Variable binding: moving a &mut or borrowing the referent?

Question

This code fails as expected at let c = a; with compile error \"use of moved value: a\":

fn main() {
    let a: &mut i32 = &mut          


        

Answer 1

So, if I just annotate b's type: no move of a into b, but instead a "re"-borrow of *a?

What am I missing?

Absolutely nothing, as in this case these two operations are semantically very similar (and equivalent if a and b belong to the same scope).

• Either you move the reference a into b, making a a moved value, and no longer available.
• Either you reborrow *a in b, making a unusable as long as b is in scope.

The second case is less definitive than the first, you can show this by putting the line defining b into a sub-scope.

This example won't compile because a is moved:

fn main() {
    let a: &mut i32 = &mut 0;
    { let b = a; }
    let c = a;
}

But this one will, because once b goes out of scope a is unlocked:

fn main() {
    let a: &mut i32 = &mut 0;
    { let b = &mut *a; }
    let c = a;
}

Now, to the question "Why does annotating the type of b change the behavior ?", my guess would be:

• When there is no type annotation, the operation is a simple and straightforward move. Nothing is needed to be checked.
• When there is a type annotation, a conversion may be needed (casting a &mut _ into a &_, or transforming a simple reference into a reference to a trait object). So the compiler opts for a re-borrow of the value, rather than a move.

For example, this code is perflectly valid:

fn main() {
    let a: &mut i32 = &mut 0;
    let b: &i32 = a;
}

and here moving a into b would not make any sense, as they are of different type. Still this code compiles: b simply re-borrows *a, and the value won't be mutably available through a as long as b is in scope.