在Rust中反转一个字符串

Question

这有什么问题:

fn main() {
    let word: &str = "lowks";
    assert_eq!(word.chars().rev(), "skwol");
}

我收到这样的错误:

error[E0369]: binary operation `==` cannot be applied to type `std::iter::Rev<std::str::Chars<'_>>`
 --> src/main.rs:4:5
  |
4 |     assert_eq!(word.chars().rev(), "skwol");
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  |
  = note: an implementation of `std::cmp::PartialEq` might be missing for `std::iter::Rev<std::str::Chars<'_>>`
  = note: this error originates in a macro outside of the current crate

这样做的正确方法是什么？

Answer 1

因为,@DK.建议,在稳定版本中.graphemes()不可用&str,你可能只是做了@huon在评论中建议的内容:

fn main() {
    let foo = "palimpsest";
    println!("{}", foo.chars().rev().collect::<String>());
}

Answer 2

第一个也是最基本的问题是,这不是你如何反转Unicode字符串.您正在颠倒代码点的顺序,您要在其中颠倒字形顺序.可能还有其他问题,我不知道.文字很难.

第二个问题由编译器指出:您正在尝试将字符串文字与char迭代器进行比较. chars并且rev不产生新的字符串,它们产生惰性序列,就像通常的迭代器一样. 以下作品:

/*!
Add the following to your `Cargo.toml`:

```cargo
[dependencies]
unicode-segmentation = "0.1.2"
```
*/
extern crate unicode_segmentation;
use unicode_segmentation::UnicodeSegmentation;

fn main() {
    let word: &str = "low?ks";
    let drow: String = word
        // Split the string into an Iterator of &strs, where each element is an
        // extended grapheme cluster.
        .graphemes(true)
        // Reverse the order of the grapheme iterator.
        .rev()
        // Collect all the chars into a new owned String.
        .collect();

    assert_eq!(drow, "skw?ol");

    // Print it out to be sure.
    println!("drow = `{}`", drow);
}

请注意,graphemes以前在标准库中作为一种不稳定的方法,因此上面的内容会破坏足够旧版本的Rust.在这种情况下,您需要使用UnicodeSegmentation::graphemes(s, true).