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regex - 迭代正则表达式捕获的生命周期问题

转载 作者:行者123 更新时间:2023-11-29 08:12:11 25 4
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我正在尝试使用正则表达式从字符串中获取所有非空白字符,但我总是遇到同样的错误。

extern crate regex; // 1.0.2

use regex::Regex;
use std::vec::Vec;

pub fn string_split<'a>(s: &'a String) -> Vec<&'a str> {
let mut returnVec = Vec::new();
let re = Regex::new(r"\S+").unwrap();

for cap in re.captures_iter(s) {
returnVec.push(&cap[0]);
}

returnVec
}

pub fn word_n(s: &String, n: i32) -> &str {
let bytes = s.as_bytes();

let mut num = 0;
let mut word_start = 0;
for (i, &item) in bytes.iter().enumerate() {
if item == b' ' || item == b'\n' {
num += 1;
if num == n {
return &s[word_start..i].trim();
}
word_start = i;
continue;
}
}

&s[..]
}

错误:

error[E0597]: `cap` does not live long enough
--> src/main.rs:11:25
|
11 | returnVec.push(&cap[0]);
| ^^^ borrowed value does not live long enough
12 | }
| - borrowed value only lives until here
|
note: borrowed value must be valid for the lifetime 'a as defined on the function body at 6:1...
--> src/main.rs:6:1
|
6 | pub fn string_split<'a>(s: &'a String) -> Vec<&'a str> {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

更多信息:

$ rustc --explain E0597

This error occurs because a borrow was made inside a variable which has a
greater lifetime than the borrowed one.

Example of erroneous code:

```
struct Foo<'a> {
x: Option<&'a u32>,
}

let mut x = Foo { x: None };
let y = 0;
x.x = Some(&y); // error: `y` does not live long enough
```
In here, `x` is created before `y` and therefore has a greater lifetime. Always
keep in mind that values in a scope are dropped in the opposite order they are
created. So to fix the previous example, just make the `y` lifetime greater than
the `x`'s one:

```
struct Foo<'a> {
x: Option<&'a u32>,
}

let y = 0;
let mut x = Foo { x: None };
x.x = Some(&y);
```

此时我已经尝试了几种延长 cap 变量生命周期的方法,但是在阅读了 Rust 书的借用和生命周期部分后,我无法得到任何工作.

最佳答案

documentation of impl<'t> Index<usize> for Captures<'t> (这是您代码中的 cap[0])说:

The text can't outlive the Captures object if this method is used, because of how Index is defined (normally a[i] is part of a and can't outlive it); to do that, use get() instead.

所以 get它有效(请注意,我已将 &'a String 参数替换为 &'a str ):

use regex::Regex;

pub fn string_split<'a>(s: &'a str) -> Vec<&'a str> {
let mut return_vec = Vec::new();
let re = Regex::new(r"\S+").unwrap();

for cap in re.captures_iter(s) {
return_vec.push(cap.get(0).unwrap().as_str());
};

return_vec
}

fn main() {
println!("{:?}", string_split("Hello, world!"));
}

关于regex - 迭代正则表达式捕获的生命周期问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51834111/

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