从字符串中删除单个尾随换行符而不克隆 [英] Remove single trailing newline from String without cloning

问题描述

我编写了一个函数来提示输入并返回结果.在此版本中，返回的字符串包括来自用户的尾随换行符.我想返回删除该换行符(以及仅该换行符)的输入:

I have written a function to prompt for input and return the result. In this version the returned string includes a trailing newline from the user. I would like to return the input with that newline (and just that newline) removed:

fn read_with_prompt(prompt: &str) -> io::Result<String> {
    let stdout = io::stdout();
    let reader = io::stdin();
    let mut input = String::new();
    print!("{}", prompt);
    stdout.lock().flush().unwrap();
    reader.read_line(&mut input)?;

    // TODO: Remove trailing newline if present
    Ok(input)
}

只删除单个尾随换行符的原因是该函数还将用于提示输入密码(适当使用 termios 停止回显)，如果某人的密码有尾随空格，则应保留.

The reason for only removing the single trailing newline is that this function will also be used to prompt for a password (with appropriate use of termios to stop echoing) and if someone's password has trailing whitespace this should be preserved.

在对如何实际删除字符串末尾的单个换行符大惊小怪之后，我最终使用了 trim_right_matches.然而，它返回一个 &str.我尝试使用 Cow 来解决这个问题，但错误仍然说 input 变量的寿命不够长.

After much fussing about how to actually remove a single newline at the end of a string I ended up using trim_right_matches. However that returns a &str. I tried using Cow to deal with this but the error still says that the input variable doesn't live long enough.

fn read_with_prompt<'a>(prompt: &str) -> io::Result<Cow<'a, str>> {
    let stdout = io::stdout();
    let reader = io::stdin();
    let mut input = String::new();
    print!("{}", prompt);
    stdout.lock().flush().unwrap();
    reader.read_line(&mut input)?;

    let mut trimmed = false;
    Ok(Cow::Borrowed(input.trim_right_matches(|c| {
        if !trimmed && c == '\n' {
            trimmed = true;
            true
        }
        else {
            false
        }
    })))
}

错误:

error[E0515]: cannot return value referencing local variable `input`
  --> src/lib.rs:13:5
   |
13 |       Ok(Cow::Borrowed(input.trim_right_matches(|c| {
   |       ^                ----- `input` is borrowed here
   |  _____|
   | |
14 | |         if !trimmed && c == '\n' {
15 | |             trimmed = true;
16 | |             true
...  |
20 | |         }
21 | |     })))
   | |________^ returns a value referencing data owned by the current function

基于之前的问题，这似乎是不可能.分配一个删除了尾随换行符的新字符串是唯一的选择吗?似乎应该有一种方法可以在不复制的情况下修剪字符串(在 C 中，您只需将 '\n' 替换为 '\0').

Based on previous questions along these lines it seems this is not possible. Is the only option to allocate a new string that has the trailing newline removed? It seems there should be a way to trim the string without copying it (in C you'd just replace the '\n' with '\0').

推荐答案

您可以使用 String::pop 或 String::truncate:

You can use String::pop or String::truncate:

fn main() {
    let mut s = "hello\n".to_string();
    s.pop();
    assert_eq!("hello", &s);

    let mut s = "hello\n".to_string();
    let len = s.len();
    s.truncate(len - 1);
    assert_eq!("hello", &s);
}

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