我想通过 channel 发送闭包:
use std::thread;
use std::sync::mpsc;
#[derive(Debug)]
struct Test {
s1: String,
s2: String,
}
fn main() {
let t = Test {
s1: "Hello".to_string(),
s2: "Hello".to_string(),
};
let (tx, rx) = mpsc::channel::<FnOnce(&mut Test)>();
thread::spawn(move || {
let mut test = t;
let f = rx.recv().unwrap();
f(&mut test);
println!("{:?}", test);
});
tx.send(move |t: &mut Test| {
let s = "test".to_string();
t.s1 = s;
});
}
( playground )
我得到了一堆错误:
error[E0277]: the trait bound `for<'r> std::ops::FnOnce(&'r mut Test): std::marker::Sized` is not satisfied
--> src/main.rs:15:20
|
15 | let (tx, rx) = mpsc::channel::<FnOnce(&mut Test)>();
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ `for<'r> std::ops::FnOnce(&'r mut Test)` does not have a constant size known at compile-time
|
= help: the trait `std::marker::Sized` is not implemented for `for<'r> std::ops::FnOnce(&'r mut Test)`
= note: required by `std::sync::mpsc::channel`
error[E0277]: the trait bound `for<'r> std::ops::FnOnce(&'r mut Test): std::marker::Sized` is not satisfied
--> src/main.rs:15:20
|
15 | let (tx, rx) = mpsc::channel::<FnOnce(&mut Test)>();
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ `for<'r> std::ops::FnOnce(&'r mut Test)` does not have a constant size known at compile-time
|
= help: the trait `std::marker::Sized` is not implemented for `for<'r> std::ops::FnOnce(&'r mut Test)`
= note: required by `std::sync::mpsc::Sender`
error[E0599]: no method named `recv` found for type `std::sync::mpsc::Receiver<for<'r> std::ops::FnOnce(&'r mut Test)>` in the current scope
--> src/main.rs:18:20
|
18 | let f = rx.recv().unwrap();
| ^^^^
|
= note: the method `recv` exists but the following trait bounds were not satisfied:
`for<'r> std::ops::FnOnce(&'r mut Test) : std::marker::Sized`
error[E0599]: no method named `send` found for type `std::sync::mpsc::Sender<for<'r> std::ops::FnOnce(&'r mut Test)>` in the current scope
--> src/main.rs:22:8
|
22 | tx.send(move |t: &mut Test| {
| ^^^^
|
= note: the method `send` exists but the following trait bounds were not satisfied:
`for<'r> std::ops::FnOnce(&'r mut Test) : std::marker::Sized`
FnOnce
似乎无法发送,但我不明白为什么。
最佳答案
是的。您的代码存在一些问题。
首先,FnOnce
是一个特征,所以你不能直接使用它。特征必须是对具体类型的约束,或者是某种间接的背后。由于您要将闭包发送到其他地方,因此您需要类似 Box<FnOnce(...)>
的东西.
其次,你不能使用Box<FnOnce(...)>
因为,由于对象安全规则,您实际上不能调用 FnOnce
通过间接寻址。
(顺便说一句,您也不想使用 FnOnce<...>
语法,这在技术上是不稳定的;请改用 FnOnce(...)
。)
要解决这个问题,您可以切换到 Fn
或 FnMut
或使用尚未稳定的FnBox
特征。我之所以走这条路,是因为它可能具有您想要的语义,并且很可能在不久的将来稳定下来。如果您对此感到不舒服,则需要适本地修改您的闭包。
以下是我和 Manishearth 的共同努力(他指出我错过了 + Send
约束):
// NOTE: Requires a nightly compiler, as of Rust 1.0.
#![feature(core)]
use std::boxed::FnBox;
use std::thread;
use std::sync::mpsc;
#[derive(Debug)]
struct Test {
s1: String,
s2: String,
}
type ClosureType = Box<FnBox(&mut Test) + Send>;
fn main() {
let t = Test { s1: "Hello".to_string(), s2: "Hello".to_string() };
let (tx, rx) = mpsc::channel::<ClosureType>();
thread::spawn(move || {
let mut test = t;
let f = rx.recv().unwrap();
f.call_box((&mut test,));
println!("{:?}", test);
});
tx.send(Box::new(move |t: &mut Test| {
let s = "test".to_string();
t.s1 = s;
})).unwrap();
// To give the output time to show up:
thread::sleep_ms(100);
}
关于rust - 是否可以通过 channel 发送闭包?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30557152/