rust/src/doc/trpl/traits.md
2015-09-03 01:36:22 -04:00

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% Traits

A trait is a language feature that tells the Rust compiler about functionality a type must provide.

Do you remember the impl keyword, used to call a function with method syntax?

struct Circle {
    x: f64,
    y: f64,
    radius: f64,
}

impl Circle {
    fn area(&self) -> f64 {
        std::f64::consts::PI * (self.radius * self.radius)
    }
}

Traits are similar, except that we define a trait with just the method signature, then implement the trait for that struct. Like this:

struct Circle {
    x: f64,
    y: f64,
    radius: f64,
}

trait HasArea {
    fn area(&self) -> f64;
}

impl HasArea for Circle {
    fn area(&self) -> f64 {
        std::f64::consts::PI * (self.radius * self.radius)
    }
}

As you can see, the trait block looks very similar to the impl block, but we dont define a body, just a type signature. When we impl a trait, we use impl Trait for Item, rather than just impl Item.

Traits bounds for generic functions

Traits are useful because they allow a type to make certain promises about its behavior. Generic functions can exploit this to constrain the types they accept. Consider this function, which does not compile:

fn print_area<T>(shape: T) {
    println!("This shape has an area of {}", shape.area());
}

Rust complains:

error: no method named `area` found for type `T` in the current scope

Because T can be any type, we cant be sure that it implements the area method. But we can add a trait constraint to our generic T, ensuring that it does:

# trait HasArea {
#     fn area(&self) -> f64;
# }
fn print_area<T: HasArea>(shape: T) {
    println!("This shape has an area of {}", shape.area());
}

The syntax <T: HasArea> means “any type that implements the HasArea trait.” Because traits define function type signatures, we can be sure that any type which implements HasArea will have an .area() method.

Heres an extended example of how this works:

trait HasArea {
    fn area(&self) -> f64;
}

struct Circle {
    x: f64,
    y: f64,
    radius: f64,
}

impl HasArea for Circle {
    fn area(&self) -> f64 {
        std::f64::consts::PI * (self.radius * self.radius)
    }
}

struct Square {
    x: f64,
    y: f64,
    side: f64,
}

impl HasArea for Square {
    fn area(&self) -> f64 {
        self.side * self.side
    }
}

fn print_area<T: HasArea>(shape: T) {
    println!("This shape has an area of {}", shape.area());
}

fn main() {
    let c = Circle {
        x: 0.0f64,
        y: 0.0f64,
        radius: 1.0f64,
    };

    let s = Square {
        x: 0.0f64,
        y: 0.0f64,
        side: 1.0f64,
    };

    print_area(c);
    print_area(s);
}

This program outputs:

This shape has an area of 3.141593
This shape has an area of 1

As you can see, print_area is now generic, but also ensures that we have passed in the correct types. If we pass in an incorrect type:

print_area(5);

We get a compile-time error:

error: the trait `HasArea` is not implemented for the type `_` [E0277]

Traits bounds for generic structs

Your generic structs can also benefit from trait constraints. All you need to do is append the constraint when you declare type parameters. Here is a new type Rectangle<T> and its operation is_square():

struct Rectangle<T> {
    x: T,
    y: T,
    width: T,
    height: T,
}

impl<T: PartialEq> Rectangle<T> {
    fn is_square(&self) -> bool {
        self.width == self.height
    }
}

fn main() {
    let mut r = Rectangle {
        x: 0,
        y: 0,
        width: 47,
        height: 47,
    };

    assert!(r.is_square());

    r.height = 42;
    assert!(!r.is_square());
}

is_square() needs to check that the sides are equal, so the sides must be of a type that implements the core::cmp::PartialEq trait:

impl<T: PartialEq> Rectangle<T> { ... }

Now, a rectangle can be defined in terms of any type that can be compared for equality.

Here we defined a new struct Rectangle that accepts numbers of any precision—really, objects of pretty much any type—as long as they can be compared for equality. Could we do the same for our HasArea structs, Square and Circle? Yes, but they need multiplication, and to work with that we need to know more about operator traits.

Rules for implementing traits

So far, weve only added trait implementations to structs, but you can implement a trait for any type. So technically, we could implement HasArea for i32:

trait HasArea {
    fn area(&self) -> f64;
}

impl HasArea for i32 {
    fn area(&self) -> f64 {
        println!("this is silly");

        *self as f64
    }
}

5.area();

It is considered poor style to implement methods on such primitive types, even though it is possible.

This may seem like the Wild West, but there are two restrictions around implementing traits that prevent this from getting out of hand. The first is that if the trait isnt defined in your scope, it doesnt apply. Heres an example: the standard library provides a Write trait which adds extra functionality to Files, for doing file I/O. By default, a File wont have its methods:

let mut f = std::fs::File::open("foo.txt").ok().expect("Couldnt open foo.txt");
let buf = b"whatever"; // byte string literal. buf: &[u8; 8]
let result = f.write(buf);
# result.unwrap(); // ignore the error

Heres the error:

error: type `std::fs::File` does not implement any method in scope named `write`
let result = f.write(buf);
               ^~~~~~~~~~

We need to use the Write trait first:

use std::io::Write;

let mut f = std::fs::File::open("foo.txt").ok().expect("Couldnt open foo.txt");
let buf = b"whatever";
let result = f.write(buf);
# result.unwrap(); // ignore the error

This will compile without error.

This means that even if someone does something bad like add methods to i32, it wont affect you, unless you use that trait.

Theres one more restriction on implementing traits: either the trait, or the type youre writing the impl for, must be defined by you. So, we could implement the HasArea type for i32, because HasArea is in our code. But if we tried to implement ToString, a trait provided by Rust, for i32, we could not, because neither the trait nor the type are in our code.

One last thing about traits: generic functions with a trait bound use monomorphization (mono: one, morph: form), so they are statically dispatched. Whats that mean? Check out the chapter on trait objects for more details.

Multiple trait bounds

Youve seen that you can bound a generic type parameter with a trait:

fn foo<T: Clone>(x: T) {
    x.clone();
}

If you need more than one bound, you can use +:

use std::fmt::Debug;

fn foo<T: Clone + Debug>(x: T) {
    x.clone();
    println!("{:?}", x);
}

T now needs to be both Clone as well as Debug.

Where clause

Writing functions with only a few generic types and a small number of trait bounds isnt too bad, but as the number increases, the syntax gets increasingly awkward:

use std::fmt::Debug;

fn foo<T: Clone, K: Clone + Debug>(x: T, y: K) {
    x.clone();
    y.clone();
    println!("{:?}", y);
}

The name of the function is on the far left, and the parameter list is on the far right. The bounds are getting in the way.

Rust has a solution, and its called a where clause:

use std::fmt::Debug;

fn foo<T: Clone, K: Clone + Debug>(x: T, y: K) {
    x.clone();
    y.clone();
    println!("{:?}", y);
}

fn bar<T, K>(x: T, y: K) where T: Clone, K: Clone + Debug {
    x.clone();
    y.clone();
    println!("{:?}", y);
}

fn main() {
    foo("Hello", "world");
    bar("Hello", "world");
}

foo() uses the syntax we showed earlier, and bar() uses a where clause. All you need to do is leave off the bounds when defining your type parameters, and then add where after the parameter list. For longer lists, whitespace can be added:

use std::fmt::Debug;

fn bar<T, K>(x: T, y: K)
    where T: Clone,
          K: Clone + Debug {

    x.clone();
    y.clone();
    println!("{:?}", y);
}

This flexibility can add clarity in complex situations.

where is also more powerful than the simpler syntax. For example:

trait ConvertTo<Output> {
    fn convert(&self) -> Output;
}

impl ConvertTo<i64> for i32 {
    fn convert(&self) -> i64 { *self as i64 }
}

// can be called with T == i32
fn normal<T: ConvertTo<i64>>(x: &T) -> i64 {
    x.convert()
}

// can be called with T == i64
fn inverse<T>() -> T
        // this is using ConvertTo as if it were "ConvertTo<i64>"
        where i32: ConvertTo<T> {
    42.convert()
}

This shows off the additional feature of where clauses: they allow bounds where the left-hand side is an arbitrary type (i32 in this case), not just a plain type parameter (like T).

Default methods

If you already know how a typical implementor will define a method, you can let your trait supply a default:

trait Foo {
    fn is_valid(&self) -> bool;

    fn is_invalid(&self) -> bool { !self.is_valid() }
}

Implementors of the Foo trait need to implement is_valid(), but they dont need to implement is_invalid(). Theyll get this default behavior. They can override the default if they so choose:

# trait Foo {
#     fn is_valid(&self) -> bool;
#
#     fn is_invalid(&self) -> bool { !self.is_valid() }
# }
struct UseDefault;

impl Foo for UseDefault {
    fn is_valid(&self) -> bool {
        println!("Called UseDefault.is_valid.");
        true
    }
}

struct OverrideDefault;

impl Foo for OverrideDefault {
    fn is_valid(&self) -> bool {
        println!("Called OverrideDefault.is_valid.");
        true
    }

    fn is_invalid(&self) -> bool {
        println!("Called OverrideDefault.is_invalid!");
        true // this implementation is a self-contradiction!
    }
}

let default = UseDefault;
assert!(!default.is_invalid()); // prints "Called UseDefault.is_valid."

let over = OverrideDefault;
assert!(over.is_invalid()); // prints "Called OverrideDefault.is_invalid!"

Inheritance

Sometimes, implementing a trait requires implementing another trait:

trait Foo {
    fn foo(&self);
}

trait FooBar : Foo {
    fn foobar(&self);
}

Implementors of FooBar must also implement Foo, like this:

# trait Foo {
#     fn foo(&self);
# }
# trait FooBar : Foo {
#     fn foobar(&self);
# }
struct Baz;

impl Foo for Baz {
    fn foo(&self) { println!("foo"); }
}

impl FooBar for Baz {
    fn foobar(&self) { println!("foobar"); }
}

If we forget to implement Foo, Rust will tell us:

error: the trait `main::Foo` is not implemented for the type `main::Baz` [E0277]