问题
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 {
// ...
}
Why is it legal? I thought implementing an external trait for an external type is illegal.
回答1:
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.
A very common example of this particular type of implementation is From:
impl From<MyType> for i32 {
// ...
}
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.
See also:
- RFC 2451, "Re-Rebalancing Coherence"
- RFC 1023, "Rebalancing Coherence"
- How do I implement a trait I don't own for a type I don't own?
1 — This isn't true, but it's a reasonable analogy.
来源：https://stackoverflow.com/questions/51236531/why-is-implementing-an-external-trait-using-my-type-as-a-parameter-for-an-extern