如何使用 Pin 而不是 Arc 通过引用异步块来传递 Vec<u8>？

Question

我想使用Arc多次对Vec<u8>进行操作：

use futures::{
executor::{block_on, ThreadPool},
task::SpawnExt,
}; // 0.3.4
use std::{pin::*, sync::Arc};
fn foo(b: Arc<Vec<u8>>) {
println!("{:?}", b);
}
#[test]
fn pin_test() {
let v = Arc::new(vec![1, 2, 3, 4, 5, 6, 7, 8, 9, 10]);
let mut pool = ThreadPool::new().unwrap();
for _ in 0..10 {
let v1 = v.clone();
let handle = pool
.spawn_with_handle(async {
foo(v1);
})
.unwrap();
block_on(handle);
}
}

我希望能够PinVec<u8>

use futures::{
executor::{block_on, ThreadPool},
task::SpawnExt,
}; // 0.3.4
use std::{pin::*, sync::Arc};
fn foo(b: &[u8]) {
println!("{:?}", b);
}
#[test]
fn pin_test() {
let v = Pin::new(vec![1, 2, 3, 4, 5, 6, 7, 8, 9, 10]);
let mut pool = ThreadPool::new().unwrap();
for _ in 0..10 {
let v1 = v.clone();
let handle = pool
.spawn_with_handle(async {
foo(&*v1);
})
.unwrap();
block_on(handle);
}
}

这给出了错误：

error[E0597]: `v1` does not live long enough
--> src/lib.rs:19:23
|
18 |                .spawn_with_handle(async {
|   ________________________________-_____-
|  |________________________________|
| ||
19 | ||                 foo(&*v1);
| ||                       ^^ borrowed value does not live long enough
20 | ||             })
| ||             -
| ||_____________|
| |______________value captured here by generator
|                argument requires that `v1` is borrowed for `'static`
...
23 |        }
|        - `v1` dropped here while still borrowed

我理解Pin应该将Vec数据固定到特定点，以便所有调用都可以引用相同的数据。使用Pin的正确方法是什么，以便我可以传递对foo()的引用？

我正在使用 Rust 1.39。

Answer 1

您缺少move.

改变

let handle = pool
.spawn_with_handle(async {
foo(&*v1);
})
.unwrap();

自

let handle = pool
.spawn_with_handle(async move {
//                   ^^^^
foo(&*v1);
})
.unwrap();