提问人:KSV 提问时间:7/28/2023 最后编辑:cafce25KSV 更新时间:7/28/2023 访问量:59
Rust 泛型函数接受原始整数和地毯整数
Rust generic function accepting both primitive and rug integers
问:
我正在尝试编写一个对整数进行操作的泛型函数,它可以接受原始整数或多精度整数(通过)。棘手的一点是,对 rug s 的引用具有惰性算术运算,在转换为整数之前返回不完整的类型。所以我认为像下面这样的东西会起作用,但我似乎无法获得正确的生存期设置(并且 rust-analyzer 在提供帮助方面并不是非常冗长):rug
Integer
use rug::Integer;
use std::ops::Mul;
fn mymul<T, S>(x: &T, y: &T) -> T
where
T: 'static + From<S>,
for<'a> &'a T: Mul<&'a T, Output = S>,
{
T::from(x * y)
}
fn main() {
let x: u64 = 3847381;
let y: u64 = 28478195;
let rug_x = Integer::from(x);
let rug_y = Integer::from(y);
println!("{} == {:?}", mymul(&x, &y), mymul(&rug_x, &rug_y));
}
有错误
error[E0308]: mismatched types
--> src/main.rs:22:43
|
22 | println!("{} == {:?}", mymul(&x, &y), mymul(&rug_x, &rug_y));
| ^^^^^^^^^^^^^^^^^^^^^ one type is more general than the other
|
= note: expected struct `integer::arith::MulIncomplete<'a>`
found struct `integer::arith::MulIncomplete<'_>`
note: the lifetime requirement is introduced here
--> src/main.rs:12:31
|
12 | for<'a> &'a T: Mul<&'a T, Output = S>,
| ^^^^^^^^^^
知道如何正确地做这种事情吗?
答:
3赞
Chayim Friedman
7/28/2023
#1
使用 HRTB 指定此函数的正确方法是:
fn mymul<T>(x: &T, y: &T) -> T
where
for<'a> &'a T: Mul<&'a T>,
for<'a> <&'a T as Mul<&'a T>>::Output: Into<T>,
{
(x * y).into()
}
评论
From
Into
T: 'static