Desugared递归异步方法

Question

我正在尝试编写已废止的递归异步方法。这是我的尝试：

use std::future::*;
use std::pin::*;

struct TestRecursiveAsync;

impl TestRecursiveAsync {
    pub fn foo(&self, value: u32) -> Pin<Box<(dyn Future<Output = u32>)>> {
        let v: Pin<Box<(dyn Future<Output = u32> + 'static)>> = Box::pin(async move {
            if value > 0 {
                self.foo(value - 100).await 
            } else {
                value
            }
        });
        v
    }
}

编译器抱怨此消息（Playground）：

   Compiling playground v0.0.1 (/playground)
error[E0759]: cannot infer an appropriate lifetime
  --> src/lib.rs:8:85
   |
7  |       pub fn foo(&self, value: u32) -> Pin<Box<(dyn Future<Output = u32>)>> {
   |                  ----- this data with an anonymous lifetime `'_`...
8  |           let v: Pin<Box<(dyn Future<Output = u32> + 'static)>> = Box::pin(async move {
   |  _____________________________________________________________________________________^
9  | |             if value > 0 {
10 | |                 self.foo(value - 100).await 
11 | |             } else {
12 | |                 value
13 | |             }
14 | |         });
   | |_________^ ...is captured here, requiring it to live as long as `'static`
   |
help: to declare that the trait object captures data from argument `self`, you can add an explicit `'_` lifetime bound
   |
7  |     pub fn foo(&self, value: u32) -> Pin<Box<(dyn Future<Output = u32> + '_)>> {
   |                                                                        ^^^^

因此，编译器要求&self的生命周期为'static，但它具有匿名名称。有办法解决这个问题吗？