Lifetime confusions. What is `'a` in this code segment?

help
1

I'm learning lifetimes. The following example (made up by me) looks confusing. Can you help on the lifetime annotation here? What does 'a resolve to in the caller side?

// This code compiles, which is confusing to me.

fn f<'a>(r: &'a &'a i32) -> &'a i32 {
    *r
}

fn main() {
    let x: i32 = 5;
    let y = &x;
    let w: &i32; // Assigned in the inner scope.
    {
        // Lifetime 'a starts here?
        let z = &y; // &'a &'a i32?
        w = f(z);
        // Lifetime 'a ends here?
    }

    // Compiles, why?
    // Why w is still valid here? Isn't `'a` gone out of scope?
    // How does the `'a` in `f` resolve?
    println!("{:?}", w);
}
2

I think the more practical interpretation of "the 'a thing" is borrow context/set of loans (instead of "lifetime"). I'll be using '{set of borrows} (invented) syntax as a "resolved" lifetime parameter. So, if I write type &'{x} T that means references of this type borrow variable x immutably.

So on your case

let y = &x;

could be "expanded" to

let y: &'{x} i32 = &x;

Similarly,

let z = &y;

expands to

let z: &'{x, y} &'{x} i32 = &y;
// (note: borrow of `x` propagates, because `y` already borrowed `x`)

So, in the function call f(z), we can see that the lifetime parametersborrow contexts on the references are different, but since the function declares them as the same (&'a&'a i32), the both get converted to the superset – '{x, y}.

We can then desugar:

let w: &'{x, y} &'{x, y} i32;
// ...
    w = f::<'{x, y}>(z);

Since the variable w borrows only from x and y, it's safe to use it after the {} block (you can see, there's not a trace of z in the borrow context) and thus the println compiles. :slight_smile:

(Your example would stop compiling, if you also put y in the {} block, because the f's signature is actually quite restrictive – w actually only points to x, despite its type that says it borrows also y. If you modified the signature to fn<'a>(r: &&'a i32) -> 'a i32), it'd compile again).

3

Ah I see. Thanks for the quick and clear explanation! Your expansion clear shows that it's about the borrow of x and y, nothing to do with z actually (z's lifetime is irrelevant here). I think that's where I went wrong! Thanks for the clear explanation again!