延长锈蚀寿命 [英] Expanding Rust Lifetime

问题描述

我有一些需要解决的代码.这是一个小辅助函数，应将Vec<&str>返回给调用函数.不过，我似乎无法正确地选择一生.

I have a bit of code that I'm fighting with. It's a little helper function that should return a Vec<&str> to the calling function. I can't seem to get the lifetime right, though.

这是代码段:

fn take_symbol<'a>(ch: &'a str, current: &'a mut String) -> &'a mut TokenList<'a> {
    let out = TokenList::<'a>::new();

    out.push(current.as_str());
    out.push(ch);

    *current = String::new();

    &mut out
}

编译器告诉我:error: 'out' does not live long enough并且该引用在'a的生存期内必须是有效的，但是在我看来，它就像为'a定义的一样.

The compiler is telling me: error: 'out' does not live long enough and that the reference must be valid for the lifetime of 'a, but it looks to me like it is defined for 'a.

我也尝试将其更改为:

let out = &mut TokenList::<'a>::new();

，它不会更改任何错误消息.或者:

which doesn't change any of the error messages. Or:

let out = &'a mut TokenList::<'a>::new();

编译器根本不喜欢的.

如何定义out的生存期为'a?

有关更多详细信息，这是我对TokenList的定义:

For further details, here is my definition of TokenList:

pub type Token<'a> = &'a str;
pub type TokenList<'a> = Vec<Token<'a>>;

推荐答案

out的生存期不是'a，因为out在函数末尾被破坏了. Rust不允许您返回对其的引用(它将允许访问释放的内存！).

The lifetime of out is not 'a, since out is destroyed at the end of the function. Rust will not allow you to return a reference to it (it would allow accessing freed memory!).

尝试将功能更改为以下内容:

Try changing your function to the following:

fn take_symbol<'a>(ch: &'a str, current: &'a mut String) -> TokenList<'a> {
    let out = TokenList::<'a>::new();

    out.push(current.as_str());
    out.push(ch);

    *current = String::new();

    out
}

通过这种方式，您会将out的所有权传递给调用方，并且它将存在足够长的时间.

This way you will pass the ownership of out to the caller and it will live long enough.

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