Rust 异步函数返回的future 生存期

试图编写递归流，其中每个流从他的父流轮询，当轮询准备好值时，它使自己的异步抓取器处理父响应，但我不知道如何存储由抓取器给定的当前future ，编译器抱怨生存期，直到我try 使用自由的异步函数(未绑定到 struct ).这里有不完整的示例，但它说明了编译器错误(在Stream实现中注释了两行):

struct AsyncFetcher {}

impl AsyncFetcher {
    async fn fetch(&self, request: String) -> String {
        format!("Response({request})")
    }
}

enum State {
    PendingParent,
    ToProcess(Option<String>),
    Processing(Pin<Box<dyn Future<Output = String>>>)
}

struct RecStream {
    parent: Option<Pin<Box<dyn Stream<Item = String>>>>,
    state: State,
    fetcher: AsyncFetcher,
}

impl RecStream {
    fn new(parent: Pin<Box<dyn Stream<Item = String>>>, fetcher: AsyncFetcher) -> Self {
        Self {
            parent: Some(parent),
            state: State::PendingParent,
            fetcher: fetcher,
        }
    }

    fn with_result(result: String, fetcher: AsyncFetcher) -> Self {
        Self {
            parent: None,
            state: State::ToProcess(Some(result)),
            fetcher: fetcher,
        }
    }
}

impl Stream for RecStream {
    type Item = String;

    fn poll_next(
        self: Pin<&mut Self>,
        cx: &mut std::task::Context<'_>,
    ) -> Poll<Option<Self::Item>> {
        let ref_mut = self.get_mut();
        let future = ref_mut.fetcher.fetch("Some str".to_string()).boxed();
        
        // let future = free_fut().boxed(); // - THIS WORKS
        ref_mut.state = State::Processing(future); // Getting error 'lifetime may not live long enough'
        
        return Poll::Pending;
    }
}

async fn free_fut() -> String {
    "Free string".to_string()
}