rust - 如何返回盒装迭代器，该盒装迭代器返回f64的盒装迭代器？

Question

我正在尝试使用CSV crate读取CSV文件并将其所有单元格懒惰地转换为f64:

use csv::ReaderBuilder; // 1.1.4

fn get_df_iterator(path: &str) {
    // Build the CSV reader and iterate over each record casting to f64
    let rdr = ReaderBuilder::new()
        .delimiter(b'\t')
        .from_path(path)
        .unwrap();

    rdr.into_records().map(|record_result| {
        record_result
            .unwrap()
            .into_iter()
            .map(|cell: &str| cell.parse::<f64>().unwrap())
    })
}

fn main() {
    let m1 = get_df_iterator("df1.csv");
}

playground
我读过要在稳定的Rust中返回带有闭包的类型，我们可以使用 Box es。我已经尝试过:

use csv::{Error, ReaderBuilder, StringRecord};

fn get_df_iterator_box(path: &str) -> Box<dyn Iterator<Item = Box<dyn Iterator<Item = f64>>>> {
    // Build the CSV reader and iterate over each record.
    let rdr = ReaderBuilder::new()
        .delimiter(b'\t')
        .from_path(path)
        .unwrap();

    // Returns Box<Iterator<Item = f64>>
    let parse_str = |cell: &str| cell.parse::<f64>().unwrap();

    let parse_record = |record_result: Result<StringRecord, Error>| {
        Box::new(record_result.unwrap().into_iter().map(parse_str))
    };

    // Box<Iterator<Item = Box<Iterator<Item = f64>>>>
    Box::new(rdr.into_records().map(parse_record))
}

fn main() {
    let m1 = get_df_iterator_box("df1.csv");
}

error[E0271]: type mismatch resolving `<[closure@src/main.rs:13:24: 15:6] as FnOnce<(std::result::Result<StringRecord, csv::Error>,)>>::Output == Box<(dyn Iterator<Item = f64> + 'static)>`
  --> src/main.rs:18:5
   |
11 |     let parse_str = |cell: &str| cell.parse::<f64>().unwrap();
   |                     ----------------------------------------- the expected closure
...
18 |     Box::new(rdr.into_records().map(parse_record))
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected struct `Map`, found trait object `dyn Iterator`
   |
   = note: expected struct `Box<Map<StringRecordIter<'_>, [closure@src/main.rs:11:21: 11:62]>>`
              found struct `Box<(dyn Iterator<Item = f64> + 'static)>`
   = note: required because of the requirements on the impl of `Iterator` for `Map<StringRecordsIntoIter<File>, [closure@src/main.rs:13:24: 15:6]>`
   = note: required for the cast to the object type `dyn Iterator<Item = Box<(dyn Iterator<Item = f64> + 'static)>>`

我不明白问题是什么。应该返回一个 Box，其中包含一个迭代器，该迭代器返回 Box的迭代器的 f64。 Here there's another playground与该更新!如果要复制相同的文件，则使用 here you can download it。

Answer 1

您可以指定闭包的返回类型，以告诉Rust您正在返回dyn trait:

let parse_record =
    |record_result: Result<StringRecord, Error>| -> Box<dyn Iterator<Item = f64>> {
        Box::new(record_result.unwrap().into_iter().map(parse_str))
    };

另一种方法是在闭包中显式转换:

let parse_record = |record_result: Result<StringRecord, Error>| {
    Box::new(record_result.unwrap().into_iter().map(parse_str)) as Box<dyn Iterator<Item = f64>>
};

您可以改为返回 impl trait 而不是使用动态调度，并完全避免这种情况:

fn get_iterator(path: &str) -> impl Iterator<Item = impl Iterator<Item = f64>> { ... }

有关返回特征的更多信息，请参见 What is the correct way to return an Iterator?。