如何在 u32 和 usize 之间进行惯用转换? [英] How to idiomatically convert between u32 and usize?

问题描述

此代码有效并打印b":

This code works and prints "b":

fn main() {
    let s = "abc";
    let ch = s.chars().nth(1).unwrap();
    println!("{}", ch);
}

另一方面，此代码会导致类型不匹配错误.

On the other hand, this code results in a mismatch type error.

fn main() {
    let s = "abc";
    let n: u32 = 1;
    let ch = s.chars().nth(n).unwrap();
    println!("{}", ch);
}

error[E0308]: mismatched types
 --> src/main.rs:5:28
  |
5 |     let ch = s.chars().nth(n).unwrap();
  |                            ^ expected usize, found u32

由于某些外部原因，我必须对变量 n 使用 u32 类型.如何将 u32 转换为 usize 并在 nth() 中使用它?

For some external reason, I have to use the u32 type for variable n. How can I convert u32 to usize and use it in nth()?

推荐答案

as 运算符适用于所有数字类型:

The as operator works for all number types:

let ch = s.chars().nth(n as usize).unwrap();

Rust 强制您强制转换整数以确保您知道有符号或溢出.

Rust forces you to cast integers to make sure you're aware of signedness or overflows.

整数常量可以有类型后缀:

Integer constants can have a type suffix:

let n = 1u32;

但是要注意负常数，比如-1i32在内部是- 1i32.

However, note that negative constants, such as -1i32 is internally - 1i32.

没有明确类型规范声明的整数变量显示为 {integer} 并且将从方法调用之一正确推断.

Integer variables declared without an explicit type specification are shown as {integer} and will be properly inferred from one of the method calls.

这篇关于如何在 u32 和 usize 之间进行惯用转换?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！