为什么我不能在带有类型参数的特征上添加一个全面的实现? [英] Why can't I add a blanket impl on a trait with a type parameter?

问题描述

考虑这两个特征:

pub trait Foo {
    fn new(arg: u32) -> Self;
}

pub trait Bar<P>: Foo {
    fn with_parameter(arg: u32, parameter: P) -> Self;
}

我想添加一揽子实现:

impl<T: Bar<P>, P: Default> Foo for T {
    fn new(arg: u32) -> Self {
        Self::with_parameter(arg, P::default())
    }
}

但我收到编译器错误:

error[E0207]: the type parameter `P` is not constrained by the impl trait, self type, or predicates
 --> src/lib.rs:9:17
  |
9 | impl<T: Bar<P>, P: Default> Foo for T {
  |                 ^ unconstrained type parameter

我想我收到这个错误是因为我违反了特征一致性规则，但我不明白这会违反什么规则.为什么不允许这种模式?而且，更重要的是，我可以在不出错的情况下实现我想要的吗?

I think I get this error because I'm violating trait coherence rules, but I don't understand exactly what rule this would break. Why is this pattern not allowed? And, more importantly, can I achieve what I want without getting an error?

推荐答案

问题是单个类型可以为 P 的多个值实现 Bar

.如果您有一个结构 Baz 实现了 Bar 和 Bar，那么哪种类型应该 Foo::new 用于 P?

The problem is that a single type could implement Bar<P> for multiple values of P. If you had a struct Baz that implemented Bar<i32> and Bar<String>, which type should Foo::new use for P?

唯一的解决方案是确保单个类型不能多次实现 Bar(如果这不是您想要的，那么您的设计就有缺陷！).为此，我们必须用关联类型替换 P 类型参数.

The only solution is to ensure that a single type cannot implement Bar more than once (if that's not what you want, then you have a flaw in your design!). To do so, we must replace the P type parameter with an associated type.

pub trait Bar: Foo {
    type Parameter;

    fn with_parameter(arg: u32, parameter: Self::Parameter) -> Self;
}

impl<T> Foo for T
where
    T: Bar,
    T::Parameter: Default,
{
    fn new(arg: u32) -> Self {
        Self::with_parameter(arg, T::Parameter::default())
    }
}

Bar 的实现如下所示:

struct Baz;

impl Bar for Baz {
    type Parameter = i32;

    fn with_parameter(arg: u32, parameter: Self::Parameter) -> Self {
        unimplemented!()
    }
}

另见:

• 为什么我得到类型参数不受约束"?在为闭包特征 (Fn) 创建全面实现时?

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