Treating Smart Pointers Like Regular References with the Deref Trait - The Rust Programming Language

## 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 real `Box<T>`: our version will not store its data on the heap. We are focusing this example on `Deref`, 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);
    }

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:

    $ 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 `std::cmp::PartialEq<&{integer}>` is not implemented for `{integer}`
      = note: this error originates in a macro outside of the current crate (in Nightly builds, run with -Z external-macro-backtrace for more info)

    error: aborting due to previous error

    For more information about this error, try `rustc --explain E0277`.
    error: could not compile `deref-example`.

    To learn more, run the command again with --verbose.

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 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() {}

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

    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);
    }

Listing 15-9: Attempting to use `MyBox<T>` in the same way we used references and `Box<T>`

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);
       |                   ^^

    error: aborting due to previous error

    For more information about this error, try `rustc --explain E0614`.
    error: could not compile `deref-example`.

    To learn more, run the command again with --verbose.

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) -> &T {
            &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);
    }

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 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() {}

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;

    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);
    }

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;

    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)[..]);
    }

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` when `T: Deref<Target=U>`
- From `&mut T` to `&mut U` when `T: DerefMut<Target=U>`
- From `&mut T` to `&U` when `T: 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.