generics - 有没有办法为智能指针中的对象通用地实现一个特征？

Question

有没有一种方法可以通用地为智能指针中的对象(例如 Box、Rc 等)的特征提供实现？

例子:

use std::borrow::Borrow;
use std::rc::Rc;

trait KindaEqual<Rhs = Self> {
    fn kinda_equal(&self, other: &Rhs) -> bool;
}

// Nice to have (still need to copy-paste for each LHS container)
impl<T, Rhs, WrappedRhs> KindaEqual<WrappedRhs> for T
where
    WrappedRhs: Borrow<Rhs>,
    T: KindaEqual<Rhs>,
{
    fn kinda_equal(&self, other: &WrappedRhs) -> bool {
        self.kinda_equal(other.borrow())
    }
}

// Best to have (handles both LHS and RHS containers)
impl<WrappedT, T, Rhs, WrappedRhs> KindaEqual<WrappedRhs> for WrappedT
where
    WrappedRhs: Borrow<Rhs>,
    WrappedT: Borrow<T>,
    T: KindaEqual<Rhs>,
{
    fn kinda_equal(&self, other: &WrappedRhs) -> bool {
        self.borrow().kinda_equal(other.borrow())
    }
}

impl KindaEqual for f64 {
    fn kinda_equal(&self, other: &Self) -> bool {
        num::abs(self - other) < 0.01f64
    }
}

impl KindaEqual<u64> for f64 {
    fn kinda_equal(&self, other: &u64) -> bool {
        num::abs(self - *other as f64) < 1f64
    }
}

fn main() {
    assert!(3.141592654.kinda_equal(&3.14));
    assert!(3.141592654.kinda_equal(&3));
    assert!(3.141592654.kinda_equal(&Rc::new(3.14)));
}

上面给出了错误:

error[E0207]: the type parameter `Rhs` is not constrained by the impl trait, self type, or predicates
 --> src/main.rs:9:9
  |
9 | impl<T, Rhs, WrappedRhs> KindaEqual<WrappedRhs> for T
  |         ^^^ unconstrained type parameter

error[E0207]: the type parameter `T` is not constrained by the impl trait, self type, or predicates
  --> src/main.rs:20:16
   |
20 | impl<WrappedT, T, Rhs, WrappedRhs> KindaEqual<WrappedRhs> for WrappedT
   |                ^ unconstrained type parameter

error[E0207]: the type parameter `Rhs` is not constrained by the impl trait, self type, or predicates
  --> src/main.rs:20:19
   |
20 | impl<WrappedT, T, Rhs, WrappedRhs> KindaEqual<WrappedRhs> for WrappedT
   |                   ^^^ unconstrained type parameter

我读到过添加关联类型可能会修复它，但这是不可取的，因为它会强制使用特征来实现它。

1. 这可以在没有关联类型的情况下解决吗？ (最好)
2. 这可以用关联类型解决吗？

Answer 1

      /-- everything here
impl<Types...>     Trait              for         Self
                     \-- has to appear here         \-- or here

作为error message表示，Types... 必须受到 Trait、Self 或 Trait 或 self 。

如果您可以为每个 LHS 容器复制粘贴，您可以将 Borrow 处理向下移动到函数本身，如下所示:

use std::borrow::Borrow;

trait KindaEqual<Rhs = Self> {
    fn kinda_equal<R: Borrow<Rhs>>(&self, other: &R) -> bool;
}

impl KindaEqual<u32> for u32 {
    fn kinda_equal<R: Borrow<u32>>(&self, other: &R) -> bool {
        self == other.borrow()
    }
}

// repeat for all supported containers
impl<T> KindaEqual<T> for Box<T>
where
    T: KindaEqual<T>,
{
    fn kinda_equal<R: Borrow<T>>(&self, other: &R) -> bool {
        (&**self).kinda_equal(other.borrow())
    }
}