Rust：dyn Fn()` 无法在线程之间安全共享

Question

我在Rust中研究多线程，发现一个关于在thread::spawn中使用Fn的错误。我的代码是这样的：

use std::thread;

fn print_hello() {
    println!("hello");
}

fn run_fn(f: &dyn Fn()) {
    let hand = thread::spawn(move || {
        f();
    });
}

fn main() {
    run_fn(&print_hello);
}

编译后，我收到以下错误：

   Compiling fntest v0.1.0 (D:\ray\source\rust\fntest)   
error[E0277]: `dyn Fn()` cannot be shared between threads safely
   --> src\main.rs:8:16
    |
8   |     let hand = thread::spawn(move || {
    |                ^^^^^^^^^^^^^ `dyn Fn()` cannot be shared between threads safely
    |
    = help: the trait `Sync` is not implemented for `dyn Fn()`
    = note: required because of the requirements on the impl of `Send` for `&dyn Fn()`
    = note: required because it appears within the type `[closure@src\main.rs:8:30: 10:6]`
note: required by a bound in `spawn`
   --> C:\Users\zhang\.rustup\toolchains\stable-x86_64-pc-windows-msvc\lib/rustlib/src/rust\library\std\src\thread\mod.rs:628:8
    |
628 |     F: Send + 'static,
    |        ^^^^ required by this bound in `spawn`

For more information about this error, try `rustc --explain 
E0277`.
error: could not compile `fntest` due to previous error 

我该如何解决这个问题？

Answer 1

根据错误注释，您无法使用，&dyn Fn因为它不是Send. 因此，您可以使用一些泛型，仅限于此处所需的内容Fn() + Send + 'static：

use std::thread;

fn print_hello() {
    println!("hello");
}

fn run_fn<F: Fn() + Send + 'static>(f: F) {
    let _hand = thread::spawn(f);
}

fn main() {
    run_fn(&print_hello);
}

操场