Can I get constraint 'b: 'a just from type &'a T<'b>?

help
1

Can I get constraint 'b: 'a just from type &'a T<'b> where T is any other Type and no other constraints around?

I found some reference says that if a type is &'a T then T: 'a, but my friend insists that T<'b> : 'adoes not imply 'b: 'a and I'm confused. Any help?

2

Yes you can. Direct quote:

T: 'a means that all lifetime parameters of T outlive 'a .

3

Depending on where the type is mentioned, yes. Rust Playground

The rules about where bounds are implied at all are complicated and confusing, but you can at least see that &'a T<'b> does imply 'b: 'a in a place where implied bounds can happen.

Trivially, for instance, with my above playground,

fn not_implied<'b>() {
    bar(&T::<'b>(PhantomData));
}

fails to compile since the bound needed by bar is not implied.

4

Here's another code-based demonstration.

5

Great example! I shall tell my friend this exmaple.

6

this is generally true, as already answered in previous posts.

if I would guess, your friend was probably trying to making a point about variance, e.g. T<'x> might be contra-variant over 'x.

but variance is really about subtyping relation between type T<'b> and type T<'a>, it is NOT about lifetime bounds (a.k.a. "outlives" relation) between type T<'b> and lifetime 'a.

for example, if we treat T<'_> as a functor, what contravariance means here is that, 'a < 'b implies T<'b> < T<'a>. i.e. if lifetime 'a outlives 'b, then type T<'b> can be coerced (or converted) into T<'a>.