为什么使用我的类型作为外部类型合法的参数来实现外部特征? [英] Why is implementing an external trait using my type as a parameter for an external type legal?
问题描述
我正在修改一些代码以依赖于 rand 0.5 版.起初，我担心如何使用 Standard 生成我自己类型的随机值，但我发现这是合法的:
I am modifying some code to depend on rand version 0.5. At first, I was worried how I would enable generating random values of my own types with Standard, but I found out this is legal:
impl ::rand::distributions::Distribution<MyType> for ::rand::distributions::Standard {
// ...
}
为什么是合法的?我认为为外部类型实现外部特征是 非法.
Why is it legal? I thought implementing an external trait for an external type is illegal.
推荐答案
这些规则(称为孤立规则或一致性规则)的全部目的是避免给定特征/类型组合的任何冲突实现.从 Rust 1.0 开始，人们认为，对于允许谁为类型实现 trait 的随意"，这不符合追求稳固稳定性的语言.
The entire purpose of these rules (called the orphan rules or coherence rules) is to avoid having any conflicting implementations for a given trait/type combination. Since Rust 1.0, it is believed that it would not be in line with a language striving for solid stability to be "willy-nilly" about who is allowed to implement a trait for a type.
这种特殊实现类型的一个非常常见的例子是From:
A very common example of this particular type of implementation is From:
impl From<MyType> for i32 {
// ...
}
通过使用本地类型参数化 trait，对于谁在实现 trait 仍然没有歧义.一种思考方式是将 Distribution 视为不是 trait 而是 trait 构造函数 1.Distribution 的每个实例都会创建一个新特征，这是为案例定制的.
By parameterizing the trait with a local type, there's still no ambiguity about who is implementing the trait. One way of thinking about it would be treating Distribution as not a trait but a trait constructor 1. Each instance of Distribution creates a new trait, one that's custom-made for the case.
另见:
1 — 这不是真的，但这是一个合理的类比.
1 — This isn't true, but it's a reasonable analogy.
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