Rust 无法将`&Vec>`转换为`&[&str]`

发布于06月18日

我的代码中有一个函数，它接受&[&'a str]:

fn user_quicksort<'a>(list: &[&'a str]) -> Vec<&'a str>

在调用方中，我有一个Box<[Box<str>]>变量，我希望将其传递给它；但是，Rust给了我以下错误:

error[E0308]: mismatched types
  --> src\main.rs:36:33
   |
36 |     let sorted = user_quicksort(&input);
   |                  -------------- ^^^^^^ expected `&[&str]`, found `&Box<[Box<str>]>`
   |                  |
   |                  arguments to this function are incorrect
   |
   = note: expected reference `&[&str]`
              found reference `&Box<[Box<str>]>`

我使用Box<str>而不是String的原因是因为Box<str>是不可变的、更快的、更小的.

我该怎么办？

这就是我得到input分的原因:

let input = fs::read_to_string("input.txt")
    .expect("Unable to read file")
    .split("\n")
    .map(|s| s.to_string().into_boxed_str())
    .collect::<Box<[Box<str>]>>();

推荐答案

我不会在排序函数中强加序列的重复，而是让它对序列inplace进行排序.

这样，在调用点，我们可以 Select 是要更改原始序列，还是要保持它不排序并构建要排序的浅层副本.

克隆这个序列的成本会很高，因为每Box<str>个序列都会被克隆. 那么我们的排序函数应该能够对原始序列的Box<str>的序列和对浅拷贝的&str的序列起作用. 在参数中使用impl AsRef<str>会有所帮助，因为Box<str>和&str(显然)都可以被认为是对str的引用.

fn user_quicksort(list: &mut [impl AsRef<str>]) {
    // rely on standard slice::sort_by() here
    list.sort_by(|a, b| a.as_ref().cmp(b.as_ref()));
}

fn main() {
    let mut input = std::fs::read_to_string("input.txt")
        .expect("Unable to read file")
        .split("\n")
        .map(|s| s.to_string().into_boxed_str())
        .collect::<Box<[Box<str>]>>();
    // if we want to keep the initial sequence as is,
    // then we create a shallow clone of the sequence and sort it
    let mut shallow_clone =
        input.iter().map(|s| s.as_ref()).collect::<Box<[&str]>>();
    for (s1, s2) in input.iter().zip(shallow_clone.iter()) {
        // just a check to be sure we didn't copy de str themselves
        if s1.as_ptr() != s2.as_ptr() {
            println!("!!! deep copy {:?} {:?}", s1, s2);
        }
    }
    user_quicksort(shallow_clone.as_mut());
    println!("sorted clone: {:?}", shallow_clone);
    println!("initial input: {:?}", input);
    // but if we want to sort the initial sequence,
    // no new allocation is needed
    user_quicksort(input.as_mut());
    println!("sorted input: {:?}", input);
}
/*
sorted clone: ["", "line 1", "line 2", "line 3", "line 4", "line 5"]
initial input: ["line 4", "line 1", "line 5", "line 2", "line 3", ""]
sorted input: ["", "line 1", "line 2", "line 3", "line 4", "line 5"]
*/