返回Vec< str>时String的生存期. [英] String's lifetime when returning Vec<&str>

问题描述

简单代码:

fn foo() -> Vec<&'static str> {

    let mut vec = Vec::new();
    let mut string = String::new();

    // doing something with string...

    vec.push(string.as_str());

    return vector; // error here: string doesn't live long enough
}

我有一个问题，我需要处理字符串并将其作为str返回到Vec中.问题在于绑定字符串的生存期不够长，因为它在foo之后超出范围.我很困惑，我真的不知道该如何解决.

I have problem that I need to process with string and return it in Vec as str. Problem is that binding string doesn't live long enough, since it goes out of scope after foo. I am confused and I don't really know how to solve that.

推荐答案

&'static str是字符串文字，例如let a : &'static str = "hello world".它存在于应用程序的整个生命周期中.

A &'static str is a string literal e.g. let a : &'static str = "hello world". It exists throughout the lifetime of the application.

如果要创建新的String，则该字符串不是静态的！

If you're creating a new String, then that string is not static!

简单地返回String的向量.

fn foo() -> Vec<String> {

    let mut vec = Vec::new();
    let mut string = String::new();

    // doing something with string...

    vec.push(string);

    return vec;
}

fn main() {
    foo();
}

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