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

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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), or it could even be left up to type inference with _ (in which case it is always inferred to be T)).

So:

let x = 0;
let y = &mut x;

let z: &mut i32 = y;  // This is an implicit reborrow (equiv. to &mut *y)
let z: &mut _ = y;  // This is an implicit reborrow
let z: _ = y;  // This is a move
let z = y;  // This is a move

Similar examples using function calls

fn takes_mut_i32(_arg: &mut i32) {}
fn takes_mut_t<T>(_arg: &mut T) {}
fn takes_any_t<T>(_arg: T) {}

takes_mut_i32(y);  // this is an implicit reborrow
takes_mut_t(y);  // this is an implicit reborrow
takes_any_t(y);  // this is a move; sometimes this surprises people and there
                 //     have been countless "bugs" filed about it
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