsabby, Oct 04 '14 at 22:36

score 9 · Accepted Answer · edited May 23 '17 at 12:32

9

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

edited May 23 '17 at 12:32

Community

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answered Oct 04 '14 at 22:24

Kyle Baran

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score 4 · Answer 2 · answered Sep 13 '19 at 06:22

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.

Making a generic type constraint on Func

2 Answers2