Questions about `&mut T` and move semantics, `&mut T` is "move-only"?

[peerreynders]

@liufei

So it's stack (re)borrow + deref coercion that makes the function call context in example 4 works!

Given:

@trentj

I agree, this is "just" an implicit reborrow, not a Deref ( Mut ) coercion.

and

@ExpHP

The compiler inserts a reborrow in any possible location where a &mut T is "moved" into a place that already has its type known to be &mut _ (where the _ is anything; it could be T , it could be some other U (in which case coercions may also be inserted),

I'm inclined to believe that it's just the implicit reborrow that is needed to get Example 4 to work. y is already a mutable reference of the matching type.

// Example 4 - modified with explicit reborrow
fn f(z: &mut i32) {
    *z += 1;
    println!("{}", *z);
}

fn main() {
    let mut x: i32 = 5;
    let y = &mut x;
    
    f(y);               // implicit reborrow
    f(&mut *y);         // explicit reborrow

    *y += 1;
    println!("{}", *y); // '8'
    x += 1;
    println!("{}", x);  // '9'
}