如何设置返回值的生命周期? [英] How do I set the return value's lifetime?

问题描述

我有以下方法检查String ID是否存在.如果没有，请生成然后返回:

I have the following method to check if an String ID exists. If it doesn't, generate and then return it:

fn generate_id(&self) -> ID<'m> {
    let id = nanoid::generate(15);
    while self[&id].is_some() {
        id = nanoid::generate(15);
    };
    id
}

ID是类型别名:type ID<'id> = &'id String;

返回值必须为&'m std::string::String，但id为std::string::String.

我尝试做:

let id: ID<'m> = nanoid::generate(15);

，但是它给出的错误与方法仅针对id的错误相同.

but then it gives the same error that the method is giving only for id.

推荐答案

生命周期是描述性的，而不是描述性的.您没有设置生命周期；它们是您编写的程序的结果.

Lifetimes are descriptive, not prescriptive. You don't set lifetimes; they are a consequence of the program you write.

您正在尝试返回对局部变量的引用.那是无效的，并且您无法编写任何生存期来使其生效.

You are trying to return a reference to a local variable. That is not valid, and there is no lifetime you can write to make it valid.

您遇到X/Y问题.真正的问题是为什么您感到需要返回引用.

You have an X/Y problem. The real question is why you feel the need to return a reference.

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