rust - 错误处理、map_err 和错误类型转换

Question

我正在学习 Rust，目前正在做练习 here .

到目前为止，我已经想出了以下内容:

impl ParsePosNonzeroError {
   ...
   ...
   fn from_parseint(err: ParseIntError) -> ParsePosNonzeroError {
       ParsePosNonzeroError::ParseInt(err)
   }

fn parse_pos_nonzero(s: &str)
-> Result<PositiveNonzeroInteger, ParsePosNonzeroError>
{
     let x: i64 = s.parse::<i64>().map_err(ParsePosNonzeroError::from_parseint);
     PositiveNonzeroInteger::new(x).map_err(ParsePosNonzeroError::from_creation)
}

当我这样做时，我收到错误消息:

! Compiling of exercises/error_handling/errors6.rs failed! Please try again. Here's the output:
error[E0308]: mismatched types
  --> exercises/error_handling/errors6.rs:34:18
   |
34 |       let x: i64 = s.parse::<i64>()
   |  ____________---___^
   | |            |
   | |            expected due to this
35 | |         .map_err(ParsePosNonzeroError::from_parseint);
   | |_____________________________________________________^ expected `i64`, found enum `Result`
   |
   = note: expected type `i64`
              found enum `Result<i64, ParsePosNonzeroError>`

浏览 map_err 的文档我看到以下内容: Maps a Result<T, E> to Result<T, F> by applying a function to a contained Err value, leaving an Ok value untouched.

我错过了什么？为什么这不起作用？

Answer 1

map_err正在返回 Result<i64, ParsePosNonzeroError>但你正试图将其影响到 x ，您将其声明为 i64多变的。您需要添加一个 ?这样一个Err将返回结果和 Ok结果将被展开:let x = s.parse::<i64>().map_err(ParsePosNonzeroError::from_parseint)?;