generics - 如何定义对可以进行按位运算的整数通用的 Rust 函数？

Question

我有以下功能:

fn f1(n: u8) -> u16 {
    1 << n
}

我可以尝试(未成功)使其在整数上通用:

extern crate num;

use num::Integer;

fn f1<T: Integer>(n: u8) -> T {
    1 << n
}

这行不通。它会产生以下错误:

error[E0308]: mismatched types
 --> src/main.rs:6:5
  |
5 | fn f1<T: Integer>(n: u8) -> T {
  |                             - expected `T` because of return type
6 |     1 << n
  |     ^^^^^^ expected type parameter, found integral variable
  |
  = note: expected type `T`
             found type `{integer}`

我知道有 Shl trait .我需要使用这个特性来完成这项工作吗？我该怎么做？

Answer 1

Do I need to use [Shl] to make this work?

是的。您还需要确保 result of the operation is the right type :

extern crate num;

use num::Integer;
use std::ops::Shl;

fn f1<T>(n: u8) -> T
where
    T: Integer + Shl<u8, Output = T>,
{
    T::one() << n
}