异步 fn 报告“‘impl Trait’的隐藏类型捕获未出现在边界中的生命周期”

Question

如果我删除'static最后一个参数的生命周期，程序就会编译。如果我把它加回去，就会失败。对我来说，似乎两者都应该有效。最小复制：

use std::io;

struct Foo {
    user: String,
    pass: String,
}

impl Foo {
    async fn not_works(
        &mut self,
        user: &str,
        pass: &str,
        app_name: &'static str,
    ) -> io::Result<()> {
        self.user = user.to_string() + app_name;
        self.pass = pass.to_string();

        self.do_stuff().await
    }

    async fn works(&mut self, user: &str, pass: &str, app_name: &str) -> io::Result<()> {
        self.user = user.to_string() + app_name;
        self.pass = pass.to_string();

        self.do_stuff().await
    }

    async fn do_stuff(&self) -> io::Result<()> {
        Ok(())
    }
}

#[tokio::main]
async fn main() {
    let mut foo = Foo {
        user: "".to_string(),
        pass: "".to_string(),
    };

    foo.not_works("test", "password", "foobar").await.unwrap();
}

error[E0700]: hidden type for `impl Trait` captures lifetime that does not appear in bounds
  --> src/main.rs:14:10
   |
14 |     ) -> io::Result<()> {
   |          ^^^^^^^^^^^^^^
   |
note: hidden type `impl Future` captures lifetime smaller than the function body
  --> src/main.rs:14:10
   |
14 |     ) -> io::Result<()> {
   |          ^^^^^^^^^^^^^^

更有可能的是，这个错误是在我的理解中，而不是编译器。我缺少什么？

Answer 1

这是实施的限制async fn。基本上，它试图将所有生命周期统一为相同，在这种情况下，它们不可能全部相同'static。

这有效：

fn workaround<'a>(
    &'a mut self,
    user: &'a str,
    pass: &'a str,
    app_name: &'static str,
) -> impl Future<Output = io::Result<()>> + 'a {
    async move {
        self.user = user.to_string() + app_name;
        self.pass = pass.to_string();

        self.do_stuff().await
    }
}