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 a similar way as references. Then we’ll look at Rust’s deref coercion feature and how it lets us work with either references or smart pointers.
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);
}
Listing 15-6: Using the dereference operator to follow a reference to an i32
value
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:
error[E0277]: the trait bound `{integer}: std::cmp::PartialEq<&{integer}>` is not satisfied --> src/main.rs:6:5 | 6 | assert_eq!(5, y); | ^^^^^^^^^^^^^^^^^ can't compare `{integer}` with `&{integer}` | = help: the trait `std::cmp::PartialEq<&{integer}>` is not implemented for `{integer}`
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);
}
Listing 15-7: Using the dereference operator on a Box<i32>
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 the value in 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 than 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
# #![allow(unused_variables)]
#fn main() {
struct MyBox<T>(T);
impl<T> MyBox<T> {
fn new(x: T) -> MyBox<T> {
MyBox(x)
}
}
#}
Listing 15-8: Defining a MyBox<T>
type
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
fn main() { let x = 5; let y = MyBox::new(x); assert_eq!(5, x); assert_eq!(5, *y); }
Listing 15-9: Attempting to use MyBox<T>
in the same way we used references and Box<T>
Here’s the resulting compilation error:
error[E0614]: type `MyBox<{integer}>` cannot be dereferenced --> src/main.rs:14:19 | 14 | assert_eq!(5, *y); | ^^
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
# #![allow(unused_variables)]
#fn main() {
use std::ops::Deref;
# struct MyBox<T>(T);
impl<T> Deref for MyBox<T> {
type Target = T;
fn deref(&self) -> &T {
&self.0
}
}
#}
Listing 15-10: Implementing Deref
on MyBox<T>
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 converts a reference to a type that implements Deref
into a reference to a type that Deref
can convert the original type into. 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
# #![allow(unused_variables)]
#fn main() {
fn hello(name: &str) {
println!("Hello, {}!", name);
}
#}
Listing 15-11: A hello
function that has the parameter name
of type &str
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;
#
# struct MyBox<T>(T);
#
# impl<T> MyBox<T> {
# fn new(x: T) -> MyBox<T> {
# MyBox(x)
# }
# }
#
# impl<T> Deref for MyBox<T> {
# type Target = T;
#
# fn deref(&self) -> &T {
# &self.0
# }
# }
#
# fn hello(name: &str) {
# println!("Hello, {}!", name);
# }
#
fn main() {
let m = MyBox::new(String::from("Rust"));
hello(&m);
}
Listing 15-12: Calling hello
with a reference to a MyBox<String>
value, which works because of deref coercion
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;
#
# struct MyBox<T>(T);
#
# impl<T> MyBox<T> {
# fn new(x: T) -> MyBox<T> {
# MyBox(x)
# }
# }
#
# impl<T> Deref for MyBox<T> {
# type Target = T;
#
# fn deref(&self) -> &T {
# &self.0
# }
# }
#
# fn hello(name: &str) {
# println!("Hello, {}!", name);
# }
#
fn main() {
let m = MyBox::new(String::from("Rust"));
hello(&(*m)[..]);
}
Listing 15-13: The code we would have to write if Rust didn’t have deref coercion
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 there is only one immutable reference to that data, and 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.