如何在不将其传递给函数的情况下重新导入可变引用？

Question

I have found a case where manually inlining a function changes the way the borrow-checker treats it, such that it no longer compiles. Presumably it is relying on the information in the function signature. How can I provide this information in the inlined version?

How I think it's working

Let 'a and 'b be lifetimes with 'a shorter than 'b (which can be written 'b: 'a ).

Suppose I have a p: &'b mut f32 . I can borrow p briefly (with &mut p ) to obtain q: &'a mut &'b mut f32 .

1. Have I correctly understood that &'a mut &'b mut f32 is equivalent to &'a mut &'a mut f32 because 'b: 'a ?

I can then dereference q (with *q ) to obtain r: &'a mut f32 . I can write to the f32 via r (with *r = something ), and I can later (outside lifetime 'a ) read back the value via p (with *p ).

With a function call

Here is some working code that I think uses the above sequence:

fn reborrow<'a, 'b: 'a>(q: &'a mut &'b mut f32) -> &'a mut f32 {
    *q
}

fn main() {
    let mut x: f32 = 3.142;
    let mut p = &mut x;
    {
        let q = &mut p;
        let r = reborrow(q);
        *r = 2.718;
    }
    assert_eq!(*p, 2.718);
}

(Replacing *q with q in the body of reborrow() also works, because Rust inserts the necessary dereference if it is missing).

Manually inlined

If I manually inline the reborrow() call, it no longer compiles:

fn main() {
    let mut x: f32 = 3.142;
    let mut p = &mut x;
    {
        let q = &mut p;
        let r = *q; <-- ERROR REPORTED HERE.
        *r = 2.718;
    }
    assert_eq!(*p, 2.718);
}

error[E0507]: cannot move out of borrowed content

2. Who took away my toys? What is the type inference thinking/missing?

3. Can I annotate the let binding somehow to make the compiler infer the same types as in the previous version?

Some other attempts

Here's another version that works, but which doesn't define the name r :

fn main() {
    let mut x: f32 = 3.142;
    let mut p = &mut x;
    {
        let q = &mut p;
        **q = 2.718;
    }
    assert_eq!(*p, 2.718);
}

Here's a work-around that defines the name r and works, but does not use the same sequence of borrows and dereferences:

fn main() {
    let mut x: f32 = 3.142;
    let mut p = &mut x;
    {
        let q = &mut p;
        let r = &mut **q;
        *r = 2.718;
    }
    assert_eq!(*p, 2.718);
}

I made a playground combining all four versions.

Answer 1

The obvious solution works, as one could expect:

fn main() {
    let mut x: f32 = 3.142;
    let mut p = &mut x;
    {
        let r: &mut f32 = p;
        *r = 2.718;
    }
    assert_eq!(*p, 2.718);
}

It seems relatively intuitive and is what I would expect a newcomer to end up with.

---

If you start thinking about it, however, it will not make sense. As described, it looks like:

• let r: &mut f32 = p; moves out of p
• and yet we use p later in assert_eq!(*p, 2.718);

A reasonable explanation would be that p is Copy , however it's not ¹ !

The answer is that, implicitly, Rust is performing a re-borrowing behind the scenes. That is, the explicit code is:

fn main() {
    let mut x: f32 = 3.142;
    let mut p = &mut x;
    {
        let r: &mut f32 = &mut *p;
        *r = 2.718;
    }
    assert_eq!(*p, 2.718);
}

We can check this by attempting to read p after re-borrowing it, and check the compiler error:

 error[E0502]: cannot borrow `p` as immutable because `*p` is also borrowed as mutable --> <anon>:6:24 | 5 | let r: &mut f32 = p; | - mutable borrow occurs here 6 | println!("{}", p); | ^ immutable borrow occurs here 7 | *r = 2.718; 8 | } | - mutable borrow ends here error: aborting due to previous error 

Which confirms that p is indeed only borrowed mutably, and not moved, cloned or copied.

¹ A mutable reference cannot be Copy or even Clone as it would violate the Aliasing XOR Mutability principle which underpins Rust safety.

Answer 2

I can't possibly begin to explain this, but you can do a similar trick as the implicit dereference and say that r is &mut f32 :

fn main() {
    let mut x: f32 = 3.142;
    let mut p = &mut x;
    {
        let q = &mut p;
        let r: &mut f32 = q;
        *r = 2.718;
    }
    assert_eq!(*p, 2.718);
}