迭代器,懒惰和所有权

Question

我开始使用Rust了,我正在玩正则表达式箱子以便我可以创建一个词法分析器.

词法分析器使用包含大量命名捕获组的大型正则表达式.我正在尝试获取正则表达式的结果并创建一个Vec<&str, &str>捕获名称和捕获值,但是在映射和过滤结果时迭代返回的值的生命周期一直存在问题.

我认为这与懒惰以及迭代器在超出范围时没有被消耗的事实有关,但我不确定如何真正解决问题.

extern crate regex;

use regex::Regex;

fn main() {
    // Define a regular expression with a bunch of named capture groups
    let expr = "((?P<num>[0-9]+)|(?P<str>[a-zA-Z]+))";
    let text = "0ab123cd";
    let re = Regex::new(&expr).unwrap();

    let tokens: Vec<(&str, &str)> = re.captures_iter(text)
        .flat_map(|t| t.iter_named())
        .filter(|t| t.1.is_some())
        .map(|t| (t.0, t.1.unwrap()))
        .collect();

    for token in tokens {
        println!("{:?}", token);
    }
}

运行上面的代码会导致以下错误:

$ cargo run
Compiling hello_world v0.0.1 (file:///Users/dowling/projects/rust_hello_world)

src/main.rs:14:23: 14:24 error: `t` does not live long enough
src/main.rs:14         .flat_map(|t| t.iter_named())
                                     ^
src/main.rs:17:19: 22:2 note: reference must be valid for the block suffix following statement 3 at 17:18...
src/main.rs:17         .collect();
src/main.rs:18 
src/main.rs:19     for token in tokens {
src/main.rs:20         println!("{:?}", token);
src/main.rs:21     }
src/main.rs:22 }
src/main.rs:14:23: 14:37 note: ...but borrowed value is only valid for the block at 14:22
src/main.rs:14         .flat_map(|t| t.iter_named())
                                     ^~~~~~~~~~~~~~
error: aborting due to previous error
Could not compile `hello_world`.

Answer 1

您的情况的限制点是.iter_named()方法:

fn iter_named(&'t self) -> SubCapturesNamed<'t>

请注意&'t self:输出的生命周期将与Captures实例的生命周期相关联.这是因为名称存储在Capture对象中,因此任何名称都&str不能超过此对象.

只有一个解决方法:您必须保持Capture实例活着:

let captures = re.captures_iter(text).collect::<Vec<_>>();
let tokens: Vec<(&str, &str)> = captures.iter()
    .flat_map(|t| t.iter_named())
    .filter(|t| t.1.is_some())
    .map(|t| (t.0, t.1.unwrap()))
    .collect();