25 KiB
% The Rust References and Lifetimes Guide
Introduction
References are one of the more flexible and powerful tools available in Rust. A reference can point anywhere: into the managed or exchange heap, into the stack, and even into the interior of another data structure. A reference is as flexible as a C pointer or C++ reference. However, unlike C and C++ compilers, the Rust compiler includes special static checks that ensure that programs use references safely. Another advantage of references is that they are invisible to the garbage collector, so working with references helps reduce the overhead of automatic memory management.
Despite their complete safety, a reference's representation at runtime is the same as that of an ordinary pointer in a C program. They introduce zero overhead. The compiler does all safety checks at compile time.
Although references have rather elaborate theoretical underpinnings (region pointers), the core concepts will be familiar to anyone who has worked with C or C++. Therefore, the best way to explain how they are used—and their limitations—is probably just to work through several examples.
By example
References, sometimes known as borrowed pointers, are only valid for a limited duration. References never claim any kind of ownership over the data that they point to: instead, they are used for cases where you would like to use data for a short time.
As an example, consider a simple struct type Point:
struct Point {x: f64, y: f64}
We can use this simple definition to allocate points in many different ways. For example, in this code, each of these three local variables contains a point, but allocated in a different place:
# struct Point {x: f64, y: f64}
let on_the_stack : Point = Point {x: 3.0, y: 4.0};
let managed_box : @Point = @Point {x: 5.0, y: 1.0};
let owned_box : ~Point = ~Point {x: 7.0, y: 9.0};
Suppose we wanted to write a procedure that computed the distance between any
two points, no matter where they were stored. For example, we might like to
compute the distance between on_the_stack and managed_box, or between
managed_box and owned_box. One option is to define a function that takes
two arguments of type Point—that is, it takes the points by value. But if we
define it this way, calling the function will cause the points to be
copied. For points, this is probably not so bad, but often copies are
expensive. Worse, if the data type contains mutable fields, copying can change
the semantics of your program in unexpected ways. So we'd like to define a
function that takes the points by pointer. We can use references to do
this:
# struct Point {x: f64, y: f64}
# fn sqrt(f: f64) -> f64 { 0.0 }
fn compute_distance(p1: &Point, p2: &Point) -> f64 {
let x_d = p1.x - p2.x;
let y_d = p1.y - p2.y;
sqrt(x_d * x_d + y_d * y_d)
}
Now we can call compute_distance() in various ways:
# struct Point {x: f64, y: f64}
# let on_the_stack : Point = Point{x: 3.0, y: 4.0};
# let managed_box : @Point = @Point{x: 5.0, y: 1.0};
# let owned_box : ~Point = ~Point{x: 7.0, y: 9.0};
# fn compute_distance(p1: &Point, p2: &Point) -> f64 { 0.0 }
compute_distance(&on_the_stack, managed_box);
compute_distance(managed_box, owned_box);
Here, the & operator takes the address of the variable
on_the_stack; this is because on_the_stack has the type Point
(that is, a struct value) and we have to take its address to get a
value. We also call this borrowing the local variable
on_the_stack, because we have created an alias: that is, another
name for the same data.
In contrast, we can pass the boxes managed_box and owned_box to
compute_distance directly. The compiler automatically converts a box like
@Point or ~Point to a reference like &Point. This is another form
of borrowing: in this case, the caller lends the contents of the managed or
owned box to the callee.
Whenever a caller lends data to a callee, there are some limitations on what the caller can do with the original. For example, if the contents of a variable have been lent out, you cannot send that variable to another task. In addition, the compiler will reject any code that might cause the borrowed value to be freed or overwrite its component fields with values of different types (I'll get into what kinds of actions those are shortly). This rule should make intuitive sense: you must wait for a borrower to return the value that you lent it (that is, wait for the reference to go out of scope) before you can make full use of it again.
Other uses for the & operator
In the previous example, the value on_the_stack was defined like so:
# struct Point {x: f64, y: f64}
let on_the_stack: Point = Point {x: 3.0, y: 4.0};
This declaration means that code can only pass Point by value to other
functions. As a consequence, we had to explicitly take the address of
on_the_stack to get a reference. Sometimes however it is more
convenient to move the & operator into the definition of on_the_stack:
# struct Point {x: f64, y: f64}
let on_the_stack2: &Point = &Point {x: 3.0, y: 4.0};
Applying & to an rvalue (non-assignable location) is just a convenient
shorthand for creating a temporary and taking its address. A more verbose
way to write the same code is:
# struct Point {x: f64, y: f64}
let tmp = Point {x: 3.0, y: 4.0};
let on_the_stack2 : &Point = &tmp;
Taking the address of fields
As in C, the & operator is not limited to taking the address of
local variables. It can also take the address of fields or
individual array elements. For example, consider this type definition
for rectangle:
struct Point {x: f64, y: f64} // as before
struct Size {w: f64, h: f64} // as before
struct Rectangle {origin: Point, size: Size}
Now, as before, we can define rectangles in a few different ways:
# struct Point {x: f64, y: f64}
# struct Size {w: f64, h: f64} // as before
# struct Rectangle {origin: Point, size: Size}
let rect_stack = &Rectangle {origin: Point {x: 1.0, y: 2.0},
size: Size {w: 3.0, h: 4.0}};
let rect_managed = @Rectangle {origin: Point {x: 3.0, y: 4.0},
size: Size {w: 3.0, h: 4.0}};
let rect_owned = ~Rectangle {origin: Point {x: 5.0, y: 6.0},
size: Size {w: 3.0, h: 4.0}};
In each case, we can extract out individual subcomponents with the &
operator. For example, I could write:
# struct Point {x: f64, y: f64} // as before
# struct Size {w: f64, h: f64} // as before
# struct Rectangle {origin: Point, size: Size}
# let rect_stack = &Rectangle {origin: Point {x: 1.0, y: 2.0}, size: Size {w: 3.0, h: 4.0}};
# let rect_managed = @Rectangle {origin: Point {x: 3.0, y: 4.0}, size: Size {w: 3.0, h: 4.0}};
# let rect_owned = ~Rectangle {origin: Point {x: 5.0, y: 6.0}, size: Size {w: 3.0, h: 4.0}};
# fn compute_distance(p1: &Point, p2: &Point) -> f64 { 0.0 }
compute_distance(&rect_stack.origin, &rect_managed.origin);
which would borrow the field origin from the rectangle on the stack
as well as from the managed box, and then compute the distance between them.
Borrowing managed boxes and rooting
We’ve seen a few examples so far of borrowing heap boxes, both managed and owned. Up till this point, we’ve glossed over issues of safety. As stated in the introduction, at runtime a reference is simply a pointer, nothing more. Therefore, avoiding C's problems with dangling pointers requires a compile-time safety check.
The basis for the check is the notion of lifetimes. A lifetime is a static approximation of the span of execution during which the pointer is valid: it always corresponds to some expression or block within the program. Code inside that expression can use the pointer without restrictions. But if the pointer escapes from that expression (for example, if the expression contains an assignment expression that assigns the pointer to a mutable field of a data structure with a broader scope than the pointer itself), the compiler reports an error. We'll be discussing lifetimes more in the examples to come, and a more thorough introduction is also available.
When the & operator creates a reference, the compiler must
ensure that the pointer remains valid for its entire
lifetime. Sometimes this is relatively easy, such as when taking the
address of a local variable or a field that is stored on the stack:
struct X { f: int }
fn example1() {
let mut x = X { f: 3 };
let y = &mut x.f; // -+ L
... // |
} // -+
Here, the lifetime of the reference y is simply L, the
remainder of the function body. The compiler need not do any other
work to prove that code will not free x.f. This is true even if the
code mutates x.
The situation gets more complex when borrowing data inside heap boxes:
# struct X { f: int }
fn example2() {
let mut x = @X { f: 3 };
let y = &x.f; // -+ L
... // |
} // -+
In this example, the value x is a heap box, and y is therefore a
pointer into that heap box. Again the lifetime of y is L, the
remainder of the function body. But there is a crucial difference:
suppose x were to be reassigned during the lifetime L? If the
compiler isn't careful, the managed box could become unrooted, and
would therefore be subject to garbage collection. A heap box that is
unrooted is one such that no pointer values in the heap point to
it. It would violate memory safety for the box that was originally
assigned to x to be garbage-collected, since a non-heap
pointer y still points into it.
Note: Our current implementation implements the garbage collector using reference counting and cycle detection.
For this reason, whenever an & expression borrows the interior of a
managed box stored in a mutable location, the compiler inserts a
temporary that ensures that the managed box remains live for the
entire lifetime. So, the above example would be compiled as if it were
written
# struct X { f: int }
fn example2() {
let mut x = @X {f: 3};
let x1 = x;
let y = &x1.f; // -+ L
... // |
} // -+
Now if x is reassigned, the pointer y will still remain valid. This
process is called rooting.
Borrowing owned boxes
The previous example demonstrated rooting, the process by which the compiler ensures that managed boxes remain live for the duration of a borrow. Unfortunately, rooting does not work for borrows of owned boxes, because it is not possible to have two references to a owned box.
For owned boxes, therefore, the compiler will only allow a borrow if the compiler can guarantee that the owned box will not be reassigned or moved for the lifetime of the pointer. This does not necessarily mean that the owned box is stored in immutable memory. For example, the following function is legal:
# fn some_condition() -> bool { true }
# struct Foo { f: int }
fn example3() -> int {
let mut x = ~Foo {f: 3};
if some_condition() {
let y = &x.f; // -+ L
return *y; // |
} // -+
x = ~Foo {f: 4};
...
# return 0;
}
Here, as before, the interior of the variable x is being borrowed
and x is declared as mutable. However, the compiler can prove that
x is not assigned anywhere in the lifetime L of the variable
y. Therefore, it accepts the function, even though x is mutable
and in fact is mutated later in the function.
It may not be clear why we are so concerned about mutating a borrowed variable. The reason is that the runtime system frees any owned box as soon as its owning reference changes or goes out of scope. Therefore, a program like this is illegal (and would be rejected by the compiler):
fn example3() -> int {
let mut x = ~X {f: 3};
let y = &x.f;
x = ~X {f: 4}; // Error reported here.
*y
}
To make this clearer, consider this diagram showing the state of
memory immediately before the re-assignment of x:
Stack Exchange Heap
x +----------+
| ~{f:int} | ----+
y +----------+ |
| &int | ----+
+----------+ | +---------+
+--> | f: 3 |
+---------+
Once the reassignment occurs, the memory will look like this:
Stack Exchange Heap
x +----------+ +---------+
| ~{f:int} | -------> | f: 4 |
y +----------+ +---------+
| &int | ----+
+----------+ | +---------+
+--> | (freed) |
+---------+
Here you can see that the variable y still points at the old box,
which has been freed.
In fact, the compiler can apply the same kind of reasoning to any memory that is (uniquely) owned by the stack frame. So we could modify the previous example to introduce additional owned pointers and structs, and the compiler will still be able to detect possible mutations:
fn example3() -> int {
struct R { g: int }
struct S { f: ~R }
let mut x = ~S {f: ~R {g: 3}};
let y = &x.f.g;
x = ~S {f: ~R {g: 4}}; // Error reported here.
x.f = ~R {g: 5}; // Error reported here.
*y
}
In this case, two errors are reported, one when the variable x is
modified and another when x.f is modified. Either modification would
invalidate the pointer y.
Borrowing and enums
The previous example showed that the type system forbids any borrowing
of owned boxes found in aliasable, mutable memory. This restriction
prevents pointers from pointing into freed memory. There is one other
case where the compiler must be very careful to ensure that pointers
remain valid: pointers into the interior of an enum.
As an example, let’s look at the following shape type that can
represent both rectangles and circles:
struct Point {x: f64, y: f64}; // as before
struct Size {w: f64, h: f64}; // as before
enum Shape {
Circle(Point, f64), // origin, radius
Rectangle(Point, Size) // upper-left, dimensions
}
Now we might write a function to compute the area of a shape. This function takes a reference to a shape, to avoid the need for copying.
# struct Point {x: f64, y: f64}; // as before
# struct Size {w: f64, h: f64}; // as before
# enum Shape {
# Circle(Point, f64), // origin, radius
# Rectangle(Point, Size) // upper-left, dimensions
# }
# static tau: f64 = 6.28;
fn compute_area(shape: &Shape) -> f64 {
match *shape {
Circle(_, radius) => 0.5 * tau * radius * radius,
Rectangle(_, ref size) => size.w * size.h
}
}
The first case matches against circles. Here, the pattern extracts the radius from the shape variant and the action uses it to compute the area of the circle. (Like any up-to-date engineer, we use the tau circle constant and not that dreadfully outdated notion of pi).
The second match is more interesting. Here we match against a
rectangle and extract its size: but rather than copy the size
struct, we use a by-reference binding to create a pointer to it. In
other words, a pattern binding like ref size binds the name size
to a pointer of type &size into the interior of the enum.
To make this more clear, let's look at a diagram of memory layout in
the case where shape points at a rectangle:
Stack Memory
+-------+ +---------------+
| shape | ------> | rectangle( |
+-------+ | {x: f64, |
| size | -+ | y: f64}, |
+-------+ +----> | {w: f64, |
| h: f64}) |
+---------------+
Here you can see that rectangular shapes are composed of five words of
memory. The first is a tag indicating which variant this enum is
(rectangle, in this case). The next two words are the x and y
fields for the point and the remaining two are the w and h fields
for the size. The binding size is then a pointer into the inside of
the shape.
Perhaps you can see where the danger lies: if the shape were somehow
to be reassigned, perhaps to a circle, then although the memory used
to store that shape value would still be valid, it would have a
different type! The following diagram shows what memory would look
like if code overwrote shape with a circle:
Stack Memory
+-------+ +---------------+
| shape | ------> | circle( |
+-------+ | {x: f64, |
| size | -+ | y: f64}, |
+-------+ +----> | f64) |
| |
+---------------+
As you can see, the size pointer would be pointing at a f64
instead of a struct. This is not good: dereferencing the second field
of a f64 as if it were a struct with two fields would be a memory
safety violation.
So, in fact, for every ref binding, the compiler will impose the
same rules as the ones we saw for borrowing the interior of a owned
box: it must be able to guarantee that the enum will not be
overwritten for the duration of the borrow. In fact, the compiler
would accept the example we gave earlier. The example is safe because
the shape pointer has type &Shape, which means "reference to
immutable memory containing a shape". If, however, the type of that
pointer were &mut Shape, then the ref binding would be ill-typed.
Just as with owned boxes, the compiler will permit ref bindings
into data owned by the stack frame even if the data are mutable,
but otherwise it requires that the data reside in immutable memory.
Returning references
So far, all of the examples we have looked at, use references in a “downward” direction. That is, a method or code block creates a reference, then uses it within the same scope. It is also possible to return references as the result of a function, but as we'll see, doing so requires some explicit annotation.
For example, we could write a subroutine like this:
struct Point {x: f64, y: f64}
fn get_x<'r>(p: &'r Point) -> &'r f64 { &p.x }
Here, the function get_x() returns a pointer into the structure it
was given. The type of the parameter (&'r Point) and return type
(&'r f64) both use a new syntactic form that we have not seen so
far. Here the identifier r names the lifetime of the pointer
explicitly. So in effect, this function declares that it takes a
pointer with lifetime r and returns a pointer with that same
lifetime.
In general, it is only possible to return references if they are derived from a parameter to the procedure. In that case, the pointer result will always have the same lifetime as one of the parameters; named lifetimes indicate which parameter that is.
In the previous examples, function parameter types did not include a lifetime name. In those examples, the compiler simply creates a fresh name for the lifetime automatically: that is, the lifetime name is guaranteed to refer to a distinct lifetime from the lifetimes of all other parameters.
Named lifetimes that appear in function signatures are conceptually
the same as the other lifetimes we have seen before, but they are a bit
abstract: they don’t refer to a specific expression within get_x(),
but rather to some expression within the caller of get_x(). The
lifetime r is actually a kind of lifetime parameter: it is defined
by the caller to get_x(), just as the value for the parameter p is
defined by that caller.
In any case, whatever the lifetime of r is, the pointer produced by
&p.x always has the same lifetime as p itself: a pointer to a
field of a struct is valid as long as the struct is valid. Therefore,
the compiler accepts the function get_x().
To emphasize this point, let’s look at a variation on the example, this time one that does not compile:
struct Point {x: f64, y: f64}
fn get_x_sh(p: @Point) -> &f64 {
&p.x // Error reported here
}
Here, the function get_x_sh() takes a managed box as input and
returns a reference. As before, the lifetime of the reference
that will be returned is a parameter (specified by the
caller). That means that get_x_sh() promises to return a reference
that is valid for as long as the caller would like: this is
subtly different from the first example, which promised to return a
pointer that was valid for as long as its pointer argument was valid.
Within get_x_sh(), we see the expression &p.x which takes the
address of a field of a managed box. The presence of this expression
implies that the compiler must guarantee that, so long as the
resulting pointer is valid, the managed box will not be reclaimed by
the garbage collector. But recall that get_x_sh() also promised to
return a pointer that was valid for as long as the caller wanted it to
be. Clearly, get_x_sh() is not in a position to make both of these
guarantees; in fact, it cannot guarantee that the pointer will remain
valid at all once it returns, as the parameter p may or may not be
live in the caller. Therefore, the compiler will report an error here.
In general, if you borrow a managed (or owned) box to create a reference, it will only be valid within the function and cannot be returned. This is why the typical way to return references is to take references as input (the only other case in which it can be legal to return a reference is if it points at a static constant).
Named lifetimes
Let's look at named lifetimes in more detail. Named lifetimes allow for grouping of parameters by lifetime. For example, consider this function:
# struct Point {x: f64, y: f64}; // as before
# struct Size {w: f64, h: f64}; // as before
# enum Shape {
# Circle(Point, f64), // origin, radius
# Rectangle(Point, Size) // upper-left, dimensions
# }
# fn compute_area(shape: &Shape) -> f64 { 0.0 }
fn select<'r, T>(shape: &'r Shape, threshold: f64,
a: &'r T, b: &'r T) -> &'r T {
if compute_area(shape) > threshold {a} else {b}
}
This function takes three references and assigns each the same
lifetime r. In practice, this means that, in the caller, the
lifetime r will be the intersection of the lifetime of the three
region parameters. This may be overly conservative, as in this
example:
# struct Point {x: f64, y: f64}; // as before
# struct Size {w: f64, h: f64}; // as before
# enum Shape {
# Circle(Point, f64), // origin, radius
# Rectangle(Point, Size) // upper-left, dimensions
# }
# fn compute_area(shape: &Shape) -> f64 { 0.0 }
# fn select<'r, T>(shape: &Shape, threshold: f64,
# a: &'r T, b: &'r T) -> &'r T {
# if compute_area(shape) > threshold {a} else {b}
# }
// -+ r
fn select_based_on_unit_circle<'r, T>( // |-+ B
threshold: f64, a: &'r T, b: &'r T) -> &'r T { // | |
// | |
let shape = Circle(Point {x: 0., y: 0.}, 1.); // | |
select(&shape, threshold, a, b) // | |
} // |-+
// -+
In this call to select(), the lifetime of the first parameter shape
is B, the function body. Both of the second two parameters a and b
share the same lifetime, r, which is a lifetime parameter of
select_based_on_unit_circle(). The caller will infer the
intersection of these two lifetimes as the lifetime of the returned
value, and hence the return value of select() will be assigned a
lifetime of B. This will in turn lead to a compilation error, because
select_based_on_unit_circle() is supposed to return a value with the
lifetime r.
To address this, we can modify the definition of select() to
distinguish the lifetime of the first parameter from the lifetime of
the latter two. After all, the first parameter is not being
returned. Here is how the new select() might look:
# struct Point {x: f64, y: f64}; // as before
# struct Size {w: f64, h: f64}; // as before
# enum Shape {
# Circle(Point, f64), // origin, radius
# Rectangle(Point, Size) // upper-left, dimensions
# }
# fn compute_area(shape: &Shape) -> f64 { 0.0 }
fn select<'r, 'tmp, T>(shape: &'tmp Shape, threshold: f64,
a: &'r T, b: &'r T) -> &'r T {
if compute_area(shape) > threshold {a} else {b}
}
Here you can see that shape's lifetime is now named tmp. The
parameters a, b, and the return value all have the lifetime r.
However, since the lifetime tmp is not returned, it would be more
concise to just omit the named lifetime for shape altogether:
# struct Point {x: f64, y: f64}; // as before
# struct Size {w: f64, h: f64}; // as before
# enum Shape {
# Circle(Point, f64), // origin, radius
# Rectangle(Point, Size) // upper-left, dimensions
# }
# fn compute_area(shape: &Shape) -> f64 { 0.0 }
fn select<'r, T>(shape: &Shape, threshold: f64,
a: &'r T, b: &'r T) -> &'r T {
if compute_area(shape) > threshold {a} else {b}
}
This is equivalent to the previous definition.
Conclusion
So there you have it: a (relatively) brief tour of the lifetime system. For more details, we refer to the (yet to be written) reference document on references, which will explain the full notation and give more examples.