Treating Smart Pointers Like Regular References with the Deref
Trait
Implementing the Deref
trait allows you to customize the behavior of the
dereference operator, *
(as opposed to the multiplication or glob
operator). By implementing Deref
in such a way that a smart pointer can be
treated like a regular reference, you can write code that operates on
references and use that code with smart pointers too.
Let’s first look at how the dereference operator works with regular references.
Then we’ll try to define a custom type that behaves like Box<T>
, and see why
the dereference operator doesn’t work like a reference on our newly defined
type. We’ll explore how implementing the Deref
trait makes it possible for
smart pointers to work in ways similar to references. Then we’ll look at
Rust’s deref coercion feature and how it lets us work with either references
or smart pointers.
Note: there’s one big difference between the
MyBox<T>
type we’re about to build and the realBox<T>
: our version will not store its data on the heap. We are focusing this example onDeref
, so where the data is actually stored is less important than the pointer-like behavior.
Following the Pointer to the Value with the Dereference Operator
A regular reference is a type of pointer, and one way to think of a pointer is
as an arrow to a value stored somewhere else. In Listing 15-6, we create a
reference to an i32
value and then use the dereference operator to follow the
reference to the data:
Filename: src/main.rs
fn main() { let x = 5; let y = &x; assert_eq!(5, x); assert_eq!(5, *y); }
The variable x
holds an i32
value, 5
. We set y
equal to a reference to
x
. We can assert that x
is equal to 5
. However, if we want to make an
assertion about the value in y
, we have to use *y
to follow the reference
to the value it’s pointing to (hence dereference). Once we dereference y
,
we have access to the integer value y
is pointing to that we can compare with
5
.
If we tried to write assert_eq!(5, y);
instead, we would get this compilation
error:
$ cargo run
Compiling deref-example v0.1.0 (file:///projects/deref-example)
error[E0277]: can't compare `{integer}` with `&{integer}`
--> src/main.rs:6:5
|
6 | assert_eq!(5, y);
| ^^^^^^^^^^^^^^^^^ no implementation for `{integer} == &{integer}`
|
= help: the trait `PartialEq<&{integer}>` is not implemented for `{integer}`
= note: this error originates in the macro `assert_eq` (in Nightly builds, run with -Z macro-backtrace for more info)
For more information about this error, try `rustc --explain E0277`.
error: could not compile `deref-example` due to previous error
Comparing a number and a reference to a number isn’t allowed because they’re different types. We must use the dereference operator to follow the reference to the value it’s pointing to.
Using Box<T>
Like a Reference
We can rewrite the code in Listing 15-6 to use a Box<T>
instead of a
reference; the dereference operator will work as shown in Listing 15-7:
Filename: src/main.rs
fn main() { let x = 5; let y = Box::new(x); assert_eq!(5, x); assert_eq!(5, *y); }
The only difference between Listing 15-7 and Listing 15-6 is that here we set
y
to be an instance of a box pointing to a copied value of x
rather than a
reference pointing to the value of x
. In the last assertion, we can use the
dereference operator to follow the box’s pointer in the same way that we did
when y
was a reference. Next, we’ll explore what is special about Box<T>
that enables us to use the dereference operator by defining our own box type.
Defining Our Own Smart Pointer
Let’s build a smart pointer similar to the Box<T>
type provided by the
standard library to experience how smart pointers behave differently from
references by default. Then we’ll look at how to add the ability to use the
dereference operator.
The Box<T>
type is ultimately defined as a tuple struct with one element, so
Listing 15-8 defines a MyBox<T>
type in the same way. We’ll also define a
new
function to match the new
function defined on Box<T>
.
Filename: src/main.rs
struct MyBox<T>(T); impl<T> MyBox<T> { fn new(x: T) -> MyBox<T> { MyBox(x) } } fn main() {}
We define a struct named MyBox
and declare a generic parameter T
, because
we want our type to hold values of any type. The MyBox
type is a tuple struct
with one element of type T
. The MyBox::new
function takes one parameter of
type T
and returns a MyBox
instance that holds the value passed in.
Let’s try adding the main
function in Listing 15-7 to Listing 15-8 and
changing it to use the MyBox<T>
type we’ve defined instead of Box<T>
. The
code in Listing 15-9 won’t compile because Rust doesn’t know how to dereference
MyBox
.
Filename: src/main.rs
struct MyBox<T>(T);
impl<T> MyBox<T> {
fn new(x: T) -> MyBox<T> {
MyBox(x)
}
}
fn main() {
let x = 5;
let y = MyBox::new(x);
assert_eq!(5, x);
assert_eq!(5, *y);
}
Here’s the resulting compilation error:
$ cargo run
Compiling deref-example v0.1.0 (file:///projects/deref-example)
error[E0614]: type `MyBox<{integer}>` cannot be dereferenced
--> src/main.rs:14:19
|
14 | assert_eq!(5, *y);
| ^^
For more information about this error, try `rustc --explain E0614`.
error: could not compile `deref-example` due to previous error
Our MyBox<T>
type can’t be dereferenced because we haven’t implemented that
ability on our type. To enable dereferencing with the *
operator, we
implement the Deref
trait.
Treating a Type Like a Reference by Implementing the Deref
Trait
As discussed in Chapter 10, to implement a trait, we need to provide
implementations for the trait’s required methods. The Deref
trait, provided
by the standard library, requires us to implement one method named deref
that
borrows self
and returns a reference to the inner data. Listing 15-10
contains an implementation of Deref
to add to the definition of MyBox
:
Filename: src/main.rs
use std::ops::Deref; impl<T> Deref for MyBox<T> { type Target = T; fn deref(&self) -> &Self::Target { &self.0 } } struct MyBox<T>(T); impl<T> MyBox<T> { fn new(x: T) -> MyBox<T> { MyBox(x) } } fn main() { let x = 5; let y = MyBox::new(x); assert_eq!(5, x); assert_eq!(5, *y); }
The type Target = T;
syntax defines an associated type for the Deref
trait
to use. Associated types are a slightly different way of declaring a generic
parameter, but you don’t need to worry about them for now; we’ll cover them in
more detail in Chapter 19.
We fill in the body of the deref
method with &self.0
so deref
returns a
reference to the value we want to access with the *
operator. The main
function in Listing 15-9 that calls *
on the MyBox<T>
value now compiles,
and the assertions pass!
Without the Deref
trait, the compiler can only dereference &
references.
The deref
method gives the compiler the ability to take a value of any type
that implements Deref
and call the deref
method to get a &
reference that
it knows how to dereference.
When we entered *y
in Listing 15-9, behind the scenes Rust actually ran this
code:
*(y.deref())
Rust substitutes the *
operator with a call to the deref
method and then a
plain dereference so we don’t have to think about whether or not we need to
call the deref
method. This Rust feature lets us write code that functions
identically whether we have a regular reference or a type that implements
Deref
.
The reason the deref
method returns a reference to a value, and that the plain
dereference outside the parentheses in *(y.deref())
is still necessary, is the
ownership system. If the deref
method returned the value directly instead of
a reference to the value, the value would be moved out of self
. We don’t want
to take ownership of the inner value inside MyBox<T>
in this case or in most
cases where we use the dereference operator.
Note that the *
operator is replaced with a call to the deref
method and
then a call to the *
operator just once, each time we use a *
in our code.
Because the substitution of the *
operator does not recurse infinitely, we
end up with data of type i32
, which matches the 5
in assert_eq!
in
Listing 15-9.
Implicit Deref Coercions with Functions and Methods
Deref coercion is a convenience that Rust performs on arguments to functions
and methods. Deref coercion works only on types that implement the Deref
trait. Deref coercion converts such a type into a reference to another type.
For example, deref coercion can convert &String
to &str
because String
implements the Deref
trait such that it returns &str
. Deref coercion happens
automatically when we pass a reference to a particular type’s value as an
argument to a function or method that doesn’t match the parameter type in the
function or method definition. A sequence of calls to the deref
method
converts the type we provided into the type the parameter needs.
Deref coercion was added to Rust so that programmers writing function and
method calls don’t need to add as many explicit references and dereferences
with &
and *
. The deref coercion feature also lets us write more code that
can work for either references or smart pointers.
To see deref coercion in action, let’s use the MyBox<T>
type we defined in
Listing 15-8 as well as the implementation of Deref
that we added in Listing
15-10. Listing 15-11 shows the definition of a function that has a string slice
parameter:
Filename: src/main.rs
fn hello(name: &str) { println!("Hello, {}!", name); } fn main() {}
We can call the hello
function with a string slice as an argument, such as
hello("Rust");
for example. Deref coercion makes it possible to call hello
with a reference to a value of type MyBox<String>
, as shown in Listing 15-12:
Filename: src/main.rs
use std::ops::Deref; impl<T> Deref for MyBox<T> { type Target = T; fn deref(&self) -> &T { &self.0 } } struct MyBox<T>(T); impl<T> MyBox<T> { fn new(x: T) -> MyBox<T> { MyBox(x) } } fn hello(name: &str) { println!("Hello, {}!", name); } fn main() { let m = MyBox::new(String::from("Rust")); hello(&m); }
Here we’re calling the hello
function with the argument &m
, which is a
reference to a MyBox<String>
value. Because we implemented the Deref
trait
on MyBox<T>
in Listing 15-10, Rust can turn &MyBox<String>
into &String
by calling deref
. The standard library provides an implementation of Deref
on String
that returns a string slice, and this is in the API documentation
for Deref
. Rust calls deref
again to turn the &String
into &str
, which
matches the hello
function’s definition.
If Rust didn’t implement deref coercion, we would have to write the code in
Listing 15-13 instead of the code in Listing 15-12 to call hello
with a value
of type &MyBox<String>
.
Filename: src/main.rs
use std::ops::Deref; impl<T> Deref for MyBox<T> { type Target = T; fn deref(&self) -> &T { &self.0 } } struct MyBox<T>(T); impl<T> MyBox<T> { fn new(x: T) -> MyBox<T> { MyBox(x) } } fn hello(name: &str) { println!("Hello, {}!", name); } fn main() { let m = MyBox::new(String::from("Rust")); hello(&(*m)[..]); }
The (*m)
dereferences the MyBox<String>
into a String
. Then the &
and
[..]
take a string slice of the String
that is equal to the whole string to
match the signature of hello
. The code without deref coercions is harder to
read, write, and understand with all of these symbols involved. Deref coercion
allows Rust to handle these conversions for us automatically.
When the Deref
trait is defined for the types involved, Rust will analyze the
types and use Deref::deref
as many times as necessary to get a reference to
match the parameter’s type. The number of times that Deref::deref
needs to be
inserted is resolved at compile time, so there is no runtime penalty for taking
advantage of deref coercion!
How Deref Coercion Interacts with Mutability
Similar to how you use the Deref
trait to override the *
operator on
immutable references, you can use the DerefMut
trait to override the *
operator on mutable references.
Rust does deref coercion when it finds types and trait implementations in three cases:
- From
&T
to&U
whenT: Deref<Target=U>
- From
&mut T
to&mut U
whenT: DerefMut<Target=U>
- From
&mut T
to&U
whenT: Deref<Target=U>
The first two cases are the same except for mutability. The first case states
that if you have a &T
, and T
implements Deref
to some type U
, you can
get a &U
transparently. The second case states that the same deref coercion
happens for mutable references.
The third case is trickier: Rust will also coerce a mutable reference to an immutable one. But the reverse is not possible: immutable references will never coerce to mutable references. Because of the borrowing rules, if you have a mutable reference, that mutable reference must be the only reference to that data (otherwise, the program wouldn’t compile). Converting one mutable reference to one immutable reference will never break the borrowing rules. Converting an immutable reference to a mutable reference would require that the initial immutable reference is the only immutable reference to that data, but the borrowing rules don’t guarantee that. Therefore, Rust can’t make the assumption that converting an immutable reference to a mutable reference is possible.