> If a function you call returns an &mut you have no way 
  > to tell

Indeed, but also consider that a function can't return a `&mut` unless a `&mut` was passed in to it somehow. This means that the only way to implicitly introduce a `&mut` is via a method call like `foo.bar()` that takes `&mut self`, but also consider that it would be impossible to call the method here unless `foo` were explicitly declared as mutable to begin with. Given that mutable variables in Rust are relatively rare, these factors conspire to broadcast any mutable references quite clearly.