Do I have to implement a trait twice when implemen

问题:

I want to implement a trait for both a for reference and non-reference type. Do I have to implement the functions twice, or this is not idiomatic to do so?

Here's the demo code:

struct Bar {}

trait Foo {
    fn hi(&self);
}

impl<'a> Foo for &'a Bar {
    fn hi(&self) {
        print!("hi")
    }
}

impl Foo for Bar {
    fn hi(&self) {
        print!("hi")
    }
}

fn main() {
    let bar = Bar {};
    (&bar).hi();
    &bar.hi();
}

回答1:

This is a good example for the Borrow trait.

use std::borrow::Borrow;

struct Bar;

trait Foo {
    fn hi(&self);
}

impl<B: Borrow<Bar>> Foo for B {
    fn hi(&self) {
        print!("hi")
    }
}

fn main() {
    let bar = Bar;
    (&bar).hi();
    &bar.hi();
}

回答2:

No, you do not have to duplicate code. Instead, you can delegate:

impl Foo for &'_ Bar {
    fn hi(&self) {
        (**self).hi()
    }
}

I would go one step further and implement the trait for all references to types that implement the trait:

impl<T: Foo> Foo for &'_ T {
    fn hi(&self) {
        (**self).hi()
    }
}

See also:

• When should I not implement a trait for references to implementors of that trait?
• Implementing a trait for reference and non reference types causes conflicting implementations

---

&bar.hi();

This code is equivalent to &(bar.hi()) and probably not what you intended.

See also:

• Why is it legal to borrow a temporary?

回答3:

You can use Cow:

use std::borrow::Cow;

#[derive(Clone)]
struct Bar;

trait Foo {
    fn hi(self) -> &'static str;
}

impl<'a, B> Foo for B where B: Into<Cow<'a, Bar>> {
    fn hi(self) -> &'static str {
        let bar = self.into();

        // bar is either owned or borrowed:
        match bar {
            Cow::Owned(_) => "Owned",
            Cow::Borrowed(_) => "Borrowed",
        }
    }
}

/* Into<Cow> implementation */

impl<'a> From<Bar> for Cow<'a, Bar> {
    fn from(f: Bar) -> Cow<'a, Bar> {
        Cow::Owned(f)
    }
}

impl<'a> From<&'a Bar> for Cow<'a, Bar> {
    fn from(f: &'a Bar) -> Cow<'a, Bar> {
        Cow::Borrowed(f)
    }
}

/* Proof it works: */

fn main() {
    let bar = &Bar;
    assert_eq!(bar.hi(), "Borrowed");

    let bar = Bar;
    assert_eq!(bar.hi(), "Owned");
}

The one advantage over Borrow is that you know if the data was passed by value or reference, if that matters to you.