What is a clean way to convert a Result into an Option?

我的未来我决定 提交于 2020-12-29 08:49:16

问题


Before updating to a more recent Rust version the following used to work:

fn example(val: &[&str]) {
    let parsed_value: Vec<usize> = val
        .iter()
        .filter_map(|e| e.parse::<usize>())
        .collect();
}

However, now the parse method returns a Result type instead of an Option and I get the error:

error[E0308]: mismatched types
 --> src/lib.rs:4:25
  |
4 |         .filter_map(|e| e.parse::<usize>())
  |                         ^^^^^^^^^^^^^^^^^^ expected enum `std::option::Option`, found enum `std::result::Result`
  |
  = note: expected type `std::option::Option<_>`
             found type `std::result::Result<usize, std::num::ParseIntError>`

I could create an Option through a conditional, but is there a better / cleaner way?


回答1:


Use Result::ok. Types added for clarity:

let res: Result<u8, ()> = Ok(42);
let opt: Option<u8> = res.ok();
println!("{:?}", opt);

For symmetry's sake, there's also Option::ok_or and Option::ok_or_else to go from an Option to a Result.


In your case, you have an iterator.

If you'd like to ignore failures, use Iterator::flat_map. Since Result (and Option) implement IntoIterator, this works:

let parsed_value: Vec<usize> = val
    .iter()
    .flat_map(|e| e.parse())
    .collect();

If you'd like to stop on the first failure, you can collect into one big Result. This is less obvious, but you can check out the implementors of FromIterator for the full list of collect-able items.

let parsed_value: Result<Vec<usize>, _> = val
    .iter()
    .map(|e| e.parse())
    .collect();

Of course, you can then convert the one big Result into an Option, as the first example shows.



来源:https://stackoverflow.com/questions/28572101/what-is-a-clean-way-to-convert-a-result-into-an-option

标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!