如何将一个元素字符串转换为char？ [英] How can I convert a one element string into a char?

问题描述

我需要将一个元素& str 转换为 char 。我能够提出这个解决方案，该解决方案也适用于 String ：

I need to convert a one element &str into char. I was able to come up with this solution that also works for String:

fn main() {
    let comma: &str = ",";
    let my_char = comma.chars().nth(0).unwrap();
   
    assert_eq!(my_char, ',');
}

有更好或更短的方法吗？

Is there a better or shorter way to do it?

推荐答案

我想不出什么大的改进，但有几点注意事项：

No huge improvements I can think of, but a few notes:

• 您可以将 .nth（0）替换为 .next（），这基本上是相同的。


• 理想情况下，您不应使用 .unwrap（），因为如果字符串为空，则程序将出现恐慌。


• 如果您真的必须惊慌，最好使用 .expect（ msg），这样可以用户对为什么感到惊慌的更好想法。

• You could replace .nth(0) with .next(), which does basically the same thing.
• You should ideally not use .unwrap(), since if the string is empty, your program will panic.
• If you really must panic, ideally use .expect("msg"), which will give users a better idea of why you panicked.

将这些因素结合在一起：

Taking those together:

fn main() {
    let comma: &str = ",";
    let my_char = comma.chars().next().expect("string is empty");

    assert_eq!(my_char, ',');
}

唯一要注意的是一个元素在某种程度上是危险的有话要说。例如，é 有一个 char ，但是é有两个（第一个是预先编写的U + 00E9，第二个是常规的 e ，然后是U + 0301结合◌́）。

The only other thing to note is that "one element" is a somewhat dangerous thing to talk about. For example, "é" has one char, but "é" has two (the first is a pre-composed U+00E9, whilst the second is a regular e followed by a U+0301 combining ◌́).

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