我有以下设置，我试图创建一个Dummy future，它旨在修改future中从f: FnMut(&mut i32) -> Fut获得的值。我传递泛型函数fun来替换它，但我得到了不同的生存期。

#![allow(unused_variables)]

pub struct Dummy<Fut, F> {
    f: F,
    futures: futures::stream::FuturesUnordered<Fut>
}

impl<Fut, F> Dummy<Fut, F>
where
    F: FnMut(&mut i32) -> Fut,
    Fut: futures::Future<Output = ()>,
{
    fn new(f: F) -> Self {
        Self { 
            f,
            futures: futures::stream::FuturesUnordered::new()
        }
    }
}

impl<Fut, F> futures::Future for Dummy<Fut, F>
where 
    F: FnMut(&mut i32) -> Fut,
    Fut: futures::Future<Output = ()>
{
    type Output = ();
    
    fn poll(self: core::pin::Pin<&mut Self>, cx: &mut core::task::Context<'_>) -> core::task::Poll<()> {
        // logic emitted
        core::task::Poll::Ready(())
    }
    
}

fn fun<'a>(f: &'a mut i32) -> impl futures::Future<Output = ()> + 'a {
    async move {
        *f += 1;
    }
}

#[tokio::main]
async fn main() {
    Dummy::new(fun).await;
}

playground link
它给出了编译错误：

error[E0308]: mismatched types
  --> src/main.rs:45:20
   |
45 |     Dummy::new(fun).await;
   |                    ^^^^^^ one type is more general than the other
   |
   = note: expected trait `for<'r> FnOnce<(&'r mut i32,)>`
              found trait `for<'a> FnOnce<(&'a mut i32,)>`
   = note: the lifetime requirement is introduced here

error[E0308]: mismatched types
  --> src/main.rs:45:20
   |
45 |     Dummy::new(fun).await;
   |                    ^^^^^^ one type is more general than the other
   |
   = note: expected trait `for<'r> FnOnce<(&'r mut i32,)>`
              found trait `for<'a> FnOnce<(&'a mut i32,)>`

For more information about this error, try `rustc --explain E0308`.
error: could not compile `playground` due to 4 previous errors

有没有可能以一种消除这些不匹配类型错误的方式来注解impl？我可以用'static来注解所有内容，就像我在这里所做的那样：playground link。但我试图避免这种解决方案，因为我想在未来的poll方法中相互借用相关数据。