我正在测试Rust-libzmq客户端
https://crates.io/crates/libzmq
https://docs.rs/libzmq/0.2.5/libzmq/struct.ClientBuilder.html
偶然发现了这种奇怪的行为。
此客户端不会发送以下消息:
use libzmq::{prelude::*, *, ServerBuilder, ClientBuilder, TcpAddr};
fn main() -> Result<(), String> {
let saddr = format!("{}:{}", "192.168.1.206","5540");
println!("Strating client for {}", saddr);
let addr2: TcpAddr = saddr.try_into().unwrap();
let client = ClientBuilder::new()
.connect(&addr2)
.build().unwrap();
client.send("test").unwrap();
println!("Finished");
Ok(())
}
这将发送:
use libzmq::{prelude::*, *, ServerBuilder, ClientBuilder, TcpAddr};
fn main() -> Result<(), String> {
let saddr = format!("{}:{}", "192.168.1.206","5540");
println!("Strating client for {}", saddr);
let addr2: TcpAddr = saddr.try_into().unwrap();
let client = ClientBuilder::new()
.connect(&addr2)
.build().unwrap();
client.send("test").unwrap();
let mut msg = Msg::new();
client.recv(&mut msg).unwrap();
println!("Finished");
Ok(())
}
对于接收,我使用python服务器(但我也在Rust上进行了测试(:
import zmq
if __name__ == '__main__':
context = zmq.Context()
socket = context.socket(zmq.SERVER)
socket.bind("tcp://0.0.0.0:5540")
print("waiting for hand shake")
m = socket.recv(copy=False)
print("got it...")
socket.send(b'READY', routing_id=m.routing_id)
print("sent reply")
socket.close()
context.term()
来自代码1:的服务器端输出
waiting for hand shake
[blocked]
来自代码2:的服务器端输出
waiting for hand shake
got it...
sent reply
将此版本用于Rust:
[dependencies]
libzmq = "0.2.5"
libzmq
机箱只是libzmq
C库的包装器。在C库的文档中,您会发现以下注释:
注意:成功调用zmq_msg_send((并不表示消息已传输到网络,只是已在其上排队"套接字"和0MQ已对该消息负责。
在第一个示例中,由于程序在调用send
后立即退出,因此消息仍在libzmq
队列中,尚未传输。在第二个示例中,当您的程序在recv
调用中等待时,消息被传输。