--- /dev/null
+---
+title: "Stacked Borrows Implemented"
+categories: internship rust
+---
+
+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.
+<!-- 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](). -->
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