Can I have a type that's both, covariant and contravariant, i.e. fully fungible/changeable with sub and super types?

Question

Can I have a type (for now forgetting its semantics) that can be covariant as well as contravariant?

for example:

public interface Foo<in out T>
{
    void DoFooWith(T arg);
}

Off to Eric Lippert's blog for the meat and potatoes of variance in C# 4.0 as there's little else anywhere that covers adequate ground on the subject.

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I tried it out anyway, not only does it not allow that, but it tells me I am missing the whole point. I need to understand the link between read-only, write-only and variance.

I guess I got some more reading to do.

But any short, epiphany inducing answers are welcome, meanwhile.

Answer 1

No, you cannot do that.

Suppose that were legal. You make an IFoo<Giraffe>. Since IFoo is covariant in T, you can convert it via typesafe reference conversion to IFoo<object>. Since it is contravariant, you can convert that to IFoo<Banana>. What possible semantics are there for IFoo<T> such that it makes sense to be able to convert an IFoo of Giraffes to an IFoo of Bananas via reference conversion? Giraffes and Bananas have nothing in common other than being reference types. You cannot possibly have a method on IFoo<Banana> that returns a Banana, because it might actually be an implementation of IFoo<Giraffe>; how would the author of the implementation know to hand out a Banana? You cannot possibly have a method on IFoo<Banana> that takes a Banana for the same reason; the implementor of IFoo<Giraffe> is expecting you to hand him a Giraffe.

Here's another way of looking at it:

• "in T" means (roughly) "T appears only in input positions".
• "out T" means (roughly) "T appears only in output positions".

Therefore "in out T" would mean... what? As we've seen already, it can only mean "T does not appear at all in any method or property." What's the point of making a generic type in T where you never use T?