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rust - 有没有办法做unwrap_or_return一个错误(任何错误)

发布于 2020-04-09 11:03:18

在下面的示例中,有什么方法可以简化返回值(最初从此处复制):

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)
}

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提问者
Omid
被浏览
163
SCappella 2020-02-01 05:10

您正在寻找问号运算符,可能会与Result::or结合使用Result::or_else,具体取决于用例的具体情况。

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)
}

(操场)

如果你知道你要返回Okor_elseResult::unwrap_or是比较合适的。查看其他类似方法Result以查看提供的方法。