recursion - 如何使用递归来反向转换Rust String？

Question

当使用递归方式反转String时，我发现很难将String的一部分移到下一个，因为切片是&str类型而不是String。
这不会运行:

fn reverse_string(s: &mut String) {
    if s.is_empty() {
        return;
    }
    // how to pass a correct parameter?
    reverse_string(&mut s[1..]);
    s.push(s.chars().nth(0).unwrap());
    s.remove(0);
}

error[E0308]: mismatched types
 --> src/lib.rs:6:20
  |
6 |     reverse_string(&mut s[1..]);
  |                    ^^^^^^^^^^^ expected struct `String`, found `str`
  |
  = note: expected mutable reference `&mut String`
             found mutable reference `&mut str`

Answer 1

Rust字符串为UTF-8，这意味着

一个代码点没有固定长度的

没有关于应该交换什么单位的定义。

如果只想交换 ASCII字符串的字符，则可以这样做:

use std::mem;

fn reverse_string_ascii(s: &mut str) {
    if !s.is_ascii() {
        return;
    }

    // Safety: We have checked that the string is ASCII,
    // so it's fine to treat it as a slice of bytes.
    unsafe {
        fn rev(b: &mut [u8]) {
            match b {
                [] => {}
                [_] => {}
                [h, rest @ .., t] => {
                    mem::swap(h, t);
                    rev(rest)
                }
            }
        }

        rev(s.as_bytes_mut());
    }
}

fn main() {
    let mut s = String::from("hello");
    reverse_string_ascii(&mut s);
    println!("{}", s);
}

尽管这里没有真正的理由使用递归，但迭代更好:

let mut todo = s.as_bytes_mut();
loop {
    match todo {
        [] => break,
        [_] => break,
        [h, rest @ .., t] => {
            mem::swap(h, t);
            todo = rest;
        }
    }
}

也可以看看:

Why is capitalizing the first letter of a string so convoluted in Rust?

Reversing a string in Rust