I've posted similar minimal examples before. The first one triggers a move (and then fails to compile), while the second one triggers a reborrow. Sure, you can explicitly write out every instance of reborrowing with `&mut *`. That would require you to understand every instance that triggers it, and would also be unbelievably noisy, since automatic dereferencing and automatic reborrowing are actually incredibly common.

  > pub fn main() {
  >     let mut x = String::from("moo");
  >     let y = &mut x;
  >     let z = y;
  >     println!("{:?}", y)
  > }
  > 
  > pub fn main() {
  >     let mut x = String::from("moo");
  >     let y = &mut x;
  >     let z: &mut _ = y;
  >     println!("{:?}", y)
  > }

Ok. So that's just triggering auto-insertion of &mut *

It's completely optional fearure. You don't have to rely on it. Like auto-insertion of semicolons in JS maybe it's better if you don't. Although likelihood of this biting you is way lower than in case of relying on semicolons autoinsertion.

It's "completely optional" in the sense that in several years of reading other people's Rust code, I have never seen anyone explicitly use it throughout a code base. It is utterly ubiquitous, rarely explained, and poorly (if at all) understood by a large majority of Rust users I've mentioned it to.

My claim is that for most Rust users, understanding the borrow checker is limited to a few heuristics that seem to work for them 80% of the time, and throwing up their hands and declaring an approach unworkable the remaining 20% of the time. To a first approximation, no one can mentally model the borrow checker beyond extremely simple examples of 4-5 lines of code.

There's a way of thinking of `Copy` which makes it not an exception: `Copy` variables/places are simply not rendered invalid/uninitialized when they are the source of a move operation, unlike non-`Copy` sources. They're still moved from bitwise like all other rust values!

I like it!