socket.timeout; 说明？

Question

I am building a port scanning program ((irrelevant to the question, just explaining the background)), and I know the IP of the host, but not what ports are open. Hence, the scan.

It is in the early stages of development, so the error handling is bad, but not bad enough to make why Python does this explainable.

It tries to connect to, say, 123.456.7.8, 1 . Obviously it's a ridiculous port to be open, so it throws an error. The error is No Route to Host or the such, right? Wrong! It is instead Operation Timed Out! .

Okay, let's increase the timeout in case my calculations were incorrect.

. .. ... ....All that did was rinse and repeat!

About 20 minutes later, the timeout is at 20 seconds, and it still is timing out. Really? Why does python raise a timed out error though, instead of No route to host! or similar?

I need to distinguish between time outs and connection failures, because there is a difference between late and nowhere. This prevents me from doing so, creating an infinite loop of hurry up and wait.

Whatever shall I do? Wherever shall I go?

Answer 1

Python socket module is a thin wrapper around your platform's socket API. The issue is unrelated to Python.

It is not necessary that you get No Route to Host error. Moreover it is common that a firewall just drops received packets (for a filtered port) that may manifest as a timeout error in your code. See Drop vs. Reject (ignore the conclusion but read the explanation of what is happening).

To workaround, make multiple concurrent connections and set a fixed timeout or use raw-sockets and send the packets yourself (you could use scapy , to investigate the behavior).