17 KiB
% Ownership
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 a simple but fundamental language-design problem.
The Tagged Union Problem
The core of the lifetime and mutability system derives from a simple problem: internal pointers to tagged unions. For instance, consider the following code:
enum Foo {
A(u32),
B(f64),
}
let mut x = B(2.0);
if let B(ref mut y) = x {
*x = A(7);
// OH NO! a u32 has been interpretted as an f64! Type-safety hole!
// (this does not actually compile)
println!("{}", y);
}
The problem here is an intersection of 3 choices:
- data in a tagged union is inline with the tag
- tagged unions are mutable
- being able to take a pointer into a tagged union
Remove any of these 3 and the problem goes away. Traditionally, functional languages have avoided this problem by removing the mutable option. This means that they can in principle keep their data inline (ghc has a pragma for this). A garbage collected imperative language like Java could alternatively solve this problem by just keeping all variants elsewhere, so that changing the variant of a tagged union just overwrites a pointer, and anyone with an outstanding pointer to the inner data is unaffected thanks to The Magic Of Garbage Collection.
Rust, by contrast, takes a subtler approach. Rust allows mutation, allows pointers to inner data, and its enums have their data allocated inline. However it prevents anything from being mutated while there are outstanding pointers to it! And this is all done at compile time.
Interestingly, Rust's std::cell
module exposes two types that offer an alternative
approach to this problem:
-
The
Cell
type allows mutation of aliased data, but instead forbids internal pointers to that data. The only way to read or write a Cell is to copy the bits in or out. -
The
RefCell
type allows mutation of aliased data and internal pointers, but manages this through runtime checks. It is effectively a thread-unsafe read-write lock.
Lifetimes
Rust's static checks are managed by the borrow checker (borrowck), which tracks mutability and outstanding loans. This analysis can in principle be done without any help locally. However as soon as data starts crossing the function boundary, we have some serious trouble. In principle, borrowck could be a massive whole-program analysis engine to handle this problem, but this would be an atrocious solution. It would be terribly slow, and errors would be horribly non-local.
Instead, Rust tracks ownership through lifetimes. Every single reference and value in Rust is tagged with a lifetime that indicates the scope it is valid for. Rust has two kinds of reference:
- Shared reference:
&
- Mutable reference:
&mut
The main rules are as follows:
- A shared reference can be aliased
- A mutable reference cannot be aliased
- A reference cannot outlive its referrent (
&'a T -> T: 'a
)
However non-mutable variables have some special rules:
- You cannot mutate or mutably borrow a non-mut variable,
Only variables marked as mutable can be borrowed mutably, though this is little more than a local lint against incorrect usage of a value.
Weird Lifetimes
Almost always, the mutability of a lifetime can be derived from the mutability of the reference it is attached to. However this is not necessarily the case. For instance in the following code:
fn foo<'a>(input: &'a mut u8) -> &'a u8 { &* input }
One would expect the output of foo to be an immutable lifetime. However we have derived it from the input, which is a mutable lifetime. So although we have a shared reference, it will have the much more limited aliasing rules of a mutable reference. As a consequence, there is no expressive benefit in a method that mutates returning a shared reference.
Lifetime Elision
In order to make common patterns more ergonomic, Rust allows lifetimes to be elided in function, impl, and type 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 thefn
definition, while output refers to result types. Sofn foo(s: &str) -> (&str, &str)
has elided one lifetime in input position and two lifetimes in output position. Note that the input positions of afn
method definition do not include the lifetimes that occur in the method'simpl
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 ofself
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 route
is to just use a small function to ensure the lifetime is bound. 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.
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: 'a : 'b
implies
'a <: b
-- if 'a' outlives
'b, then
'ais a subtype of
'b. This is a very 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.
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 covariant if
T <: U
impliesF<T> <: F<U>
- F is contravariant if
T <: U
impliesF<U> <: F<T>
- F is invariant otherwise (no subtyping relation can be derived)
Some important variances:
&
is covariant (as is *const by metaphor)&mut
is invariant (as is *mut by metaphor)Fn(T)
is contravariant with respect toT
Box
,Vec
, and all other collections are covariantUnsafeCell
,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 covariant.
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
covariant, and 'static
is a subtype of any 'a', so
&'static stris a subtype of
&'a str. Therefore, if
&mutwas *also* covariant, then the lifetime of the
&'static strwould successfully be "shrunk" down to the shorter lifetime of the string, and
replacewould 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 covariance vs
invariance: if covariance 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 covariant, 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 covariant allows them to be covariant 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> {
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 confusing case, largely because contravariance is easily the
most confusing kind of variance, and basically never comes up. To understand it,
consider a function that takes a function len
that takes a function F
.
fn len<F>(func: F) -> usize
where F: Fn(&'static str) -> usize
{
func("hello")
}
We require that F is a Fn that can take an &'static str
and print a usize. Now
say we have a function that can take an &'a str
(for some 'a). Such a function actually
accepts more inputs, since &'static str
is a subtype of &'a str
. Therefore
len
should happily accept such a function!
So a Fn(&'a str)
is a subtype of a Fn(&'static str)
because
&'static str
is a subtype of &'a str
. Exactly contravariance.
The variance of *const
and *mut
is basically arbitrary as they're not at all
type or memory safe, so their variance is determined in analogy to & and &mut
respectively.
PhantomData and PhantomFn
This is all well and good for the types the standard library provides, but how is variance determined for type that you define? The variance of a type over its generic arguments is determined by how they're stored.
struct Foo<'a, 'b, A, B, C, D, E, F, G, H> {
a: &'a A, // covariant over 'a and A
b: &'b mut B, // invariant over 'b and B
c: *const C, // covariant over C
d: *mut D, // invariant over D
e: Vec<E>, // covariant over E
f: Cell<F>, // invariant over F
g: G // covariant over G
h1: H // would also be covariant over H except...
h2: Cell<H> // invariant over H, because invariance wins
}
However 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
reachable. 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.
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 variance. This was deemed to be less error-prone than explicitly telling the type-system the kind of variance that you want.
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>,
}
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