有没有办法做 unwrap_or_return 一个错误（任何错误）

Question

有什么方法可以简化以下示例中的 returns（最初是从 here 复制的）：

use std::num::ParseIntError;

fn multiply(first_number_str: &str, second_number_str: &str) -> Result<i32, ParseIntError> {
    let first_number = match first_number_str.parse::<i32>() {
        Ok(first_number)  => first_number,
        Err(e) => return Err(e),
    };

    let second_number = match second_number_str.parse::<i32>() {
        Ok(second_number)  => second_number,
        Err(e) => return Err(AnotherError::ParseError("error")),
    };

    Ok(first_number * second_number)
}

我的意思是：

use std::num::ParseIntError;

fn multiply(first_number_str: &str, second_number_str: &str) -> Result<i32, ParseIntError> {
    let first_number = first_number_str.parse::<i32>()
        .unwrap_or_return(|e| Err(e));

    let second_number = second_number_str.parse::<i32>()
        .unwrap_or_return(|e| Err(AnotherError::ParseError("error"));

    Ok(first_number * second_number)
}

Answer 1

您正在寻找 question mark operator, possibly combined with Result::or or Result::or_else，具体取决于您的用例的具体情况。

该代码示例可以重写为

use std::num::ParseIntError;

pub fn multiply(first_number_str: &str, second_number_str: &str) -> Result<i32, ParseIntError> {
    let first_number = first_number_str.parse::<i32>()?;
    let second_number = second_number_str.parse::<i32>().or_else(|e| Err(e))?;
    // The closure in `or_else` could also return `Ok` or some different error with type `ParseIntError`.
    // let second_number = second_number_str.parse::<i32>().or_else(|_e| Ok(27))?;

    Ok(first_number * second_number)
}

(playground)

如果你知道你要在or_elsereturnOk，Result::unwrap_or是比较合适的。查看 Result 上的其他类似方法以查看提供的内容。

有没有办法做 unwrap_or_return 一个错误（任何错误）

Is there any way of doing unwrap_or_return an Error (any error)

return

rust

unwrap