Some clarification on taking a reference from a return value

[sirlan-ff00ff]

with this following code (in lisette 0.12.0)

struct A {}
impl A {
  fn n() -> A { A{} }
  fn m(self: mut Ref<A>) { }
}

fn main() {
  let mut a1 = &A.n()
  a1.m()

  let mut a2 = A.n()
  let mut a2r = &a2
  a2r.m()
}

it gives the following error

  ✕ Missing write permission
    ╭─[src/main.lis:10:3]
  9 │   let mut a1 = &A.n()
 10 │   a1.m()
    ·   ──┬─
    ·     ╰── expected mut Ref<A>, found Ref<A>
 11 │ 
    ╰────
  help: mut Ref<A> permits writes, Ref<A> does not. Make the value writable where it is created,
 or pass a .clone() · code: [infer.needs_writable]

Which makes me wonder, why does it consider the the &A.n() to not be mutable, and if there is a way to make it read as mutable, as lisette doesn't have a &mut, and trying to write the type or using as will also be errors

Although doing

  let mut a1 = {let mut a = A.n(); &a}
  a1.m() 

seems valid, but looks like a hack

Is the intended way to prefer having the function itself return a mut Ref<A> and discouraging taking the reference of return values? if so would be nice to have a specific hint about that

[ivov]

Hi there! This is another manifestation of the same bug reported by Ryan at #1374, fixed at #1375, releasing this weekend. The checker was looking for a writable place behind the & but call results were being excluded, so it fell back to read-only Ref<A>. No need to return mut Ref<A> from n or use the block form, &A.n() is the intended way. Starting on v0.12.1 your first snippet will work as written, tested just now against main.