New submission from Piotr Jurkiewicz:
After setting socket.settimeout(5.0), socket.send() returns immediately,
instead of returning after specified timeout.
Steps to reproduce:
Open two python interpreters.
In the first one (the receiver) execute:
>>> import socket
>>> r = socket.socket(socket.AF_UNIX, socket.SOCK_DGRAM)
>>> r.bind("test.sock")
In the second one (the sender) execute:
>>> import socket
>>> s = socket.socket(socket.AF_UNIX, socket.SOCK_DGRAM)
Then run the following command 11 times:
>>> s.sendto("msg", "test.sock")
On the 12 run command will block. This happens because datagram sockets queue
on Linux is 11 messages long. Interrupt the command.
So far so good.
Then set sender socket timeout:
>>> s.settimeout(5.0)
Expected behavior:
s.sendto() should block for a 5 seconds and THEN raise error 11
(EAGAIN/EWOULDBLOCK).
Actual behavior:
s.sendto() raises the error IMMEDIATELY.
>>> s.sendto("msg", "test.sock")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
socket.error: [Errno 11] Resource temporarily unavailable
So, in fact, s.settimeout(5.0) does not have any effect.
I think that problem is that settimeout() sets the socket to the non-blocking
mode (docs say: "Timeout mode internally sets the socket in non-blocking
mode.").
As described [here](http://stackoverflow.com/q/13556972/2178047) setting
timeout on non-blocking sockets is impossible.
In fact, when I set timeout manually with setsockopt(), everything works as
expected:
>>> s.setblocking(1) #go back to blocking mode
>>> tv = struct.pack("ll", 5, 0)
>>> s.setsockopt(socket.SOL_SOCKET, socket.SO_SNDTIMEO, tv)
Now s.sendto() raises the error after 5 seconds, as expected.
----------
components: IO, Library (Lib)
messages: 235013
nosy: piotrjurkiewicz
priority: normal
severity: normal
status: open
title: socket.settimeout(5.0) does not have any effect
type: behavior
versions: Python 2.7, Python 3.2, Python 3.3, Python 3.4, Python 3.5, Python 3.6
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Python tracker <[email protected]>
<http://bugs.python.org/issue23351>
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