modify confusing variance example

[lcnr]

Given fn(T) -> T the type T is only invariant while wrapped inside of an item. If you have fn(T) -> T directly you can freely "shrink" the T in return position or "grow" the T in argument position. E.g.

fn split_lt<'a, 'short, 'long>(
    value: fn(&'a ()) -> &'a (),
) -> fn(&'long ()) -> &'short ()
where
    'a: 'short,
    'long: 'a,
{
    value
}

[ehuss]

This was added in #874 where Niko reviewed it as being correct. @nikomatsakis Do you perhaps have any comments on this case?

[lcnr]

Extended that section to talk about fn(T) -> T a bit more in-depth

[nikomatsakis]

Seems right. To be 100% pedantic, you would talk about the "variance of a (struct/enum/union) type Foo<T, U> with respect to its type parameters T (or U)". i.e., if Foo is covariant with respect to T, then if you have Foo<A> and Foo<B> for some types A and B, Foo<A> <: Foo<B> iff A <: B.

Given that fn is not a struct/enum/union type that is parameterized by T, it's not really correct to say that fn(T) is contravariant with respect to T. You say that fn is contravariant with respect to its arguments (and covariant with respect to its return type).

[ehuss]

Merging per Niko's approval.