如何在Rust中绑定`Output`类型的运算符特征？ [重复]

Question

这个问题在这里已有答案：

• How do I require a generic type implement an operation like Add, Sub, Mul, or Div in a generic function? 2回答
• How do I use floating point number literals when using generic types? 3回答

我想使用以下代码：

use std::ops::Rem;

fn modulo<T: PartialOrd + Rem>(num: T, det: T) -> T {
    let num = num % det;
    if num < 0.0 {
        num + det.abs()
    } else {
        num
    }
}

这只是取自实验性euclidean_division特征的通用代码。提供的错误如下：

error[E0369]: binary operation `<` cannot be applied to type `<T as std::ops::Rem>::Output`
 --> src/lib.rs:5:8
  |
5 |     if num < 0.0 {
  |        ^^^^^^^^^
  |
  = note: an implementation of `std::cmp::PartialOrd` might be missing for `<T as std::ops::Rem>::Output`

error[E0599]: no method named `abs` found for type `T` in the current scope
 --> src/lib.rs:6:19
  |
6 |         num + det.abs()
  |                   ^^^

error[E0369]: binary operation `+` cannot be applied to type `<T as std::ops::Rem>::Output`
 --> src/lib.rs:6:9
  |
6 |         num + det.abs()
  |         ^^^^^^^^^^^^^^^
  |
  = note: an implementation of `std::ops::Add` might be missing for `<T as std::ops::Rem>::Output`

error[E0308]: mismatched types
 --> src/lib.rs:8:9
  |
3 | fn modulo<T: PartialOrd + Rem>(num: T, det: T) -> T {
  |                                                   - expected `T` because of return type
...
8 |         num
  |         ^^^ expected type parameter, found associated type
  |
  = note: expected type `T`
             found type `<T as std::ops::Rem>::Output`

显然Rem::rem的输出应该与T的输出类型相同，但我认为编译器并不知道。

有没有办法解决这个问题，或者我应该按照每晚的版本以同样的方式实现它？

Answer 1

具有基元的泛型并非如此简单。修复第一个错误会显示其他错误，因为您将T与0.0进行比较。有很多方法可以解决这些问题;这是使用num的一种方式：

use num::{Signed, Zero};
use std::ops::Rem;

fn modulo<T>(num: T, det: T) -> T
where
    T: Rem<Output = T>,
    T: PartialOrd,
    T: Zero,
    T: Signed,
    T: Copy,
{
    let num = num % det;
    if num < T::zero() {
        num + det.abs()
    } else {
        num
    }
}