Rust 将trait方法和关联类型标记为专门化的默认值时,预期输出类型会发生更改
我想为Rust中的大多数Rust 将trait方法和关联类型标记为专门化的默认值时,预期输出类型会发生更改,rust,type-inference,specialization,Rust,Type Inference,Specialization,我想为Rust中的大多数Rem类型实现一个操作: #![feature(specialization)] use std::ops::{Add, Rem}; /// Define a modulo operation, in the mathematical sense. /// This differs from Rem because the result is always non-negative. pub trait Modulo<T> { type Outpu
Rem
类型实现一个操作:
#![feature(specialization)]
use std::ops::{Add, Rem};
/// Define a modulo operation, in the mathematical sense.
/// This differs from Rem because the result is always non-negative.
pub trait Modulo<T> {
type Output;
#[inline]
fn modulo(self, other: T) -> Self::Output;
}
/// Implement modulo operation for types that implement Rem, Add and Clone.
// Add and Clone are needed to shift the value by U if it is below zero.
impl<U, T> Modulo<T> for U
where
T: Clone,
U: Rem<T>,
<U as Rem<T>>::Output: Add<T>,
<<U as Rem<T>>::Output as Add<T>>::Output: Rem<T>
{
default type Output = <<<U as Rem<T>>::Output as Add<T>>::Output as Rem<T>>::Output;
#[inline]
default fn modulo(self, other: T) -> Self::Output {
((self % other.clone()) + other.clone()) % other
}
}
我不明白为什么会这样。我需要默认值
s,因为我想将其专门用于复制
类型
我每晚都在使用Rust 1.29.0。这里是问题的一个较小的复制品(an): 在这种情况下,
foo
的原始实现将不再有意义-它不再返回类型为Self::Output
的值
当前的专门化实现要求您同时考虑本地和全局,这是您必须阅读错误消息的上下文。这并不理想,但像这样的问题(我相信还有许多更复杂的事情)是它还不稳定的部分原因 FWIW,你可以说
U::Output:Add
-一个级别的deep不需要解决它的
@Shepmaster,但是我不知道从错误消息中得出的结论是什么
#![feature(specialization)]
trait Example {
type Output;
fn foo(&self) -> Self::Output;
}
impl<T> Example for T {
default type Output = i32;
default fn foo(&self) -> Self::Output {
42
}
}
fn main() {}
impl<T> Example for T
where
T: Copy,
{
type Output = bool;
}