为引用和非引用类型实现特征时，是否必须两次实现特征？

Question

我想为供参考和非参考类型实现特征。我必须两次实施这些功能，还是这样做不是习惯？

这是演示代码：

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();
}

Answer 1

这是Borrow特性的一个很好的例子。

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();
}

Answer 2

否，您不必重复代码。相反，您可以委派：

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

我将更进一步，为实现该特征的类型的所有引用实现该特征：

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

另请参阅：

• 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();

此代码等效于&(bar.hi())，可能不是您想要的。

另请参阅：

• Why is it legal to borrow a temporary?

Answer 3

您可以使用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");
}

与Borrow相比，优点之一是您知道数据是通过值还是引用传递，如果这对您来说很重要。