Question

a) Compiles

        Func<string, bool> f1 = (Func<object, bool>)null;

b) Not

        Func<int, bool> f2 = (Func<object, bool>)null;

Why value types are special here? Is contravariance broken with value types?

Answer 1

Generic variance only works with reference types, yes. (This is so that the CLR knows that everything’s still just a reference, so the JITted code is still the same… the bits involved in a reference are the same whatever type you’re talking about, whereas treating int as object requires a boxing conversion. Basically you can keep representational identity with reference types).

From the C# 4 spec, section 13.1.3.2:

A type T<A1, …, An> is variance-convertible to a type T<B1, …, Bn> if T is either an interface or a delegate type declared with the variant type parameters T<X1, …, Xn>, and for each variant type parameter Xi one of the following holds:

• Xi is covariant and an implicit reference or identity conversion exists from Ai to Bi
• Xi is contravariant and an implicit reference or identity conversion exists from Bi to Ai
• Xi is invariant and an identity conversion exists from Ai to Bi

It’s the “implicit reference conversion” rather than just “implicit conversion” bit which is a problem for value types.

For much more detail around generic variance, see Eric Lippert’s blog series on the topic.