From 021466ba05c160cfa526c7946a14c6ac4763b480 Mon Sep 17 00:00:00 2001 From: Ralf Jung Date: Sun, 15 Apr 2018 12:17:06 +0200 Subject: [PATCH] clarify &&T = &Pin --- ...04-10-safe-intrusive-collections-with-pinning.md | 13 +++++++++---- 1 file changed, 9 insertions(+), 4 deletions(-) diff --git a/personal/_posts/2018-04-10-safe-intrusive-collections-with-pinning.md b/personal/_posts/2018-04-10-safe-intrusive-collections-with-pinning.md index 37da65c..a3f0cca 100644 --- a/personal/_posts/2018-04-10-safe-intrusive-collections-with-pinning.md +++ b/personal/_posts/2018-04-10-safe-intrusive-collections-with-pinning.md @@ -400,10 +400,9 @@ Since all the mutation we perform there happens inside a `Cell`, why shouldn't w That's a good question! It seems perfectly fine to change `insert` to take `&Pin>`. I can't come up with any counter-example. -However, the formal model also cannot justify this variant of `insert`: As we defined previously, `Pin<'a, T>.shr` just uses `T.shr`. +However, the formal model also cannot justify this variant of `insert`: Our model as defind previously defines `Pin<'a, T>.shr` in terms of `T.shr`. That made it easy to justify [`Pin::deref`](https://doc.rust-lang.org/nightly/std/mem/struct.Pin.html#method.deref). -However, it also means `&Pin` and `&&T` *are the same type*. -The two invariants are equivalent. +However, as a consequence, `Pin<'a, T>.shr` and `(&'a T).shr` are literally the same invariant, and hence `&Pin` and `&&T` *are the same type*. We could literally write functions transmuting between the two, and we could justify them in my model. Clearly, `insert` taking `entry: &&T` would be unsound as nothing stops the entry from being moved later: {% highlight rust %} @@ -421,7 +420,13 @@ fn main() { {% endhighlight %} This shows that the version of `insert` that takes a shared reference cannot be sound in the model. -I do have an idea for how to solve this problem: Introduce a *fourth* "mode" or typestate, the "shared pinned" state, with an accompanying invariant. +Notice that this is a consequence of a choice I made when building the model, namely the choice to define `Pin<'a, T>.shr` in terms of `T.shr`. +This does *not* show that `&Pin` and `&&T` have to be the same type given the public API and the contract they provide. +Different choices in the model could lead to a different situation. +The problem is, how else *could* we define `Pin<'a, T>.shr`, if we do not want to use `T.shr`? +What *is* the invariant of a shared reference to a pinned reference? + +I do have an idea for how to answer this question: Introduce a *fourth* "mode" or typestate, the "shared pinned" state, with an accompanying invariant. However, I previously argued strongly against introducing such a fourth state, on the grounds that three typestates is already complicated enough. In fact, an earlier version of the `Pin` API used to have two kinds of pinned references (shared and mutable) reflecting the two distinct "shared pinned" and "(owned) pinned" typestates. The shared variant got subsequently removed, and I feel somewhat guilty about that now because I strongly pushed for it. -- 2.30.2