Nested borrows and invariance

Now let's consider nested references:

  • A &'medium &'long U coerces to a &'short &'short U
  • A &'medium mut &'long mut U coerces to a &'short mut &'long mut U...
    • ...but not to a &'short mut &'short mut U

We say that &mut T is invariant in T, which means any lifetimes in T cannot change (grow or shrink) at all. In the example, T is &'long mut U, and the 'long cannot be changed.

Why not? Consider this:

#![allow(unused)]
fn main() {
fn bar(v: &mut Vec<&'static str>) {
    let w: &mut Vec<&'_ str> = v; // call the lifetime 'w
    let local = "Gottem".to_string();
    w.push(&*local);
} // `local` drops
}

If 'w was allowed to be shorter than 'static, we'd end up with a dangling reference in *v after bar returns.

You will inevitably end up with a feel for covariance from using references with their flexible outer lifetimes, but eventually hit a use case where invariance matters and causes some borrow check errors, because it's (necessarily) so much less flexible. It's just part of the Rust learning experience.


Let's look at one more property of nested references you may run into:

  • You can get a &'long U from a &'short &'long U
    • Just copy it out!
  • But you cannot get a &'long mut U from a &'short mut &'long mut U
    • You can only reborrow a &'short mut U

The reason is again to prevent memory unsafety.

Additionally,

  • You cannot get a &'long U or any &mut U from a &'short &'long mut U
    • You can only reborrow a &'short U

Recall that once a shared reference exist, any number of copies of it could simultaneously exist. Therefore, so long as the outer shared reference exists (and could be used to observe U), the inner &mut must not be usable in a mutable or otherwise exclusive fashion.

And once the outer reference expires, the inner &mut is active and must again be exclusive, so it must not be possible to obtain a &'long U either.