如何通过特征委派具有非静态参数的异步函数？

Question

类似于此代码：

use std::future::Future;
use std::pin::Pin;

trait A {
    fn handle<'a>(&'a self, data: &'a i32) -> Pin<Box<dyn 'a + Future<Output = ()>>>;
}

impl<'b, Fut> A for fn(&'b i32) -> Fut
where
    Fut: 'b + Future<Output = ()>,
{
    fn handle<'a>(&'a self, data: &'a i32) -> Pin<Box<dyn 'a + Future<Output = ()>>> {
        Box::pin(self(data))
    }
}

如何为所有A实现async fn(&i32)？

Answer 1

此代码应该有效：

use std::future::Future;
use std::pin::Pin;

trait A<'a> {
    fn handle(&'a self, data: &'a i32) -> Pin<Box<dyn 'a + Future<Output=()>>>;
}

impl <'a, F, Fut> A<'a> for F
where F: 'static + Fn(&'a i32) -> Fut,
      Fut: 'a + Future<Output=()>
{
    fn handle(&'a self, data: &'a i32) -> Pin<Box<dyn 'a + Future<Output=()>>> {
        Box::pin(self(data))
    }
}

然后我们可以使用for<'a> A<'a>在所有地方委派异步函数。

async fn processor(data: &i32) {
}

fn consume(a: impl for<'a> A<'a>) {
}

fn main() {
    consume(processor);
}