Here is a (much) more complete draft of the tutorial on borrowed pointers. It is becoming more in-depth than I intended. I hope to later extract a much shorter subset. But I thought I’d post what I’ve got so far.

Borrowed pointers

Borrowed pointers are one of the more flexible and powerful tools available in Rust. A borrowed pointer can be used to point anywhere: into the shared and exchange heaps, into the stack, and even into the interior of another data structure. With regard to flexibility, it is comparable to a C pointer or C++ reference. However, unlike C and C++, the Rust compiler includes special checks that ensure that borrowed pointers are being used safely. Another advantage of borrowed pointers is that they are invisible to the garbage collector, so working with borrowed pointers helps keep things efficient.

We have done our best to ensure that these checks mostly occur behind the scenes; but to get the most out of the system, you will have to understand a bit about how the compiler reasons about your program.

This document can be considered a reference guide to borrowed pointers. I have tried to explain the ideas in a sort-of tutorial style but at a level of depth that is inappropriate for a tutorial. This is my first attempt at explaining these concepts in written form, so feedback is welcome. I hope later to revisit this document and extract a subset that will be more suitable for a general tutorial.

Why are they called borrowed pointers?

We call borrowed pointers borrowed because, unlike shared and unique boxes, they never imply ownership. A borrowed pointer is always a pointer into memory owned by someone else, generally a caller to the current function. In this context, owning memory basically means taking responsibility for freeing it. The fact that borrowed pointers never imply ownership is also why they can be ignored by the garbage collector: it is never the responsibility of the borrowed pointer to keep memory alive.

Borrowed pointers can be created in two ways. The first is to explicitly take the address of something, such as a local variable or a field:

let r: {x: int, y; int} = {x: 10, y: 10}; let p: &{x: int, y; int} = &r; let q: &int = &r.x;

In the previous example, the memory is a record r stored on the stack. The variable p is a borrow of the record as a whole. The variable q is a borrow of the field x specifically.

One convenience: in C, the & operator can only be applied to lvalues (assignable locations). In Rust, the & operator can also be applied to rvalues, in which case it means “allocate some temporary space on the stack and copy the value in there”. So, for example, the following code create a new record, store it on the stack, and take its address all in one step:

let p = &{x: 10, y: 10};

The second way to create a borrowed pointer is by converting a shared or unique box into a borrowed pointer. These conversions happen implicitly on function and method calls (we may later expand the places where such conversions can occur). This is useful for writing routines that operate over data no matter where it is stored. For example:

type point = {x: int, y: int}; fn distance_from_origin(r: &point) { sqrt(r.x * r.x + r.y * r.y) } ... // Record stored on stack: let r1 = &{x: 10, y: 10}; let d1 = distance_from_origin(r1); ... // Record stored in shared box: let r2 = @{x: 10, y: 10}; let d2 = distance_from_origin(r2); ... // Record stored in unique box: let r3 = ~{x: 10, y: 10}; let d3 = distance_from_origin(r3);

One rather bizarre looking expression that you might see from time to time is something like &*r2 (where r2 is the shared box defined in the previous example). This has the effect of converting a type like @point into a &point (it also works for a ~point ). It works because we are taking the address of the dereference of the pointer. In effect, the & and * “cancel each other out” and so you are left with a borrowed copy of r2 .

Some examples of using borrowed pointers

(Warning: This section is particularly incomplete)

Storing data on the stack

The stack is a very convenient and efficient place to store data with a limited lifetime. For example, imagine that we were going to be drawing a scene; we might have some context that will be required at each point:

type point = {x: int, y: int}; type rect = {upper_left: point, lower_right: point}; type draw_context = { canvas: &canvas, bounds: rect };

Here the type draw_context is defined with two fields. The first, canvas , is presumably some sort of OS drawing layer. We don’t need to own the canvas, just use it, so draw_context contains a borrowed pointer to the canvas. The second field is a rectangle that contains the drawing bounds.

Now let’s look at the top-level routine for drawing the scene:

fn draw_scene(canvas: &canvas, bounds: rect) { let ctxt = &{canvas: canvas, bounds: bounds}; draw_starry_sky(ctxt); draw_tree(ctxt, 15, 25); reticulate_splines(ctxt); }

The first line which defines the variable ctxt uses the & operator to allocate space on the stack and create the context record. Using the & operator in this way (that is, with an rvalue, or non-assignable expression) is actually a shorthand for something like the following:

let temp = {canvas: canvas, bounds: bounds}; let ctxt = &temp;

Lifetimes

In the compiler, each borrowed pointer is associated with a lifetime (you may also have heard the term region). A lifetime is a block or an expression during the pointer may be safely used. The compiler reports an error if a borrowed pointer is used outside of its lifetime.

So far we have always written the type of a borrowed pointer as &T . In fact, the full type is </T where lt is a the name of a lifetime. You will rarely need to write this form but it may appear in error messages, and we will use it in the tutorial to clarify what’s going on. To make the idea of lifetimes more concrete, let’s look at two examples.

Concrete lifetimes

The first example is a function that declares an integer variable and takes its address. The resulting pointer is passed to a function proc_int() .

fn lifetime_example(cond: bool) { // --+ Lifetime A if cond { // |--+ Lifetime B let x: int = 10; // | | // | |--+ Lifetime C proc_int(&x); // | | | // | |--+ } // |--+ } // --+ fn proc_int(x: &int) { /*...*/ }

Alongside the example is a depiction of three of the relevant lifetimes. The first, lifetime A, corresponds to the body of the method. The second, lifetime B, corresponds to the then block of the if statement. Finally, the third, lifetime C, corresponds to the call to proc_int() .

When it sees the expression &x , the compiler will automatically determine the lifetime for the resulting pointer. The compiler attempts to select the shortest lifetime that it can. In this case, the lifetime will be Lifetime C, because the pointer is only used during the call expression itself. The maximum lifetime that the compiler would permit would be Lifetime B, because that is the lifetime of the variable x itself.

This example shows that lifetimes have a hierarchical relationship derived from the code itself. Every expression has a corresponding lifetime, and the lifetimes of subexpressions are nested inside of the lifetimes for the outer expressions.

Lifetime parameters

In the previous example, all of the relevant pointer lifetimes corresponded to expressions within the method itself. However, this is not always the case. Consider the function proc_int() that we saw in the previous example:

fn proc_int(x: &int) { // --+ Lifetime D #debug["x=%d", *x]; // | } // --+

What is the lifetime of the parameter x ? You might think it would be D , the lifetime of the method body, but that is not correct. After all, x was given to us from the caller, so its lifetime is going to be some expression that the caller knows about, but the callee (in this case, proc_int() ) does not.

To handle this case, functions can be parameterized by lifetimes. In fact, this happens implicitly whenever a parameter is a borrowed pointer. This means that the compiler invents a synthetic lifetime, let’s call if X, and says “the lifetime of the parameter x is X”. All the compiler knows about X is that it is some expression in the caller which is at least as long as the method call itself. So, in fact, this is the set of lifetimes that the compiler thinks about:

// --+ Lifetime X fn proc_int(x: &int) { // |--+ Lifetime D #debug["x=%d", *x]; // | | } // |--+ // --+

Here the lifetime D for the method body is seen as a sublifetime of this lifetime parameter X. Each time that proc_int() is called, the lifetime X will refer to some lifetime in the caller. So, in in the case of our first example, the lifetime X would refer to the lifetime C that corresponded to the call to proc_int() .

Multiple lifetime parameters

By default, every borrowed pointer that appears inside the function signature is assigned to the same lifetime parameter. So if you had a function like select_ints() , each of the parameters and also the return type are all assigned the same lifetime, X:

// --+ Lifetime X fn max(a: &int, b: &int) -> &int { // | if *a > *b {a} else {b} // | } // | // --+

Just because max() considers each of those borrowed pointers to have the same lifetime does not mean that they must have the same lifetime in the caller. For example, consider this function that calls max() :

fn calls_max(cond: bool) { // --+ Lifetime A let x: int = 10; // | if cond { // |--+ Lifetime B let y: int = 20; // | | let z: &int = max(&x, &y); // | | assert *z == 20; // | | } // |--+ } // --+

In this case, the lifetime of &x is A and the lifetime of &y is B. When calling max() , the lifetime of the parameter Y would be selected as B, which is the shorter of the two. This is also the lifetime of the result z .

Named lifetime parameters

You can also opt to give lifetime parameters explicit names. For example, the max() function could be equivalently written as:

fn max(a: &X/int, b: &X/int) -> &X/int { if *a > *b {a} else {b} }

This can sometimes be useful if you want to distinguish between the lifetimes of certain parameters. For example, the function select() takes a third parameter c but that parameter is never returned:

fn select(a: &X/int, b: &X/int, c: &Y/int) -> &X/int { if *z > 0 {x} else {y} }

Because c is never returned, select() assigns it to a named lifetime parameter Y (vs a , b , and the return type, which are given the lifetime X ). This means that c will be considered to have a distinct lifetime from a , b , and the return type.

Pointers with no designated lifetime are considered distinct from all other names, so select() could also be written with a , b , and the return type using the anonymous lifetime:

fn select(a: &int, b: &int, c: &Y/int) -> &int { if *z > 0 {x} else {y} }

How do lifetimes help ensure safety?

Lifetimes are important for safety because they limit how long a borrowed pointer can be used. Without this guarantee, it would not be possible to lend out unique and shared boxes. Consider the case of a unique box u that gets lent out as a parameter and is later sent to another task:

type point = {mut x: int, mut y: int}; fn example() { let u = ~{mut x: 10, mut y: 10}; process_point(u); send_to_other_task(u); } fn process_point(p: &point) { /* ... */ } fn send_to_other_task(p: ~point) { /* ... */ }

Without lifetimes, we could not rule out the possibility that process_point() squirelled away a copy of p , perhaps in a data structure somewhere. Then, after the point was sent to another task, we might still make use of that copy, causing data races and generally wreaking havoc. Similar problems can occur with shared boxes, since borrowed pointers are invisible to the garbage collector. Lifetimes ensure that these scenarios can’t happen because they guarantee that borrowed pointers like p will only be used during their lifetimes.

Borrowing and safety

The previous section explained pointer lifetimes and showed how they prevent a borrowed pointer from “leaking out” and being used after the loan has expired. In a sense, lifetimes protect the lender from a misbehaving borrower, by ensuring that the borrower does not keep ahold of the pointer longer than they are supposed to. However, there is another side to this coin: the borrower must also be protected from a misbehaving lender.

Lenders can misbehave by invalidating memory that has been lent out. Just to push the loan metaphor to the breaking point, we call this overleveraging. As an example, consider the following program, which creates a unique value, lends it out, and then—before the loan has expired—tries to send the unique value to another task:

fn example() { // --+ Lifetime A let u = ~{mut x: 10, mut y: 10}; // | let b = &*u; // | send_to_other_task(u); // | b.x += 1; // | } // --+ // (Note: this program is invalid and will be rejected // by the compiler)

Here, the variable b is a borrowed pointer pointing at the interior of u . The lifetime of this pointer will be the method body, shown as lifetime A. Immediately after creating b , the value in u is sent to another task. But then the program uses b to modify the fields of that value. If this program were allowed to execute, it would result in an data race.

But of course it does not execute. In fact the compiler reports an error:

example.rs:4:27: 4:28 error: moving out of immutable local variable prohibited due to outstanding loan example.rs:4 send_to_other_task(u); ^ example.rs:3:18: 3:19 note: loan of immutable local variable granted here example.rs:3 let b = &*u; ^

What this message is telling you is that the variable u cannot be moved because it was lent out to create b , and the lifetime of b has not expired.

There are actually three ways that lenders could potentially invalid borrowed pointers. The first way is to move the memory, as we have seen. The second way is to reassign a unique box. The third way is to mutate an enum. Let’s look at those two cases in more detail.

Reassigning unique boxes

Unlike shared boxes, unique boxes are not garbage collected. Instead, they are eagerly freed as soon as the owning reference to them goes out of scope or is mutated. So, if we have a program like the following:

fn incr_unique() { let mut u = ~0; for 10.times { u = ~(*u + 1); } }

Each iteration around the loop, the variable u is reassigned with a new unique value. The previous unique value will be freed. Without extra safety checks, these eager frees could result in the dreaded “dangling pointer” that haunts the nightmares of C programmers everywhere:

fn incr_unique() { let mut u = ~0; for 10.times { let b = &*u; u = ~(*u + 1); // here, the memory pointed at in `b` has been freed use(*b); } } // (Note: this program is invalid and will be rejected // by the compiler)

Reassigning enums

Enums have the interesting behavior that when they are reassigned they can change the types of their contents or even cause their contents not to exist altogether. For example, consider this program:

fn incr_some() { let v = some(~3); alt v { none => {} some(ref p) => { v = none; use(*p); } } } // (Note: this program is invalid and will be rejected // by the compiler)

Here, the value v is initially assigned a some value. We then match against this value and, using a ref binding, create a pointer p to the inside of v . However, then, on the next line we reassign v with none —now the pointer p is invalidated, because it was supposed to point to the argument of the some variant, but v is no longer a some variant.

To better understand what’s going on, consider the following depiction which shows the stack and heap for a call to incr_some() right before the (second) assignment to v . The value v is on the stack with a tag of some and a data field of type ~int , which points into the exchange heap. The value p is a pointer into this data field.

Stack Exchange heap ----- ------------- v +------------+ | tag: some | +-->| ~int |------>+---+ | p +------------+ | 3 | | | &~int |--+ +---+ | +------------+ | | | +-------------------+

Now this is the situation after the assignment occurs. Note that the tag of v has changed to none and the data field has been invalidated, as none has no arguments. This in turn causes the data in the exchange heap to be freed. But now the pointer p still exists, and it points into the (now invalidated) data field of v . This is bad.

Stack Exchange heap ----- ------------- v +------------+ | tag: none | +-->| xxx | | p +------------+ | | &~int |--+ | +------------+ | | | +-------------------+

How the compiler prevents overleverage

The goal of the compiler is to guarantee that each time a borrowed pointer with lifetime L is created, the compiler will ensure that the memory which is being borrowed is valid for the entirety of the lifetime L. We can currently make this guarantee in one of three ways, depending on the owner of the data being borrowed.

Data ownership

Before getting into the precise rules that the compiler uses to prevent overleverage, it is important to understand how Rust defines data ownership. In Rust, owning data essentially means that you are responsible for freeing it. In C and C++, ownership is typically by convention (though the various smart pointers in C++ help to formalize these conventions). In Rust, ownership is built into the type system.

Rust types basically fall into four categories: by value ( T ), unique boxes ( ~T ), shared boxes ( @T ), and borrowed pointers ( &T ):

Value types and unique boxes both imply exclusive ownership. This means that if a variable with value type or a unique box goes out of scope, its contents will be dropped (I’ll define drop more specifically below, but it essentially means “freed”). The same applies if a mutable location of value/unique box type is overwritten with a new value.

and both imply exclusive ownership. This means that if a variable with value type or a unique box goes out of scope, its contents will be dropped (I’ll define drop more specifically below, but it essentially means “freed”). The same applies if a mutable location of value/unique box type is overwritten with a new value. Shared boxes imply shared- or co-ownership: that is, a shared box is collected only when all references to it are overwritten or go out of scope. Each reference is therefore a kind of equal owner: any one is enough to keep the shared box alive. In the same sense, no reference is an owner, as there is no way to guarantee the absence of other owners. Therefore, we say that shared boxes are not owned at all, but rather co-owned.

imply shared- or co-ownership: that is, a shared box is collected only when all references to it are overwritten or go out of scope. Each reference is therefore a kind of equal owner: any one is enough to keep the shared box alive. In the same sense, no reference is an owner, as there is no way to guarantee the absence of other owners. Therefore, we say that shared boxes are not owned at all, but rather co-owned. Finally, borrowed pointers never imply ownership. Borrowed pointers must always have a lifetime that is a subset of some other owning type.

To help clarify, consider this example:

fn ownership_example(x: @int) { let mut y = {f: ~4, g: x}; let z = &y.f; y = {f: ~5, g: @5}; // (*) }

Here, the parameter x is a shared box. It therefore shares ownership of its referent with all other references.

The variable y has a value type {f: ~int, g: @int} . This is an interesting type because it shows how ownership is transitive: that is, because the variable y owns its referent, it also owns anything that is owned by the record itself. In this case, that means that y.f is also owned, and y.g is co-owned (it’s a shared box).

Finally, the variable z is a borrowed pointer pointing at the interior of y . Its lifetime, as we’ll see shortly, will be limited by the compiler to the lifetime of the current stack frame, as the current stack frame owns y which in turn owns the field y.f .

Here is a graphical depiction of the state of the program immediately before the assignment marked with (*) :

Stack Exchange Heap Shared Heap x +---------+ | @int | ---------------------+ y +---------+ +---+ | +---+ +-> | f: ~int | ---> | 4 | +-->| _ | | | g: @int | ---+ +---+ | +---+ | +---------+ | | | z | &~int | -+ | | | +---------+ | +-----------------+ | | +----------------+

Once the assignment occurs, everything that was owned by y will be freed. Therefore, the heap afterwards looks like:

Stack Exchange Heap Shared Heap x +---------+ | @int | ---------------------+ y +---------+ +---+ | +---+ +-> | f: ~int | ----------> | 5 | +-->| _ | | | g: @int | ---+ +---+ +---+ | +---------+ | | z | &~int | -+ | +---+ | +---------+ | +-------------------> | 5 | | | +---+ +----------------+

Here you can see that the unique box was freed (that was (indirectly) owned by y ). The shared box is no longer referenced by y , but it is not freed because it was co-owned by x . Meanwhile, the variable z is unaffected, though of course the value *z will be different.

The rules for preventing overleverage

The rules for preventing overleverage depend on the owner of the data being borrowed. Basically the compiler tries to accept as many programs as it can, so it takes advantage of whatever knowledge is available.

The best case is when the data to be borrowed is exclusively owned by the stack frame. This does not necessarily mean that the data is stored on the stack frame. For example, the data might be found in a unique box stored in a local variable: in that case, the stack frame owns the local, which then owns the box, which owns its contents, and so the data is ultimately owned by the stack frame. As described below, data owned by the stack frame can be closely tracked by the compiler and hence it can be very flexible with what it accepts.

For data which is not owned by the stack frame—or for which the compiler cannot determine a concrete owner—a stricter set of rules is applied. If the data being borrowed is unstable—meaning that mutating the container would invalidate the borrowed pointer—then the data must reside in an immutable field, so as to guarantee that no mutations occur.

Finally, if the program is borrowing unstable data in a mutable location, the final fallback is to accept the borrow but only for pure actions. Pure actions are actions which do not modify any data unless it is owned by the current stack frame.

The next three subsections examine the three cases in more detail.

Loaning out data owned by the current stack frame

When the data to be borrowed is owned by the current stack frame, the compiler can track it quite precisely. The compiler internally tracks all loans of values owned by the stack frame that are in scope at each point. A loan is in scope for the entire lifetime of the borrowed pointer.

The compiler will check that all actions are compatible with the set of loans that are in scope. This means that it is not legal to move data out a variable or path when a loan for that path (or some enclosing) is in scope. Similarly, if the borrowed pointer is an immutable pointer (i.e., &T and not &const T or &mut T ), then any assignment to the value which was lent out is prohibited.

This last point concerning mutability is subtle. In general, it is legal to create an immutable pointer to mutable data so long as the compiler can guarantee that the data will not be mutated for the lifetime of the pointer. Currently we can guarantee immutability in two cases: one is when the data that is lent out is owned by the current stack frame, and the other is when only pure actions are taken for the lifetime of the reference. Purity is discussed below.

Most of the prior examples have made use of loans, so I will just give one more example that shows loans of subfields as well as immutable loans of mutable data.

fn example() { let u = ~{mut x: 10, mut y: 10}; let b = &u.x; // (0) u.x += 11; // (1) send_to_other_task(u); // (2) #debug["%d", *b]; } // (Note: this program is invalid and will be rejected // by the compiler)

Here the b has type &int —that is, a pointer to immutable memory containing an int . However, b points at the field u.x , which is declared as mutable. The compiler notes this discrepancy and records that u.x is lent out as immutable for the lifetime of b (here, the remainder of the method body). This means that the assignment to u.x marked as (1) will be considered illegal. The compiler will report a message like “assigning to mutable field prohibited due to outstanding loan” followed by a note indicating that the loan was granted on the line marked (0) .

The attempt to move u on the line marked (2) will be prohibited in a similar fashion. The compiler will report “moving out of immutable local variable prohibited due to outstanding loan”. This is somewhat interesting, because it shows that although we borrowed u.x this also implicitly borrows u , as u is the owner of u.x . In general, borrowing a value also borrows its owner.

Data owned by a shared box or with an unknown owner

The previous section focused on data which was owned by the current stack frame, but said nothing about other cases. Very often, of course, we would like to borrow data owned by a shared box. Unless the data being borrowed is part of a unique box or the interior of an enum, this is not generally an issue.

For example, the following program does not cause any sort of error:

type T = @{mut f: {g: int}}; fn foo(v: T) -> int { let x = &mut v.f.g; *x }

The reason is that even though the field g is part of a record stored in a mutable location, it is not harmful if the field f is reassigned, as it will not change the type of the field g . Field g will remain an integer. Note however that when we took the address of v.f.g we used the operator &mut , which means that the type of x will be &mut int : basically, we had to explicitly acknowledge that the value we borrowed is stored in mutable memory.

However, if there is an attempt to borrow unstable memory then the compiler must be more careful:

type T = @{mut f: {g: ~int}}; fn foo(v: T) -> int { let x = &*v.f.g; ... *x }

Here we attempted to borrow the integer found inside the unique box v.f.g . This is dangerous because if v.f.g were reassigned, then the pointer x would be invalidated. Furthermore, it is not enough for the compiler to prevent you from modifying v.f.g : because v is a shared box, there may some alias to v that you could modify instead and still cause the same effect. Therefore, this program is illegal.

Basically borrowing unstable memory that is owned by a shared box is only safe if it is stored in an immutable field. For example, if we modify the type T to remove the mut qualified:

type T = @{f: {g: ~int}};

Now, so long as the shared box remains live, there is no way for the fields f or g to be modified, and hence the unique box will remain live.

It is still true that the compiler must guarantee that the shared box remains live. Because of the nature of shared boxes, however, this is not a problem: all we must do is guarantee that some reference to the shared box exists for the duration of the borrow. In some cases, the compiler can see that this is already the case in your program: for example, in the function foo , the parameter v is immutable and refers to the shared box, so so long as v is in scope the shared box will not be freed.

However, even if v were a mutable local, there is no problem: the compiler would just create a temporary with the current value so as to ensure that the shared box is not collected while the borrowed pointer exists. Here is an example bar() that makes use of this feature:

fn bar() -> int { let mut x = @{f: 3}; let y = &x.f; assert *y == 3; x = @{f: 4}; // overwrite x with new value assert *y == 3; // but y still points at the old memory }

What happens here is that the field f of the shared box is borrowed, but then the reference x to the shared box is mutated. If x were the last reference to the box, and the compiler did nothing, then the box might be freed at this point. But in fact the compiler sees that x is declared as a mutable local variable and so it inserts a temporary that refers to x . This is called rooting x , because it ensures that there is a root for the garbage collector to find. In other words, the function bar() is compiled to something like:

fn bar() -> int { let mut x = @{f: 3}; let _x = x; // compiler-inserted root let y = &x.f; assert *y == 3; x = @{f: 4}; // overwrite x with new value assert *y == 3; // but y still points at the old memory }

Purity

Sometimes neither of the two techniques described above apply. In that case, the compiler will accept any borrow, so long as all code for the lifetime of the borrow is pure. Pure code is code which does not modify any data that is not owned by the current stack frame. Basically you can borrow anything so long as you promise not to make any mutation (except for local variables).