rust: 线程通信代码-编译有问题(寻求帮助)

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引

经过与 rust 编译器长时间较量，无奈败下阵来。寻找 v 友助阵。

编译不通过代码如下

下面的代码想实现，单生产者，多消费者模型。通过 channel 传递 String 变量

use std::sync::mpsc::{channel, Receiver, Sender}; use std::sync::{Arc, Mutex}; use std::thread; fn main() { let mut txs = vec![]; let mut rxs: Vec<Arc<Mutex<Receiver<String>>>> = Vec::new(); let mut all_recv = vec![]; for i in 0..5 { let (tx, rx) = channel(); txs.push(tx); rxs.push(Arc::new(Mutex::new(rx))); } let sender = thread::spawn(move || { for i in 0..1000 { let tx = &txs[i % 5]; tx.send(i.to_string()).expect("send fail"); } }); for i in 0..5 { let rx = &'static rxs[i]; let receiver = thread::spawn(move || loop { let value = match rx.lock().unwrap().recv() { Ok(value) => value, Err(_) => break, }; }); all_recv.push(receiver); } sender.join(); for (_, v) in all_recv.iter().enumerate() { v.join(); } }

报错信息如下

error: expected `:`, found `rxs` --> src/main.rs:24:27 | 24 | let rx = &'static rxs[i]; | ^^^ expected `:`

Supplement 1 Nov 22, 2019

好了

重温了rust程序设计，使用多所有权可以解决编译错误，完整代码如下。如有不完美的地方，欢迎指正

use std::sync::mpsc::{channel, Receiver, Sender}; use std::sync::{Arc, Mutex}; use std::thread; fn main() { let mut txs = vec![]; let mut rxs: Vec<Arc<Mutex<Receiver<String>>>> = Vec::new(); let mut all_recv = vec![]; for i in 0..5 { let (tx, rx) = channel(); txs.push(tx); rxs.push(Arc::new(Mutex::new(rx))); } let sender = thread::spawn(move || { for i in 0..1000 { let tx = &txs[i % 5]; tx.send(i.to_string()).expect("send fail"); } }); for i in 0..5 { //let rx = &'static rxs[i]; let rx = &rxs[i]; let rx = Arc::clone(rx); let receiver = thread::spawn(move || loop { let value = match rx.lock().unwrap().recv() { Ok(value) => value, Err(_) => break, }; println!("{}", value); }); all_recv.push(receiver); } sender.join(); for v in all_recv { v.join(); } }

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Let rxs Receiver vec