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

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?

like image 490
vise Avatar asked Feb 17 '15 22:02

vise


1 Answers

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.

like image 166
Shepmaster Avatar answered Nov 20 '22 15:11

Shepmaster