我将拥有多个需要访问一个主套接字的功能。
更好的是:
有人能提供一个最佳方法的例子吗?
我来自Python / Nim背景,这样的事情很容易完成。
修改 如何将套接字作为arg传递给线程中调用的函数。 实施例
fn main() {
let mut s = BufferedStream::new((TcpStream::connect(server).unwrap()));
let thread = Thread::spawn(move || {
func1(s, arg1, arg2);
});
while true {
func2(s, arg1);
}
}
答案 0 :(得分:1)
我们可以使用TcpStream::try_clone
:
use std::io::Read;
use std::net::{TcpStream, Shutdown};
use std::thread;
fn main() {
let mut stream = TcpStream::connect("127.0.0.1:34254").unwrap();
let stream2 = stream.try_clone().unwrap();
let _t = thread::spawn(move || {
// close this stream after one second
thread::sleep_ms(1000);
stream2.shutdown(Shutdown::Read).unwrap();
});
// wait for some data, will get canceled after one second
let mut buf = [0];
stream.read(&mut buf).unwrap();
}
如果你可以提供帮助,通常(比如99.9%的话)让任何全局可变状态成为一个坏主意。就像你说的那样:将套接字传递给需要它的函数。
use std::io::{self, Write};
use std::net::TcpStream;
fn send_name(stream: &mut TcpStream) -> io::Result<()> {
stream.write(&[42])?;
Ok(())
}
fn send_number(stream: &mut TcpStream) -> io::Result<()> {
stream.write(&[1, 2, 3])?;
Ok(())
}
fn main() {
let mut stream = TcpStream::connect("127.0.0.1:31337").unwrap();
let r = send_name(&mut stream).and_then(|_| send_number(&mut stream));
match r {
Ok(..) => println!("Yay, sent!"),
Err(e) => println!("Boom! {}", e),
}
}
您还可以将TcpStream
传递给管理它的结构,从而为您提供放置类似方法的位置。
use std::io::{self, Write};
use std::net::TcpStream;
struct GameService {
stream: TcpStream,
}
impl GameService {
fn send_name(&mut self) -> io::Result<()> {
self.stream.write(&[42])?;
Ok(())
}
fn send_number(&mut self) -> io::Result<()> {
self.stream.write(&[1, 2, 3])?;
Ok(())
}
}
fn main() {
let stream = TcpStream::connect("127.0.0.1:31337").unwrap();
let mut service = GameService { stream: stream };
let r = service.send_name().and_then(|_| service.send_number());
match r {
Ok(..) => println!("Yay, sent!"),
Err(e) => println!("Boom! {}", e),
}
}
这些都不是特定于Rust的,这些通常是适用的编程实践。