Question

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
    }
}

这只是从实验性euclidean_division功能中提取的通用代码。提供的错误大致如下：

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`

很显然，Rem::rem的输出应该与T的类型相同，但是我认为编译器并不知道这一点。

是否有解决此问题的方法，还是应该按夜间版本中的相同方式按类型实现它？

Answer 1

带有原语的泛型不是那么简单。修正第一个错误会揭示其他错误，因为您将T与0.0进行了比较。有很多方法可以解决这些问题。这是使用num的一种方法：

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
    }
}

如何在Rust中绑定“输出”类型的运算符特征？

1 个答案: