I am experimenting with ZeroMQ. And I found it really interesting that in ZeroMQ, it does not matter whether either connect or bind happens first. I tried looking into the source code of ZeroMQ but it was too big to find anything.
The code is as follows.
# client side
import zmq
ctx = zmq.Context()
socket = ctx.socket(zmq.PAIR)
socket.connect('tcp://*:2345') # line [1]
# make it wait here
# server side
import zmq
ctx = zmq.Context()
socket = ctx.socket(zmq.PAIR)
socket.bind('tcp://localhost:2345')
# make it wait here
If I start client side first, the server has not been started yet, but magically the code is not blocked at line [1]. At this point, I checked with ss and made sure that the client is not listening on any port. Nor does it have any open connection. Then I start the server. Now the server is listening on port 2345, and magically the client is connected to it. My question is how does the client know the server is now online?
The best place to ask your question is the ZMQ mailing list, as many of the developers (and founders!) of the library are active there and can answer your question directly, but I'll give it a try. I'll admit that I'm not a C developer so my understanding of the source is limited, but here's what I gather, mostly from src/tcp_connector.cpp (other transports are covered in their respective files and may behave differently).
Line 214 starts the open() method, and here looks to be the meat of what's going on.
To answer your question about why the code is not blocked at Line [1], see line 258. It's specifically calling a method to make the socket behave asynchronously (for specifics on how unblock_socket() works you'll have to talk to someone more versed in C, it's defined here).
On line 278, it attempts to make the connection to the remote peer. If it's successful immediately, you're good, the bound socket was there and we've connected. If it wasn't, on line 294 it sets the error code to EINPROGRESS and fails.
To see what happens then, we go back to the start_connecting() method on line 161. This is where the open() method is called from, and where the EINPROGRESS error is used. My best understanding of what's happening here is that if at first it does not succeed, it tries again, asynchronously, until it finds its peer.