rust/lifetimes.md

% Ownership and Lifetimes

Ownership is the breakout feature of Rust. It allows Rust to be completely memory-safe and efficient, while avoiding garbage collection. Before getting into the ownership system in detail, we will consider the motivation of this design.

TODO: Interior Mutability section

Living Without Garbage Collection

We will assume that you accept that garbage collection is not always an optimal solution, and that it is desirable to manually manage memory to some extent. If you do not accept this, might I interest you in a different language?

Regardless of your feelings on GC, it is pretty clearly a massive boon to making code safe. You never have to worry about things going away too soon (although whether you still wanted to be pointing at that thing is a different issue...). This is a pervasive problem that C and C++ need to deal with. Consider this simple mistake that all of us who have used a non-GC'd language have made at one point:

fn as_str(data: &u32) -> &str {
    // compute the string
    let s = format!("{}", data);

    // OH NO! We returned a reference to something that
    // exists only in this function!
    // Dangling pointer! Use after free! Alas!
    // (this does not compile in Rust)
    &s
}

This is exactly what Rust's ownership system was built to solve. Rust knows the scope in which the &s lives, and as such can prevent it from escaping. However this is a simple case that even a C compiler could plausibly catch. Things get more complicated as code gets bigger and pointers get fed through various functions. Eventually, a C compiler will fall down and won't be able to perform sufficient escape analysis to prove your code unsound. It will consequently be forced to accept your program on the assumption that it is correct.

This will never happen to Rust. It's up to the programmer to prove to the compiler that everything is sound.

Of course, rust's story around ownership is much more complicated than just verifying that references don't escape the scope of their referrent. That's because ensuring pointers are always valid is much more complicated than this. For instance in this code,

let mut data = vec![1, 2, 3];
// get an internal reference
let x = &data[0];

// OH NO! `push` causes the backing storage of `data` to be reallocated.
// Dangling pointer! User after free! Alas!
// (this does not compile in Rust)
data.push(4);

println!("{}", x);

naive scope analysis would be insufficient to prevent this bug, because data does in fact live as long as we needed. However it was changed while we had a reference into it. This is why Rust requires any references to freeze the referrent and its owners.

References

There are two kinds of reference:

• Shared reference: &
• Mutable reference: &mut

Which obey the following rules:

• A reference cannot outlive its referrent
• A mutable reference cannot be aliased

To define aliasing, we must define the notion of paths and liveness.

Paths

If all Rust had were values, then every value would be uniquely owned by a variable or composite structure. From this we naturally derive a tree of ownership. The stack itself is the root of the tree, with every variable as its direct children. Each variable's direct children would be their fields (if any), and so on.

From this view, every value in Rust has a unique path in the tree of ownership. References to a value can subsequently be interpretted as a path in this tree. Of particular interest are prefixes: x is a prefix of y if x owns y

However much data doesn't reside on the stack, and we must also accomodate this. Globals and thread-locals are simple enough to model as residing at the bottom of the stack. However data on the heap poses a different problem.

If all Rust had on the heap was data uniquely by a pointer on the stack, then we can just treat that pointer as a struct that owns the value on the heap. Box, Vec, String, and HashMap, are examples of types which uniquely own data on the heap.

Unfortunately, data on the heap is not always uniquely owned. Rc for instance introduces a notion of shared ownership. Shared ownership means there is no unique path. A value with no unique path limits what we can do with it. In general, only shared references can be created to these values. However mechanisms which ensure mutual exclusion may establish One True Owner temporarily, establishing a unique path to that value (and therefore all its children).

The most common way to establish such a path is through interior mutability, in contrast to the inherited mutability that everything in Rust normally uses. Cell, RefCell, Mutex, and RWLock are all examples of interior mutability types. These types provide exclusive access through runtime restrictions. However it is also possible to establish unique ownership without interior mutability. For instance, if an Rc has refcount 1, then it is safe to mutate or move its internals.

Liveness

Roughly, a reference is live at some point in a program if it can be dereferenced. Shared references are always live unless they are literally unreachable (for instance, they reside in freed or leaked memory). Mutable references can be reachable but not live through the process of reborrowing.

A mutable reference can be reborrowed to either a shared or mutable reference. Further, the reborrow can produce exactly the same reference, or point to a path it is a prefix of. For instance, a mutable reference can be reborrowed to point to a field of its referrent:

let x = &mut (1, 2);
{
    // reborrow x to a subfield
    let y = &mut x.0;
    // y is now live, but x isn't
    *y = 3;
}
// y goes out of scope, so x is live again
*x = (5, 7);

It is also possible to reborrow into multiple mutable references, as long as they are to disjoint: no reference is a prefix of another. Rust explicitly enables this to be done with disjoint struct fields, because disjointness can be statically proven:

let x = &mut (1, 2);
{
    // reborrow x to two disjoint subfields
    let y = &mut x.0;
    let z = &mut x.1;
    // y and z are now live, but x isn't
    *y = 3;
    *z = 4;
}
// y and z go out of scope, so x is live again
*x = (5, 7);

However it's often the case that Rust isn't sufficiently smart to prove that multiple borrows are disjoint. This does not mean it is fundamentally illegal to make such a borrow, just that Rust isn't as smart as you want.

To simplify things, we can model variables as a fake type of reference: owned references. Owned references have much the same semantics as mutable references: they can be re-borrowed in a mutable or shared manner, which makes them no longer live. Live owned references have the unique property that they can be moved out of (though mutable references can be swapped out of). This is only given to live owned references because moving its referrent would of course invalidate all outstanding references prematurely.

As a local lint against inappropriate mutation, only variables that are marked as mut can be borrowed mutably.

It is also interesting to note that Box behaves exactly like an owned reference. It can be moved out of, and Rust understands it sufficiently to reason about its paths like a normal variable.

Aliasing

With liveness and paths defined, we can now properly define aliasing:

A mutable reference is aliased if there exists another live reference to it or one of its prefixes.

That's it. Super simple right? Except for the fact that it took us two pages to define all of the terms in that defintion. You know: Super. Simple.

Actually it's a bit more complicated than that. In addition to references, Rust has raw pointers: *const T and *mut T. Raw pointers have no inherent ownership or aliasing semantics. As a result, Rust makes absolutely no effort to track that they are used correctly, and they are wildly unsafe.

It is an open question to what degree raw pointers have alias semantics. However it is important for these definitions to be sound that the existence of a raw pointer does not imply some kind of live path.

Lifetimes

Rust enforces these rules through lifetimes. Lifetimes are effectively just names for scopes on the stack, somewhere in the program. Each reference, and anything that contains a reference, is tagged with a lifetime specifying the scope it's valid for.

Within a function body, Rust generally doesn't let you explicitly name the lifetimes involved. This is because it's generally not really necessary to talk about lifetimes in a local context; rust has all the information and can work out everything.

However once you cross the function boundary, you need to start talking about lifetimes. Lifetimes are denoted with an apostrophe: 'a, 'static. To dip our toes with lifetimes, we're going to pretend that we're actually allowed to label scopes with lifetimes, and desugar the examples from the start of this chapter.

Our examples made use of aggressive sugar around scopes and lifetimes, because writing everything out explicitly is extremely noisy. All rust code relies on aggressive inference and elision of "obvious" things.

One particularly interesting piece of sugar is that each let statement implicitly introduces a scope. For the most part, this doesn't really matter. However it does matter for variables that refer to each other. As a simple example, let's completely desugar this simple piece of Rust code:

let x = 0;
let y = &x;
let z = &y;

becomes:

// NOTE: `'a:` and `&'a x` is not valid syntax!
'a: {
    let x: i32 = 0;
    'b: {
        let y: &'a i32 = &'a x;
        'c: {
            let z: &'b &'a i32 = &'b y;
        }
    }
}

Wow. That's... awful. Let's all take a moment to thank Rust for being a huge pile of sugar with sugar on top.

Anyway, let's look at some of those examples from before:

fn as_str(data: &u32) -> &str {
    let s = format!("{}", data);
    &s
}

desugars to:

fn as_str<'a>(data: &'a u32) -> &'a str {
    'b: {
        let s = format!("{}", data);
        return &'b s
    }
}

This signature of as_str takes a reference to a u32 with some lifetime, and promises that it can produce a reference to a str that can live just as long. Already we can see why this signature might be trouble. That basically implies that we're going to find a str somewhere in the scope that u32 originated in, or somewhere even earlier. That's uh... a big ask.

We then proceed to compute the string s, and return a reference to it. Unfortunately, since s was defined in the scope 'b, the reference we're returning can only live for that long. From the perspective of the compiler, we've failed twice here. We've failed to fulfill the contract we were asked to fulfill ('b is unrelated to 'a); and we've also tried to make a reference outlive its referrent by returning an &'b, where 'b is in our function.

Shoot!

Of course, the right way to right this function is as follows:

fn to_string(data: &u32) -> String {
    format!("{}", data)
}

We must produce an owned value inside the function to return it! The only way we could have returned an &'a str would have been if it was in a field of the &'a u32, which is obviously not the case.

(Actually we could have also just returned a string literal, though this limits the behaviour of our function just a bit.)

How about the other example:

let mut data = vec![1, 2, 3];
let x = &data[0];
data.push(4);
println!("{}", x);

'a: {
    let mut data: Vec<i32> = vec![1, 2, 3];
    'b: {
        let x: &'a i32 = Index::index(&'a data, 0);
        'c: {
            // Exactly what the desugar for Vec::push is is up to Rust.
            // This particular desugar is a decent approximation for our
            // purpose. In particular methods oft invoke a temporary borrow.
            let temp: &'c mut Vec = &'c mut data;
            // NOTE: Vec::push is not valid syntax
            Vec::push(temp, 4);
        }
        println!("{}", x);
    }
}

Here the problem is that we're trying to mutably borrow the data path, while we have a reference into something it's a prefix of. Rust subsequently throws up its hands in disgust and rejects our program. The correct way to write this is to just re-order the code so that we make x after we push:

TODO: convince myself of this.

let mut data = vec![1, 2, 3];
data.push(4);

let x = &data[0];
println!("{}", x);

Lifetime Elision

In order to make common patterns more ergonomic, Rust allows lifetimes to be elided in function signatures.

A lifetime position is anywhere you can write a lifetime in a type:

&'a T
&'a mut T
T<'a>

Lifetime positions can appear as either "input" or "output":

• For fn definitions, input refers to the types of the formal arguments in the fn definition, while output refers to result types. So fn foo(s: &str) -> (&str, &str) has elided one lifetime in input position and two lifetimes in output position. Note that the input positions of a fn method definition do not include the lifetimes that occur in the method's impl header (nor lifetimes that occur in the trait header, for a default method).

• In the future, it should be possible to elide impl headers in the same manner.

Elision rules are as follows:

• Each elided lifetime in input position becomes a distinct lifetime parameter.

• If there is exactly one input lifetime position (elided or not), that lifetime is assigned to all elided output lifetimes.

• If there are multiple input lifetime positions, but one of them is &self or &mut self, the lifetime of self is assigned to all elided output lifetimes.

• Otherwise, it is an error to elide an output lifetime.

Examples:

fn print(s: &str);                                      // elided
fn print<'a>(s: &'a str);                               // expanded

fn debug(lvl: uint, s: &str);                           // elided
fn debug<'a>(lvl: uint, s: &'a str);                    // expanded

fn substr(s: &str, until: uint) -> &str;                // elided
fn substr<'a>(s: &'a str, until: uint) -> &'a str;      // expanded

fn get_str() -> &str;                                   // ILLEGAL

fn frob(s: &str, t: &str) -> &str;                      // ILLEGAL

fn get_mut(&mut self) -> &mut T;                        // elided
fn get_mut<'a>(&'a mut self) -> &'a mut T;              // expanded

fn args<T:ToCStr>(&mut self, args: &[T]) -> &mut Command                  // elided
fn args<'a, 'b, T:ToCStr>(&'a mut self, args: &'b [T]) -> &'a mut Command // expanded

fn new(buf: &mut [u8]) -> BufWriter;                    // elided
fn new<'a>(buf: &'a mut [u8]) -> BufWriter<'a>          // expanded

Unbounded Lifetimes

Unsafe code can often end up producing references or lifetimes out of thin air. Such lifetimes come into the world as unbounded. The most common source of this is derefencing a raw pointer, which produces a reference with an unbounded lifetime. Such a lifetime becomes as big as context demands. This is in fact more powerful than simply becoming 'static, because for instance &'static &'a T will fail to typecheck, but the unbound lifetime will perfectly mold into &'a &'a T as needed. However for most intents and purposes, such an unbounded lifetime can be regarded as 'static.

Almost no reference is 'static, so this is probably wrong. transmute and transmute_copy are the two other primary offenders. One should endeavour to bound an unbounded lifetime as quick as possible, especially across function boundaries.

Given a function, any output lifetimes that don't derive from inputs are unbounded. For instance:

fn get_str<'a>() -> &'a str;

will produce an &str with an unbounded lifetime. The easiest way to avoid unbounded lifetimes is to use lifetime elision at the function boundary. If an output lifetime is elided, then it must be bounded by an input lifetime. Of course, it might be bounded by the wrong lifetime, but this will usually just cause a compiler error, rather than allow memory safety to be trivially violated.

Within a function, bounding lifetimes is more error-prone. The safest and easiest way to bound a lifetime is to return it from a function with a bound lifetime. However if this is unacceptable, the reference can be placed in a location with a specific lifetime. Unfortunately it's impossible to name all lifetimes involved in a function. To get around this, you can in principle use copy_lifetime, though these are unstable due to their awkward nature and questionable utility.

Higher-Rank Trait Bounds

// TODO: make aturon less mad

Generics in Rust generally allow types to be instantiated with arbitrary associated lifetimes, but this fixes the lifetimes they work with once instantiated. For almost all types, this is exactly the desired behaviour. For example slice::Iter can work with arbitrary lifetimes, determined by the slice that instantiates it. However once Iter is instantiated the lifetimes it works with cannot be changed. It returns references that live for some particular 'a.

However some types are more flexible than this. In particular, a single instantiation of a function can process arbitrary lifetimes:

fn identity(input: &u8) -> &u8 { input }

What is the lifetime that identity works with? There is none. If you think this is "cheating" because functions are statically instantiated, then you need only consider the equivalent closure:

let identity = |input: &u8| input;

These functions are higher ranked over the lifetimes they work with. This means that they're generic over what they handle after instantiation. For most things this would pose a massive problem, but because lifetimes don't exist at runtime, this is really just a compile-time mechanism. The Fn traits contain sugar that allows higher-rank lifetimes to simply be expressed by simply omitting lifetimes:

fn main() {
    foo(|input| input);
}

fn foo<F>(f: F)
    // F is higher-ranked over the lifetime these references have
    where F: Fn(&u8) -> &u8
{
    f(&0);
    f(&1);
}

The desugaring of this is actually unstable:

#![feature(unboxed_closures)]

fn main() {
    foo(|input| input);
}

fn foo<F>(f: F)
    where F: for<'a> Fn<(&'a u8,), Output=&'a u8>
{
    f(&0);
    f(&1);
}

for<'a> is how we declare a higher-ranked lifetime. Unfortunately higher-ranked lifetimes are still fairly new, and are missing a few features to make them maximally useful outside of the Fn traits.

Subtyping and Variance

Although Rust doesn't have any notion of inheritance, it does include subtyping. In Rust, subtyping derives entirely from lifetimes. Since lifetimes are derived from scopes, we can partially order them based on an outlives relationship. We can even express this as a generic bound: T: 'a specifies that T outlives 'a.

We can then define subtyping on lifetimes in terms of lifetimes: if 'a : 'b ("a outlives b"), then 'a is a subtype of b. This is a large source of confusion, because a bigger scope is a sub type of a smaller scope. This does in fact make sense. The intuitive reason for this is that if you expect an &'a u8, then it's totally fine for me to hand you an &'static u8 in the same way that if you expect an Animal in Java, it's totally fine for me to hand you a Cat.

(Note, the subtyping relationship and typed-ness of lifetimes is a fairly arbitrary construct that some disagree with. I just find that it simplifies this analysis.)

TODO: higher rank lifetime subtyping

Variance is where things get really harsh.

Variance is a property that type constructors have. A type constructor in Rust is a generic type with unbound arguments. For instance Vec is a type constructor that takes a T and returns a Vec<T>. & and &mut are type constructors that take a lifetime and a type.

A type constructor's variance is how the subtypes of its inputs affects the subtypes of its outputs. There are three kinds of variance:

• F is variant if T being a subtype of U implies F<T> is a subtype of F<U>
• F is invariant otherwise (no subtyping relation can be derived)

(For those of you who are familiar with variance from other languages, what we refer to as "just" variant is in fact covariant. Rust does not have contravariance. Historically Rust did have some contravariance but it was scrapped due to poor interactions with other features.)

Some important variances:

• & is variant (as is *const by metaphor)
• &mut is invariant (as is *mut by metaphor)
• Fn(T) -> U is invariant with respect to T, but variant with respect to U
• Box, Vec, and all other collections are variant
• UnsafeCell, Cell, RefCell, Mutex and all "interior mutability" types are invariant

To understand why these variances are correct and desirable, we will consider several examples. We have already covered why & should be variant when introducing subtyping: it's desirable to be able to pass longer-lived things where shorter-lived things are needed.

To see why &mut should be invariant, consider the following code:

fn main() {
    let mut forever_str: &'static str = "hello";
    {
        let string = String::from("world");
        overwrite(&mut forever_str, &mut &*string);
    }
    println!("{}", forever_str);
}

fn overwrite<T: Copy>(input: &mut T, new: &mut T) {
    *input = *new;
}

The signature of overwrite is clearly valid: it takes mutable references to two values of the same type, and replaces one with the other. We have seen already that & is variant, and 'static is a subtype of any 'a, so &'static str is a subtype of &'a str. Therefore, if &mut was also variant, then the lifetime of the &'static str would successfully be "shrunk" down to the shorter lifetime of the string, and replace would be called successfully. The string would subsequently be dropped, and forever_str would point to freed memory when we print it!

Therefore &mut should be invariant. This is the general theme of variance vs invariance: if variance would allow you to store a short-lived value in a longer-lived slot, then you must be invariant.

Box and Vec are interesting cases because they're variant, but you can definitely store values in them! This is fine because you can only store values in them through a mutable reference! The mutable reference makes the whole type invariant, and therefore prevents you from getting in trouble.

Being variant allows them to be variant when shared immutably (so you can pass a &Box<&'static str> where a &Box<&'a str> is expected). It also allows you to forever weaken the type by moving it into a weaker slot. That is, you can do:

fn get_box<'a>(&'a u8) -> Box<&'a str> {
    // string literals are `&'static str`s
    Box::new("hello")
}

which is fine because unlike the mutable borrow case, there's no one else who "remembers" the old lifetime in the box.

The variance of the cell types similarly follows. & is like an &mut for a cell, because you can still store values in them through an &. Therefore cells must be invariant to avoid lifetime smuggling.

Fn is the most subtle case, because it has mixed variance. To see why Fn(T) -> U should be invariant over T, consider the following function signature:

// 'a is derived from some parent scope
fn foo(&'a str) -> usize;

This signature claims that it can handle any &str that lives at least as long as 'a. Now if this signature was variant with respect to &str, that would mean

fn foo(&'static str) -> usize;

could be provided in its place, as it would be a subtype. However this function has a stronger requirement: it says that it can only handle &'static strs, and nothing else. Therefore functions are not variant over their arguments.

To see why Fn(T) -> U should be variant over U, consider the following function signature:

// 'a is derived from some parent scope
fn foo(usize) -> &'a str;

This signature claims that it will return something that outlives 'a. It is therefore completely reasonable to provide

fn foo(usize) -> &'static str;

in its place. Therefore functions are variant over their return type.

*const has the exact same semantics as &, so variance follows. *mut on the other hand can dereference to an &mut whether shared or not, so it is marked as invariant in analogy to cells.

This is all well and good for the types the standard library provides, but how is variance determined for type that you define? A struct, informally speaking, inherits the variance of its fields. If a struct Foo has a generic argument A that is used in a field a, then Foo's variance over A is exactly a's variance. However this is complicated if A is used in multiple fields.

• If all uses of A are variant, then Foo is variant over A
• Otherwise, Foo is invariant over A

struct Foo<'a, 'b, A, B, C, D, E, F, G, H> {
    a: &'a A,     // variant over 'a and A
    b: &'b mut B, // invariant over 'b and B
    c: *const C,  // variant over C
    d: *mut D,    // invariant over D
    e: Vec<E>,    // variant over E
    f: Cell<F>,   // invariant over F
    g: G          // variant over G
    h1: H         // would also be variant over H except...
    h2: Cell<H>   // invariant over H, because invariance wins
}

PhantomData

When working with unsafe code, we can often end up in a situation where types or lifetimes are logically associated with a struct, but not actually part of a field. This most commonly occurs with lifetimes. For instance, the Iter for &'a [T] is (approximately) defined as follows:

pub struct Iter<'a, T: 'a> {
    ptr: *const T,
    end: *const T,
}

However because 'a is unused within the struct's body, it's unbound. Because of the troubles this has historically caused, unbound lifetimes and types are illegal in struct definitions. Therefore we must somehow refer to these types in the body. Correctly doing this is necessary to have correct variance and drop checking.

We do this using PhantomData, which is a special marker type. PhantomData consumes no space, but simulates a field of the given type for the purpose of static analysis. This was deemed to be less error-prone than explicitly telling the type-system the kind of variance that you want, while also providing other useful information.

Iter logically contains &'a T, so this is exactly what we tell the PhantomData to simulate:

pub struct Iter<'a, T: 'a> {
    ptr: *const T,
    end: *const T,
    _marker: marker::PhantomData<&'a T>,
}

Dropck

When a type is going out of scope, Rust will try to Drop it. Drop executes arbitrary code, and in fact allows us to "smuggle" arbitrary code execution into many places. As such additional soundness checks (dropck) are necessary to ensure that a type T can be safely instantiated and dropped. It turns out that we really don't need to care about dropck in practice, as it often "just works".

However the one exception is with PhantomData. Given a struct like Vec:

struct Vec<T> {
    data: *const T, // *const for variance!
    len: usize,
    cap: usize,
}

dropck will generously determine that Vec does not own any values of type T. This will unfortunately allow people to construct unsound Drop implementations that access data that has already been dropped. In order to tell dropck that we do own values of type T, and may call destructors of that type, we must add extra PhantomData:

struct Vec<T> {
    data: *const T, // *const for covariance!
    len: usize,
    cap: usize,
    _marker: marker::PhantomData<T>,
}

Raw pointers that own an allocation is such a pervasive pattern that the standard library made a utility for itself called Unique<T> which:

• wraps a *const T,
• includes a PhantomData,
• auto-derives Send/Sync as if T was contained
• marks the pointer as NonZero for the null-pointer optimization

Splitting Lifetimes

The mutual exclusion property of mutable references can be very limiting when working with a composite structure. Borrowck understands some basic stuff, but will fall over pretty easily. Borrowck understands structs sufficiently to understand that it's possible to borrow disjoint fields of a struct simultaneously. So this works today:

struct Foo {
    a: i32,
    b: i32,
    c: i32,
}

let mut x = Foo {a: 0, b: 0, c: 0};
let a = &mut x.a;
let b = &mut x.b;
let c = &x.c;
*b += 1;
let c2 = &x.c;
*a += 10;
println!("{} {} {} {}", a, b, c, c2);

However borrowck doesn't understand arrays or slices in any way, so this doesn't work:

let x = [1, 2, 3];
let a = &mut x[0];
let b = &mut x[1];
println!("{} {}", a, b);

<anon>:3:18: 3:22 error: cannot borrow immutable indexed content `x[..]` as mutable
<anon>:3     let a = &mut x[0];
                          ^~~~
<anon>:4:18: 4:22 error: cannot borrow immutable indexed content `x[..]` as mutable
<anon>:4     let b = &mut x[1];
                          ^~~~
error: aborting due to 2 previous errors

While it was plausible that borrowck could understand this simple case, it's pretty clearly hopeless for borrowck to understand disjointness in general container types like a tree, especially if distinct keys actually do map to the same value.

In order to "teach" borrowck that what we're doing is ok, we need to drop down to unsafe code. For instance, mutable slices expose a split_at_mut function that consumes the slice and returns two mutable slices. One for everything to the left of the index, and one for everything to the right. Intuitively we know this is safe because the slices don't alias. However the implementation requires some unsafety:

fn split_at_mut(&mut self, mid: usize) -> (&mut [T], &mut [T]) {
    unsafe {
        let self2: &mut [T] = mem::transmute_copy(&self);

        (ops::IndexMut::index_mut(self, ops::RangeTo { end: mid } ),
         ops::IndexMut::index_mut(self2, ops::RangeFrom { start: mid } ))
    }
}

This is pretty plainly dangerous. We use transmute to duplicate the slice with an unbounded lifetime, so that it can be treated as disjoint from the other until we unify them when we return.

However more subtle is how iterators that yield mutable references work. The iterator trait is defined as follows:

trait Iterator {
    type Item;

    fn next(&mut self) -> Option<Self::Item>;
}

Given this definition, Self::Item has no connection to self. This means that we can call next several times in a row, and hold onto all the results concurrently. This is perfectly fine for by-value iterators, which have exactly these semantics. It's also actually fine for shared references, as they admit arbitrarily many references to the same thing (although the iterator needs to be a separate object from the thing being shared). But mutable references make this a mess. At first glance, they might seem completely incompatible with this API, as it would produce multiple mutable references to the same object!

However it actually does work, exactly because iterators are one-shot objects. Everything an IterMut yields will be yielded at most once, so we don't actually ever yield multiple mutable references to the same piece of data.

In general all mutable iterators require some unsafe code somewhere, though. Whether it's raw pointers, or safely composing on top of another IterMut.

For instance, VecDeque's IterMut:

pub struct IterMut<'a, T:'a> {
    // The whole backing array. Some of these indices are initialized!
    ring: &'a mut [T],
    tail: usize,
    head: usize,
}

impl<'a, T> Iterator for IterMut<'a, T> {
    type Item = &'a mut T;

    fn next(&mut self) -> Option<&'a mut T> {
        if self.tail == self.head {
            return None;
        }
        let tail = self.tail;
        self.tail = wrap_index(self.tail.wrapping_add(1), self.ring.len());

        unsafe {
            // might as well do unchecked indexing since wrap_index has us
            // in-bounds, and many of the "middle" indices are uninitialized
            // anyway.
            let elem = self.ring.get_unchecked_mut(tail);

            // round-trip through a raw pointer to unbound the lifetime from
            // ourselves
            Some(&mut *(elem as *mut _))
        }
    }
}

A very subtle but interesting detail in this design is that it relies on privacy to be sound. Borrowck works on some very simple rules. One of those rules is that if we have a live &mut Foo and Foo contains an &mut Bar, then that &mut Bar is also live. Since IterMut is always live when next can be called, if ring were public then we could mutate ring while outstanding mutable borrows to it exist!

Weird Lifetimes

Given the following code:

struct Foo;

impl Foo {
    fn mutate_and_share(&mut self) -> &Self { &*self }
    fn share(&self) {}
}

fn main() {
    let mut foo = Foo;
    let loan = foo.mutate_and_share();
    foo.share();
}

One might expect it to compile. We call mutate_and_share, which mutably borrows foo temporarily, but then returns only a shared reference. Therefore we would expect foo.share() to succeed as foo shouldn't be mutably borrowed.

However when we try to compile it:

<anon>:11:5: 11:8 error: cannot borrow `foo` as immutable because it is also borrowed as mutable
<anon>:11     foo.share();
              ^~~
<anon>:10:16: 10:19 note: previous borrow of `foo` occurs here; the mutable borrow prevents subsequent moves, borrows, or modification of `foo` until the borrow ends
<anon>:10     let loan = foo.mutate_and_share();
                         ^~~
<anon>:12:2: 12:2 note: previous borrow ends here
<anon>:8 fn main() {
<anon>:9     let mut foo = Foo;
<anon>:10     let loan = foo.mutate_and_share();
<anon>:11     foo.share();
<anon>:12 }
          ^

What happened? Well, the lifetime of loan is derived from a mutable borrow. This makes the type system believe that foo is mutably borrowed as long as loan exists, even though it's a shared reference. This isn't a bug, although one could argue it is a limitation of the design. In particular, to know if the mutable part of the borrow is really expired we'd have to peek into implementation details of the function. Currently, type-checking a function does not need to inspect the bodies of any other functions or types.