如何将泛型转换为原始类型

Question

我是 Rust 的新手。我不知道如何将通用类型 <T> 转换为原始类型。

有一个小例子，一个泛型类型的函数 sum:

fn sum<T: std::ops::Add<Output = T>, U>(x:T, y: U) -> T {
    // is there any line of code similar to:
    // x + y as T
    
    x + y as T
    
    // or check the type
    // match type(x) {
    //    i32 => x + y as i32,
    //    i64 => x + y as i64,
    //    f32 => x + y as f32,
    //    _ => 0
    // }
}

fn main() {
    let a = 1;
    let b = 22.22;
    println!("{}", sum(a, b));
    
    let a = 11.11;
    let b = 2;
    println!("{}", sum(a, b));
}

Answer 1

最初的想法可能是要求 U: Into<T>. However, there is no e.g. impl Into<i32> for f32 so that won't work. Might also think of requiring T: Add<U, Output = T>，但是出于同样的原因，这在您的情况下也行不通。

相反，您可以使用 num crate, specifically the AsPrimitive 特征。

use std::ops::Add;

// num = "0.3"
use num::cast::AsPrimitive;

fn sum<T, U>(x: T, y: U) -> T
where
    T: Copy + 'static,
    T: Add<Output = T>,
    U: AsPrimitive<T>,
{
    x + y.as_()
}

使用 sum() 的实现，然后执行 main() 将输出以下内容：

23
13.11

如何将泛型转换为原始类型

How to cast a generic type to primitive type

generics

rust

casting

primitive-types