当T无法复制时,如何将Option <T>列表转换为T列表？

Question

如何获取a Vec<Option<T>>,T无法复制的位置以及展开所有Some值？

我在这map一步中遇到了错误.我很高兴移动原始列表的所有权并"扔掉" Nones.

#[derive(Debug)]
struct Uncopyable {
    val: u64,
}

fn main() {
    let num_opts: Vec<Option<Uncopyable>> = vec![
        Some(Uncopyable { val: 1 }),
        Some(Uncopyable { val: 2 }),
        None,
        Some(Uncopyable { val: 4 }),
    ];

    let nums: Vec<Uncopyable> = num_opts
        .iter()
        .filter(|x| x.is_some())
        .map(|&x| x.unwrap())
        .collect();
    println!("nums: {:?}", nums);
}

操场

这给出了错误

error[E0507]: cannot move out of borrowed content
  --> src/main.rs:17:15
   |
17 |         .map(|&x| x.unwrap())
   |               ^-
   |               ||
   |               |hint: to prevent move, use `ref x` or `ref mut x`
   |               cannot move out of borrowed content

Answer 1

在Rust中,当您需要值时,通常需要移动元素或克隆它们.

由于移动更为通用,因此只需要进行两项更改:

let nums: Vec<Uncopyable> = num_opts
    .into_iter()
//  ^~~~~~~~~~~~-------------- Consume vector, and iterate by value
    .filter(|x| x.is_some())
    .map(|x| x.unwrap())
//       ^~~------------------ Take by value
    .collect();

正如llogiq指出的那样,filter_map已经专门过滤掉None了:

let nums: Vec<Uncopyable> = num_opts
    .into_iter()
//  ^~~~~~~~~~~~-------- Consume vector, and iterate by value
    .filter_map(|x| x)
//              ^~~----- Take by value
    .collect();

然后它工作(消费num_opts).