在以下示例(最初从here复制)中,有什么方法可以简化返回值:
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)
}
最佳答案
您正在寻找question mark operator,可能与 Result::or
或 Result::or_else
结合使用,具体取决于您的用例。
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)
如果您知道要在
Ok
中返回or_else
,则 Result::unwrap_or
更合适。在Result
上查看其他类似方法,以了解提供了什么。
关于rust - 有没有办法做unwrap_or_return一个错误(任何错误),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60011028/