Question

我想在将该类型的变量移至线程的函数中使用泛型：

use std::thread;

fn ret_x<T: Send>(x: T) -> T {
    let handle = thread::spawn(move || return x);
    return handle.join().unwrap();
}

执行此操作时，会出现错误：

error[E0310]: the parameter type `T` may not live long enough
 --> src/lib.rs:4:18
  |
3 | fn ret_x<T: Send>(x: T) -> T {
  |          -- help: consider adding an explicit lifetime bound `T: 'static`...
4 |     let handle = thread::spawn(move || return x);
  |                  ^^^^^^^^^^^^^
  |
note: ...so that the type `[closure@src/lib.rs:4:32: 4:48 x:T]` will meet its required lifetime bounds
 --> src/lib.rs:4:18
  |
4 |     let handle = thread::spawn(move || return x);
  |                  ^^^^^^^^^^^^^

error[E0310]: the parameter type `T` may not live long enough
 --> src/lib.rs:4:18
  |
3 | fn ret_x<T: Send>(x: T) -> T {
  |          -- help: consider adding an explicit lifetime bound `T: 'static`...
4 |     let handle = thread::spawn(move || return x);
  |                  ^^^^^^^^^^^^^
  |
note: ...so that the type `T` will meet its required lifetime bounds
 --> src/lib.rs:4:18
  |
4 |     let handle = thread::spawn(move || return x);
  |                  ^^^^^^^^^^^^^

如果我给T赋予生命周期限制'static，则它将编译而不会出错，但是由于该函数需要在函数内部声明的变量上使用，因此不能为'static。 / p>

T将被拥有，而不是引用。有什么办法可以指定吗？

如何在通用线程上指定生存期

0 个答案: