[web.git] / personal / _posts / 2018-11-16-stacked-borrows-implementation.md

---
title: "Stacked Borrows Implemented"
categories: internship rust
forum: https://internals.rust-lang.org/t/stacked-borrows-implemented/8847
---

Three months ago, I proposed [Stacked Borrows]({% post_url
2018-08-07-stacked-borrows %}) as a model for defining what kinds of aliasing
are allowed in Rust, and the idea of a [validity invariant]({% post_url
2018-08-22-two-kinds-of-invariants %}) that has to be maintained by all code at
all times.  Since then I have been busy implementing both of these, and
developed Stacked Borrows further in doing so.  This post describes the latest
version of Stacked Borrows, and reports my findings from the implementation
phase: What worked, what did not, and what remains to be done.  There will also
be an opportunity for you to help the effort!

<!-- MORE -->

What Stacked Borrows does is that it defines a semantics for Rust programs such
that some things about references always hold true for every valid execution
(meaning executions where no [undefined behavior]({% post_url
2017-07-14-undefined-behavior %}) occurred): `&mut` references are unique (we
can rely on no accesses by other functions happening to the memory they point
to), and `&` references are immutable (we can rely on no writes happening to the
memory they point to, unless there is an `UnsafeCell`).  Usually we have the
borrow checker guarding us against such nefarious violations of reference type
guarantees, but alas, when we are writing unsafe code, the borrow checker cannot
help us.  We have to define a set of rules that makes sense even for unsafe
code.

I will explain these rules again in this post.  The explanation is not going to
be the same as last time, not only because it changed a bit, but also because I
think I understand the model better myself now so I can do a better job
explaining it.

Ready?  Let's get started.  I hope you brought some time, because this is a
rather lengthy post.  If you are not interested in a detailed description of
Stacked Borrows, you can skip most of this post and go right to [section 4].  If
you only want to know how to help, jump to [section 6].

## 1 Enforcing Uniqueness

Let us first ignore the part about `&` references being immutable and focus on
uniqueness of mutable references.  Namely, we want to define our model in a way
that calling the following function will trigger undefined behavior:

{% highlight rust %}
fn demo0() {
  let x = &mut 1u8;
  let y = &mut *x;
  *y = 5;
  // Write through a pointer aliasing `y`
  *x = 3;
  // Use `y` again, asserting it is still exclusive
  let _val = *y;
}
{% endhighlight %}

We want this function to be disallowed because between two uses of `y`, there is
a use of another pointer for the same location, violating the fact that `y`
should be unique.

Notice that this function does not compile, the borrow checker won't allow it.
That's great!  It is undefined behavior, after all.  But the entire point of
this exercise is to explain *why* we have undefined behavior here *without*
referring to the borrow checker, because we want to have rules that also work
for unsafe code.  In fact, you could say that retroactively, these rules explain
why the borrow checker works the way it does:  We can pretend that the model came
first, and the borrow checker is merely doing compile-time checks to make sure
we follow the rules of the model.

To be able to do this, we have to pretend our machine has two things which real
CPUs do not have.  This is an example of adding "shadow state" or "instrumented
state" to the "virtual machine" that we [use to specify Rust]({% post_url
2017-06-06-MIR-semantics %}).  This is not an uncommon approach, often times
source languages make distinctions that do not appear in the actual hardware.  A
related example is
[valgrind's memcheck](http://valgrind.org/docs/manual/mc-manual.html) which
keeps track of which memory is initialized to be able to detect memory errors:
During a normal execution, uninitialized memory looks just like all other
memory, but to figure out whether the program is violating C's memory rules, we
have to keep track of some extra state.

For stacked borrows, the extra state looks as follows:

1. For every pointer, we keep track of an extra "tag" that records when and how
   this pointer was created.
2. For every location in memory, we keep track of a stack of "items", indicating
   which tag a pointer must have to be allowed to access this location.

These exist separately, i.e., when a pointer is stored in memory, then we both
have a tag stored as part of this pointer value (remember,
[bytes are more than `u8`]({% post_url 2018-07-24-pointers-and-bytes %})), and
every byte occupied by the pointer has a stack regulating access to this
location.  Also these two do not interact, i.e., when loading a pointer from
memory, we just load the tag that was stored as part of this pointer.  The stack
of a location, and the tag of a pointer stored at some location, do not have any
effect on each other.

In our example, there are two pointers (`x` and `y`) and one location of
interest (the one both of these pointers point to, initialized with `1u8`).
When we initially create `x`, it gets tagged `Uniq(0)` to indicate that it is a
unique reference, and the location's stack has `Uniq(0)` at its top to indicate
that this is the latest reference allowed to access said location.  When we
create `y`, it gets a new tag, `Uniq(1)`, so that we can distinguish it from
`x`.  We also push `Uniq(1)` onto the stack, indicating not only that `Uniq(1)`
is the latest reference allow to access, but also that it is "derived from"
`Uniq(0)`: The tags higher up in the stack are descendants of the ones further
down.

So after both references are created, we have: `x` tagged `Uniq(0)`, `y` tagged
`Uniq(1)`, and the stack contains `[Uniq(0), Uniq(1)]`. (Top of the stack is on
the right.)

When we use `y` to access the location, we make sure its tag is at the top of
the stack: check, no problem here.  When we use `x`, we do the same thing: Since
it is not at the top yet, we pop the stack until it is, which is easy.  Now the
stack is just `[Uniq(0)]`.  Now we use `y` again and... blast!  Its tag is not
on the stack.  We have undefined behavior.

In case you got lost, here is the source code with comments indicating the tags
and the stack of the one location that interests us:

{% highlight rust %}
fn demo0() {
  let x = &mut 1u8; // tag: `Uniq(0)`
  // stack: [Uniq(0)]

  let y = &mut *x; // tag: `Uniq(1)`
  // stack: [Uniq(0), Uniq(1)]

  // Pop until `Uniq(1)`, the tag of `y`, is on top of the stack:
  // Nothing changes.
  *y = 5;
  // stack: [Uniq(0), Uniq(1)]

  // Pop until `Uniq(0)`, the tag of `x`, is on top of the stack:
  // We pop `Uniq(1)`.
  *x = 3;
  // stack: [Uniq(0)]

  // Pop until `Uniq(1)`, the tag of `y`, is on top of the stack:
  // That is not possible, hence we have undefined behavior.
  let _val = *y;
}
{% endhighlight %}

Well, actually having undefined behavior here is good news, since that's what we
wanted from the start!  And since there is an implementation of the model in
[miri](https://github.com/solson/miri/), you can try this yourself: The amazing
@shepmaster has integrated miri into the playground, so you can
[put the example there](https://play.rust-lang.org/?version=stable&mode=debug&edition=2015&gist=d15868687f79072688a0d0dd1e053721)
(adjusting it slightly to circumvent the borrow checker), then select "Tools -
Miri" and it will complain (together with a rather unreadable backtrace, we sure
have to improve that one):

```
error[E0080]: constant evaluation error: Borrow being dereferenced (Uniq(1037)) does not exist on the stack
 --> src/main.rs:6:14
  |
6 |   let _val = *y;
  |              ^^ Borrow being dereferenced (Uniq(1037)) does not exist on the stack
  |
```

## 2 Enabling Sharing

If we just had unique pointers, Rust would be a rather dull language.  Luckily
enough, there are also two ways to have shared access to a location: through
shared references (safely), and through raw pointers (unsafely).  Moreover,
shared references *sometimes* (but not when they point to an `UnsafeCell`)
assert an additional guarantee: Their destination is immutable.

For example, we want the following code to be allowed -- not least because this
is actually safe code accepted by the borrow checker, so we better make sure
this is not undefined behavior:

{% highlight rust %}
fn demo1() {
  let x = &mut 1u8;
  // Create several shared references, and we can also still read from `x`
  let y1 = &*x;
  let _val = *x;
  let y2 = &*x;
  let _val = *y1;
  let _val = *y2;
}
{% endhighlight %}

However, the following code is *not* okay:

{% highlight rust %}
fn demo2() {
  let x = &mut 1u8;
  let y = &*x;
  // Create raw reference aliasing `y` and write through it
  let z = x as *const u8 as *mut u8;
  unsafe { *z = 3; }
  // Use `y` again, asserting it still points to the same value
  let _val = *y;
}
{% endhighlight %}

If you
[try this in miri](https://play.rust-lang.org/?version=stable&mode=debug&edition=2015&gist=1bc8c2f432941d02246fea0808e2e4f4),
you will see it complain:

```
 --> src/main.rs:6:14
  |
6 |   let _val = *y;
  |              ^^ Location is not frozen long enough
  |
```

How is it doing that, and what is a "frozen" location?

To explain this, we have to extend the "shadow state" of our "virtual machine" a
bit.  First of all, we introduce a new kind of tag that a pointer can carry: A
*shared* tag.  The following Rust type describes the possible tags of a pointer:

{% highlight rust %}
pub type Timestamp = u64;
pub enum Borrow {
    Uniq(Timestamp),
    Shr(Option<Timestamp>),
}
{% endhighlight %}

You can think of the timestamp as a unique ID, but as we will see, for shared
references, it is also important to be able to determine which of these IDs was
created first.  The timestamp is optional in the shared tag because that tag is
also used by raw pointers, and for raw pointers, we are often not able to track
when and how they are created (for example, when raw pointers are converted to
integers and back).

We use a separate type for the items on our stack, because there we do not need
a timestamp for shared pointers:

{% highlight rust %}
pub enum BorStackItem {
    Uniq(Timestamp),
    Shr,
}
{% endhighlight %}

And finally, a "borrow stack" consists of a stack of `BorStackItem`, together
with an indication of whether the stack (and the location it governs) is
currently *frozen*, meaning it may only be read, not written:

{% highlight rust %}
pub struct Stack {
    borrows: Vec<BorStackItem>, // used as a stack; never empty
    frozen_since: Option<Timestamp>, // virtual frozen "item" on top of the stack
}
{% endhighlight %}

### 2.1 Executing the Examples

Let us now look at what happens when we execute our two example programs.  To
this end, I will embed comments in the source code.  There is only one location
of interest here, so whenever I talk about a "stack", I am referring to the
stack of that location.

{% highlight rust %}
fn demo1() {
  let x = &mut 1u8; // tag: `Uniq(0)`
  // stack: [Uniq(0)]; not frozen
  
  let y1 = &*x; // tag: `Shr(Some(1))`
  // stack: [Uniq(0), Shr]; frozen since 1

  // Access through `x`.  We first check whether its tag `Uniq(0)` is in the
  // stack (it is).  Next, we make sure that either our item *or* `Shr` is on
  // top *or* the location is frozen.  The latter is the case, so we go on.
  let _val = *x;
  // stack: [Uniq(0), Shr]; frozen since 1

  // This is not an access, but we still dereference `x`, so we do the same
  // actions as on a read.  Just like in the previous line, nothing happens.
  let y2 = &*x; // tag: `Shr(Some(2))`
  // stack: [Uniq(0), Shr]; frozen since 1

  // Access through `y1`.  Since the shared tag has a timestamp (1) and the type
  // (`u8`) does not allow interior mutability (no `UnsafeCell`), we check that
  // the location is frozen since (at least) that timestamp.  It is.
  let _val = *y1;
  // stack: [Uniq(0), Shr]; frozen since 1

  // Same as with `y2`: The location is frozen at least since 2 (actually, it
  // is frozen since 1), so we are good.
  let _val = *y2;
  // stack: [Uniq(0), Shr]; frozen since 1
}
{% endhighlight %}

This example demonstrates a few new aspects.  First of all, there are actually
two operations that perform tag-related checks in this model (so far):
Dereferencing a pointer (whenever you have a `*`, also implicitly), and actual
memory accesses.  Operations like `&*x` are an example of operations that
dereference a pointer without accessing memory.  Secondly, *reading* through a
mutable reference is actually okay *even when that reference is not exclusive*.
It is only *writing* through a mutable reference that "re-asserts" its
exclusivity.  I will come back to these points later, but let us first go
through another example.

{% highlight rust %}
fn demo2() {
  let x = &mut 1u8; // tag: `Uniq(0)`
  // stack: [Uniq(0)]; not frozen
  
  let y = &*x; // tag: `Shr(Some(1))`
  // stack: [Uniq(0), Shr]; frozen since 1

  // The `x` here really is a `&*x`, but we have already seen above what
  // happens: `Uniq(0)` must be in the stack, but we leave it unchanged.
  let z = x as *const u8 as *mut u8; // tag irrelevant because raw
  // stack: [Uniq(0), Shr]; frozen since 1

  // A write access through a raw pointer: Unfreeze the location and make sure
  // that `Shr` is at the top of the stack.
  unsafe { *z = 3; }
  // stack: [Uniq(0), Shr]; not frozen

  // Access through `y`.  There is a timestamp in the `Shr` tag, and the type
  // `u8` does not allow interior mutability, but the location is not frozen.
  // This is undefined behavior.
  let _val = *y;
}
{% endhighlight %}

### 2.2 Dereferencing a Pointer
[section 2.2]: #22-dereferencing-a-pointer

As we have seen, we consider the tag of a pointer already when dereferencing it,
before any memory access happens.  The operation on a dereference never mutates
the stack, but it performs some basic checks that might declare the program UB.
The reason for this is twofold: First of all, I think we should require some
basic validity for pointers that are dereferenced even when they do not access
memory. Secondly, there is the practical concern for the implementation in miri:
When we dereference a pointer, we are guaranteed to have type information
available (crucial for things that depend on the presence of an `UnsafeCell`),
whereas having type information on every memory access would be quite hard to
achieve in miri.

Notice that on a dereference, we have *both* a tag at the pointer *and* the type
of a pointer, and the two might not agree, which we do not always want to rule
out (after a `transmute`, we might have raw or shared pointers with a unique
tag, for example).

The following checks are done on every pointer dereference, for every location
covered by the pointer (`size_of_val` tells us how many bytes the pointer
covers):

1. If this is a raw pointer, do nothing and reset the tag used for the access to
   `Shr(None)`.  Raw accesses are checked as little as possible.
2. If this is a unique reference and the tag is `Shr(Some(_))`, that's an error.
3. If the tag is `Uniq`, make sure there is a matching `Uniq` item with the same
   ID on the stack.
4. If the tag is `Shr(None)`, make sure that either the location is frozen or
   else there is a `Shr` item on the stack.
5. If the tag is `Shr(Some(t))`, then the check depends on whether the location
   is inside an `UnsafeCell` or not, according to the type of the reference.
    - Locations outside `UnsafeCell` must have `frozen_since` set to `t` or an
      older timestamp.
    - `UnsafeCell` locations must either be frozen or else have a `Shr` item in
      their stack (same check as if the tag had no timestamp).

### 2.3 Accessing Memory
[section 2.3]: #23-accessing-memory

On an actual memory access, we know the tag of the pointer that was used to
access (unless it was a raw pointer, in which case the tag we see is
`Shr(None)`), and we know whether we are reading from or writing to the current
location.  We perform the following operations on all locations affected by the
access:

1. If the location is frozen and this is a read access, nothing happens (even
   if the tag is `Uniq`).
2. Unfreeze the location (set `frozen_since` to `None`).  Either the location is
   already unfrozen, or this is a write.
3. Pop the stack until the top item matches the tag of the pointer.
    - A `Uniq` item matches a `Uniq` tag with the same ID.
    - A `Shr` item matches any `Shr` tag (with or without timestamp).
    - When we are reading, a `Shr` item matches a `Uniq` tag.

    If we pop the entire stack without finding a match, then we have undefined
    behavior.

To understand these rules better, try going back through the three examples we
have seen so far and applying these rules for dereferencing pointers and
accessing memory to understand how they interact.

The most subtle point here is that we make a `Uniq` tag match a `Shr` item and
also accept `Uniq` reads on frozen locations.  This is required to make `demo1`
work: Rust permits read accesses through mutable references even when they are
not currently actually unique.  Our model hence has to do the same.

## 3 Retagging and Creating Raw Pointers

We have talked quite a bit about what happens when we *use* a pointer.  It is
time we take a close look at *how pointers are created*.  However, before we go
there, I would like us to consider one more example:

{% highlight rust %}
fn demo3(x: &mut u8) -> u8 {
    some_function();
    *x
}
{% endhighlight %}

The question is: Can we move the load of `x` to before the function call?
Remember that the entire point of Stacked Borrows is to enforce a certain
discipline when using references, in particular, to enforce uniqueness of
mutable references.  So we should hope that the answer to that question is "yes"
(and that, in turn, is good because we might use it for optimizations).
Unfortunately, things are not so easy.

The uniqueness of mutable references entirely rests on the fact that the pointer
has a unique tag: If our tag is at the top of the stack (and the location is not
frozen), then any access with another tag will pop our item from the stack (or
cause undefined behavior).  This is ensured by the memory access checks.  Hence,
if our tag is *still* on the stack after some other accesses happened (and we
know it is still on the stack every time we dereference the pointer, as per the
dereference checks described above), we know that no access through a pointer
with a different tag can have happened.

### 3.1 Guaranteed Freshness

However, what if `some_function` has an exact copy of `x`?  We got `x` from our
caller (whom we do not trust), maybe they used that same tag for another
reference (copied it with `transmute_copy` or so) and gave that to
`some_function`?  There is a simple way we can circumvent this concern: Generate
a new tag for `x`.  If *we* generate the tag (and we know generation never emits
the same tag twice, which is easy), we can be sure this tag is not used for any
other reference.  So let us make this explicit by putting a `Retag` instruction
into the code where we generate new tags:

{% highlight rust %}
fn demo3(x: &mut u8) -> u8 {
    Retag(x);
    some_function();
    *x
}
{% endhighlight %}

These `Retag` instructions are inserted by the compiler pretty much any time
references are copied: At the beginning of every function, all inputs of
reference type get retagged.  On every assignment, if the assigned value is of
reference type, it gets retagged.  Moreover, we do this even when the reference
value is inside the field of a `struct` or `enum`, to make sure we really cover
all references.  (This recursive descent is already implemented, but the
implementation has not landed yet.)  However, we do *not* descend recursively
through references: Retagging a `&mut &mut u8` will only retag the *outer*
reference.

Retagging is the *only* operation that generates fresh tags.  Taking a reference
simply forwards the tag of the pointer we are basing this reference on.

Here is our very first example with explicit retagging:

{% highlight rust %}
fn demo0() {
  let x = &mut 1u8; // nothing interesting happens here
  Retag(x); // tag of `x` gets changed to `Uniq(0)`
  // stack: [Uniq(0)]; not frozen
  
  let y = &mut *x; // nothing interesting happens here
  Retag(y); // tag of `y` gets changed to `Uniq(1)`
  // stack: [Uniq(0), Uniq(1)]; not frozen

  // Check that `Uniq(1)` is on the stack, then pop to bring it to the top.
  *y = 5;
  // stack: [Uniq(0), Uniq(1)]; not frozen

  // Check that `Uniq(0)` is on the stack, then pop to bring it to the top.
  *x = 3;
  // stack: [Uniq(0)]; not frozen

  // Check that `Uniq(1)` is on the stack -- it is not, hence UB.
  let _val = *y;
}
{% endhighlight %}

For each reference, `Retag` does the following (we will slightly refine these
instructions later) on all locations covered by the reference (again, according
to `size_of_val`):

1. Compute a fresh tag, `Uniq(_)` for a mutable reference and `Shr(Some(_))` for
   a shared reference.
2. Perform the checks that would also happen when we dereference this reference.
3. Perform the actions that would also happen when an actual access happens
   through this reference (for shared references a read access, for mutable
   references a write access).
4. If the new tag is `Uniq`, push it onto the stack.  (The location cannot be
   frozen: `Uniq` tags are only created for mutable references, and we just
   performed the actions of a write access to memory, which unfreezes
   locations.)
5. If the new tag is `Shr`:
    - If the location is already frozen, we do nothing.
    - Otherwise:
      1. Push a `Shr` item to the stack.
      2. If the location is outside of `UnsafeCell`, it gets frozen with the
         timestamp of the new reference.

One high-level way to think about retagging is that it computes a fresh tag, and
then performs a reborrow of the old reference with the new tag.

### 3.2 When Pointers Escape

Creating a shared reference is not the only way to share a location: We can also
create raw pointers, and if we are careful enough, use them to access a location
from different aliasing pointers.  (Of course, "careful enough" is not very
precise, but the precise answer is the very model I am describing here.)

To account for this, we need one final ingredient in our model: a special
instruction that indicates that a reference was cast to a raw pointer, and may
thus be accessed from these raw pointers in a shared way.  Consider the
[following example](https://play.rust-lang.org/?version=stable&mode=debug&edition=2015&gist=253868e96b7eba85ef28e1eabd557f66):

{% highlight rust %}
fn demo4() {
  let x = &mut 1u8;
  Retag(x); // tag of `x` gets changed to `Uniq(0)`
  // stack: [Uniq(0)]; not frozen

  // Make sure what `x` points to is accessible through raw pointers.
  EscapeToRaw(x);
  // stack: [Uniq(0), Shr]; not frozen
  
  let y1 = x as *mut u8;
  let y2 = y1;
  unsafe {
    // All of these first dereference a raw pointer (no checks, tag gets
    // ignored) and then perform a read or write access with `Shr(None)` as
    // the tag, which is already the top of the stack so nothing changes.
    *y1 = 3;
    *y2 = 5;
    *y2 = *y1;
  }

  // Writing to `x` again pops `Shr` off the stack, as per the rules for
  // write accesses.
  *x = 7;
  // stack: [Uniq(0)]; not frozen

  // Any further access through the raw pointers is undefined behavior, even
  // reads: The write to `x` re-asserted that `x` is the unique reference for
  // this memory.
  let _val = unsafe { *y1 };
}
{% endhighlight %}

The behavior of `EscapeToRaw` is best described as "reborrowing for a raw
pointer": The steps are the same as for `Retag` above, except that the new
pointer's tag is `Shr(None)` and we do not freeze (i.e., we behave as if the
entire pointee was inside an `UnsafeCell`).

Knowing about both `Retag` and `EscapeToRaw`, you can now go back to `demo2` and
should be able to fully explain why the stack changes the way it does in that
example.

### 3.3 The Case of the Aliasing References

Everything I described so far was pretty much in working condition as of about a
week ago.  However, there was one thorny problem that I only discovered fairly
late, and as usual it is best demonstrated by an example -- entirely in safe
code:

{% highlight rust %}
fn demo_refcell() {
  let rc = &mut RefCell::new(23u8);
  Retag(rc); // tag gets changed to `Uniq(0)`
  // We will consider the stack of the location where `23` is stored; the
  // `RefCell` bookkeeping counters are not of interest.
  // stack: [Uniq(0)]

  // Taking a shared reference shares the location but does not freeze, due
  // to the `UnsafeCell`.
  let rc_shr = &*rc;
  Retag(rc_shr); // tag gets changed to `Shr(Some(1))`
  // stack: [Uniq(0), Shr]; not frozen

  // Lots of stuff happens here but it does not matter for this example.
  let mut bmut = rc_shr.borrow_mut();
  
  // Obtain a mutable reference into the `RefCell`.
  let mut_ref = &mut *bmut;
  Retag(mut_ref); // tag gets changed to `Uniq(2)`
  // stack: [Uniq(0), Shr, Uniq(2)]; not frozen
  
  // And at the same time, a fresh shared reference to its outside!
  // This counts as a read access through `rc`, so we have to pop until
  // at least a `Shr` is at the top of the stack.
  let shr_ref = &*rc; // tag gets changed to `Shr(Some(3))`
  Retag(shr_ref);
  // stack: [Uniq(0), Shr]; not frozen

  // Now using `mut_ref` is UB because its tag is no longer on the stack.  But
  // that is bad, because it is usable in safe code.
  *mut_ref += 19;
}
{% endhighlight %}

Notice how `mut_ref` and `shr_ref` alias!  And yet, creating a shared reference
to the memory already covered by our unique `mut_ref` must not invalidate
`mut_ref`.  If we follow the instructions above, when we retag `shr_ref` after
it got created, we have no choice but pop the item matching `mut_ref` off the
stack.  Ouch.

This made me realize that creating a shared reference has to be very weak inside
`UnsafeCell`.  In fact, it is entirely equivalent to `EscapeToRaw`: We just have
to make sure some kind of shared access is possible, but we have to accept that
there might be active mutable references assuming exclusive access to the same
locations.  That on its own is not enough, though.

I also added a new check to the retagging procedure: Before taking any action
(i.e., before step 3, which could pop items off the stack), we check if the
reborrow is redundant: If the new reference we want to create is already
dereferencable (because its item is already on the stack and, if applicable, the
stack is already frozen), *and* if the item that justifies this is moreover
"derived from" the item that corresponds to the old reference, then we just do
nothing.  Here, "derived from" means "further up the stack".  Basically, the
reborrow has already happened and the new reference is ready for use; *and*
because of that "derived from" check, we know that using the new reference will
*not* pop the item corresponding to the old reference off the stack.  In that
case, we avoid popping anything, to keep other references valid.

It may seem like this rule can never apply, because how can our fresh tag match
something that's already on the stack?  This is indeed impossible for `Uniq`
tags, but for `Shr` tags, matching is more liberal.  For example, this rule
applies in our example above when we create `shr_ref` from `mut_ref`.  We do not
require freezing (because there is an `UnsafeCell`), there is already a `Shr` on
the stack (so the new reference is dereferencable) and the item matching the old
reference (`Uniq(0)`) is below that `Shr` (so after using the new reference, the
old one remains dereferencable).  Hence we do nothing, keeping the `Uniq(2)` on
the stack, such that the access through `mut_ref` at the end remains valid.

This may sound like a weird rule, and it is.  I would surely not have thought of
this if `RefCell` would not force our hands here.  However, as we shall see in
[section 5], it also does not to break any of the important properties of the
model (mutable references being unique and shared references being immutable
except for `UnsafeCell`).  Moreover, when pushing an item to the stack (at the
end of the retag action), we can now be sure that the stack is not yet frozen:
if it were frozen, the reborrow would be redundant.

With this extension, the instructions for retagging and `EscapeToRaw` now look
as follows (again executed on all locations covered by the reference, according
to `size_of_val`):

1. Compute a fresh tag: `Uniq(_)` for a mutable reference, `Shr(Some(_))` for a
   shared reference, `Shr(None)` if this is `EscapeToRaw`.
2. Perform the checks that would also happen when we dereference this reference.
   Remember the position of the item matching the tag in the stack.
3. Redundancy check: If the new tag passes the checks performed on a
   dereference, and if the item that makes this check succeed is *above* the one
   we remembered in step 2 (where the "frozen" state is considered above every
   item in the stack), then we stop.  We are done for this location.
4. Perform the actions that would also happen when an actual access happens
   through this reference (for shared references a read access, for mutable
   references a write access).<br>
   Now the location cannot be frozen any more: If the fresh tag is `Uniq`, we
   just unfroze; if the fresh tag is `Shr` and the location was already frozen,
   then the redundancy check (step 3) would have kicked in.
5. If the new tag is `Uniq`, push it onto the stack.
6. If the new tag is `Shr`, push a `Shr` item to the stack.  Then, if the
   location is outside of `UnsafeCell`, it gets frozen with the timestamp of the
   new reference.

The one thing I find slightly unsatisfying about the redundancy check is that it
seems to overlap a bit with the rule that on a *read* access, a `Shr` item
matches a `Uniq` tag.  Both of these together enable the read-only use of
mutable references that have already been shared; I would prefer to have a
single condition enabling that instead of two working together.  Still, overall
I think this is a pleasingly clean model; certainly much cleaner than what I
proposed last year and at the same time much more compatible with existing code.

## 4 Differences to the Original Proposal
[section 4]: #4-differences-to-the-original-proposal

The key differences to the original proposal is that the check performed on a
dereference, and the check performed on an access, are not the same check.  This
means there are more "moving parts" in the model, but it also means we do not
need a weird special exception (about reads from frozen locations) for `demo1`
any more like the original proposal did.  The main reason for this change,
however, is that on an access, we just do not know if we are inside an
`UnsafeCell` or not, so we cannot do all the checks we would like to do.
Accordingly, I also rearranged terminology a bit.  There is no longer one
"reactivation" action, instead there is a "deref" check and an "access" action,
as described above in sections [2.2][section 2.2] and [2.3][section 2.3].

Beyond that, I made the behavior of shared references and raw pointers more
uniform.  This helped to fix test failures around `iter_mut` on slices, which
first creates a raw reference and then a shared reference: In the original
model, creating the shared reference invalidates previously created raw
pointers.  As a result of the more uniform treatment, this no longer happens.
(Coincidentally, I did not make this change with the intention of fixing
`iter_mut`.  I did this change because I wanted to reduce the number of case
distinctions in the model.  Then I realized the relevant test suddenly passed
even with the full model enabled, investigated what happened, and realized I
accidentally had had a great idea. :D )

The tag is now "typed" (`Uniq` vs `Shr`) to be able to support `transmute`
between references and shared pointers.  Such `transmute` were an open question
in the original model and some people raised concerns about it in the ensuing
discussion.  I invite all of you to come up with strange things you think you
should be able to `transmute` and throw them at miri so that we can see if your
use-cases are covered. :)

Creating a shared reference now always pushes a `Shr` item onto the stack, even
when there is no `UnsafeCell`. This means that starting with a mutable reference
`x`, `&*x as *const _ as *mut _` is pretty much equivalent to `x as *mut _`; the
fact that we have an intermediate shared reference does not matter (not for the
aliasing model, anyway).  During the implementation, I realized that in `x as
*const _` on a mutable reference, `x` actually first gets coerced to shared
reference, which then gets cast to a raw pointer.  This happens in
`NonNull::from`, so if you later write to that `NonNull`, you end up writing to
a raw pointer that was created from a shared reference.  Originally I intended
this to be strictly illegal.  This is writing to a shared reference after all,
how dare you!  However, it turns out it's actually no big deal *if the shared
reference does not get used again later*.  This is an access-based model after
all, if a reference never gets used again we do not care much about enforcing
any guarantees for it.  (This is another example of a coincidental fix, where I
had a surprisingly passing test case and then investigated what happened.)

The redundancy check during retagging can be seen as refining a similar check
that the original model did whenever a new reference was created (where we
wouldn't change the state if the new borrow is already active).

Finally, the notion of "function barriers" from the original Stacked Borrows has
not been implemented yet.  This is the next item on my todo list.

## 5 Key Properties
[section 5]: #5-key-properties

Let us look at the two key properties that I set out as design goals, and see
how the model guarantees that they hold true in all valid (UB-free) executions.

### 5.1 Mutable References are Unique

The property I would like to establish here is that: After creating (retagging,
really) a `&mut`, if we then run some unknown code *that does not get passed the
reference*, and then we use the reference again (reading or writing), we can be
sure that this unknown code did not access the memory behind our mutable
reference at all (or we have UB).  For example:

{% highlight rust %}
fn demo_mut_unique(our: &mut i32) -> i32 {
  Retag(our); // So we can be sure the tag is unique

  *our = 5;

  unknown_code();

  // We know this will return 5, and moreover if `unknown_code` does not panic
  // we know we could do the write after calling `unknown_code` (because it
  // cannot even read from `our`).
  *our
}
{% endhighlight %}

The proof sketch goes as follows: After retagging the reference, we know it is
at the top of the stack and the location is not frozen.  (The "redundant
reborrow" rule does not apply because a fresh `Uniq` tag can never be
redundant.)  For any access performed by the unknown code, we know that access
cannot use the tag of our reference because the tags are unique and not
forgeable.  Hence if the unknown code accesses our locations, that would pop our
tag from the stack.  When we use our reference again, we know it is on the
stack, and hence has not been popped off.  Thus there cannot have been an access
from the unknown code.

Actually this theorem applies *any time* we have a reference whose tag we can be
sure has not been leaked to anyone else, and which points to locations which
have this tag at the top of the (unfrozen) stack.  This is not just the case
immediately after retagging.  We know our reference is at the top of the stack
after writing to it, so in the following example we know that `unknown_code_2`
cannot access `our`:

{% highlight rust %}
fn demo_mut_advanced_unique(our: &mut u8) -> u8 {
  Retag(our); // So we can be sure the tag is unique

  unknown_code_1(&*our);

  // This "re-asserts" uniqueness of the reference: After writing, we know
  // our tag is at the top of the stack.
  *our = 5;

  unknown_code_2();

  // We know this will return 5
  *our
}
{% endhighlight %}

### 5.2 Shared References (without `UnsafeCell)` are Immutable

The key property of shared references is that: After creating (retagging,
really) a shared reference, if we then run some unknown code (it can even have
our reference if it wants), and then we use the reference again, we know that
the value pointed to by the reference has not been changed.  For example:

{% highlight rust %}
fn demo_shr_frozen(our: &u8) -> u8 {
  Retag(our); // So we can be sure the tag actually carries a timestamp

  // See what's in there.
  let val = *our;
  
  unknown_code(our);

  // We know this will return `val`
  *our
}
{% endhighlight %}

The proof sketch goes as follows: After retagging the reference, we know the
location is frozen (this is the case even if the "redundant reborrow" rule
applies).  If the unknown code does any write, we know this will unfreeze the
location.  The location might get re-frozen, but only at the then-current
timestamp.  When we do our read after coming back from the unknown code, this
checks that the location is frozen *at least* since the timestamp given in its
tag, so if the location is unfrozen or got re-frozen by the unknown code, the
check would fail.  Thus the unknown code cannot have written to the location.

One interesting observation here for both of these proofs is that all we rely on
when the unknown code is executed are the actions performed on every memory
access.  The additional checks that happen when a pointer is dereferenced only
matter in *our* code, not in the foreign code.  Hence we have no problem
reasoning about the case where we call some code via FFI that is written in a
language without a notion of "dereferencing", all we care about is the actual
memory accesses performed by that foreign code.  This also indicates that we
could see the checks on pointer dereference as another "shadow state operation"
next to `Retag` and `EscapeToRaw`, and then these three operations plus the
actions on memory accesses are all that there is to Stacked Borrows.  This is
difficult to implement in miri because dereferences can happen any time a path
is evaluated, but it is nevertheless interesting and might be useful in a
"lower-level MIR" that does not permit dereferences in paths.

## 6 Evaluation, and How You Can Help
[section 6]: #6-evaluation-and-how-you-can-help

I have implemented both the validity invariant and the model as described above
in miri. This [uncovered](https://github.com/rust-lang/rust/issues/54908) two
[issues](https://github.com/rust-lang/rust/issues/54957) in the standard
library, but both were related to validity invariants, not Stacked Borrows.
With these exceptions, the model passes the entire test suite.  There were some
more test failures in earlier versions (as mentioned in [section 4]), but the
final model accepts all the code covered by miri's test suite.  (If you look
close enough, you can see that three libstd methods are currently whitelisted
and what they do is not checked.  However, even before I ran into these cases,
[efforts](https://github.com/rust-lang/rust/pull/54668) were already
[underway](https://github.com/rust-lang/rfcs/pull/2582) that would fix all of
them, so I am not concerned about them.)  Moreover I wrote a bunch of
compile-fail tests to make sure the model catches various violations of the key
properties it should ensure.

I am quite happy with this!  I was expecting much more trouble, expecting to run
into cases where libstd does strange things that are common or otherwise hard to
declare illegal and that my model could not reasonably allow.  I see the test
suite passing as an indication that this model may be well-suited for Rust.

However, miri's test suite is tiny, and I have but one brain to come up with
counterexamples!  In fact I am quite a bit worried because I literally came up
with `demo_refcell` less than two weeks ago, so what else might I have missed?
This where you come in.  Please test this model!  Come up with something funny
you think should work (I am thinking about funny `transmute` in particular,
using type punning through unions or raw pointers if you prefer that), or maybe
you have some crate that has some unsafe code and a test suite (you do have a
test suite, right?) that might run under miri.

The easiest way to try the model is the
[playground](https://play.rust-lang.org/): Type the code, select "Tools - Miri",
and you'll see what it does.

For things that are too long for the playground, you have to install miri on
your own computer.  miri depends on rustc nightly and has to be updated
regularly to keep working, so it is not well-suited for crates.io.  Instead,
installation instructions for miri are provided
[in the README](https://github.com/solson/miri/#running-miri).  We are still
working on making installing miri easier.  Please let me know if you are having
trouble with anything.  You can report issues, comment on this post or find me
in chat (as of recently, I am partial to Zulip where we have an
[unsafe code guidelines stream](https://rust-lang.zulipchat.com/#narrow/stream/136281-wg-unsafe-code-guidelines)).

With miri installed, you can `cargo miri` a project with a binary to run it in
miri.  Dependencies should be fully supported, so you can use any crate you
like.  It is not unlikely, however, that you will run into issues because miri
does not support some operation.  In that case please search the
[issue tracker](https://github.com/solson/miri/issues) and report the issue if
it is new.  We cannot support everything, but we might be able to do something
for your case.

Unfortunately, `cargo miri test` is currently broken; if you want to help with
that [here are some details](https://github.com/solson/miri/issues/479).
Moreover, wouldn't it be nice if we could
[run the entire libcore, liballoc and libstd test suite in miri](https://github.com/rust-lang/rust/issues/54914)?
There are tons of interesting cases of Rust's core data structures being
exercise there, and the comparatively tiny miri test suite has already helped to
find two soundness bugs, so there are probably more.  Once `cargo miri test`
works again, it would be great to find a way to run it on the standard library
test suites, and set up something so that this happens automatically on a
regular basis (so that we notice regressions).

As you can see, there is more than enough work for everyone.  Don't be shy!  I
have a mere two weeks left on this internship, after which I will have to
significantly reduce my Rust activities in favor of finishing my PhD.  I won't
disappear entirely though, don't worry -- I will still be able to mentor you if
you want to help with any of the above tasks. :)

Thanks to @nikomatsakis for feedback on a draft of this post, to @shepmaster for
making miri available on the playground, and to @oli-obk for reviewing all my
PRs at unparalleled speed. <3

If you want to
help or report results of your experiments, if you have any questions or
comments, please join the
[discussion in the forums](https://internals.rust-lang.org/t/stacked-borrows-implemented/8847).