#425 Remove UNIX socket from FS before binding.#441
#425 Remove UNIX socket from FS before binding.#4411st1 merged 1 commit intopython:masterfrom socketpair:unix_rm
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| if path[0] not in (0, '\x00'): | ||
| try: | ||
| if stat.S_ISSOCK(os.stat(path).st_mode): | ||
| os.remove(path) |
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What if the file exists and is not a socket?
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if stat.S_ISSOCK(os.stat(path).st_mode)
As much as we can. Let's kernel to decide what to do during bind()
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I don't understand: can stat.S_ISSOCK return True if it's not a socket?
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No, it can not. If this path refers to plain file, bind syscall will fail. We must not erase non-sockets in any case.
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Ah, OK, I now understand what was @ZhukovAlexander's question about.
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Do you still have a questions ?
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No, no further questions about this line :)
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I think we should also remove the socket after the server is closed. |
| os.remove(path) | ||
| except OSError: | ||
| # Directory may have permissions only to create socket. | ||
| pass |
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Sure, but we should introduce something like log = logging.getLogger(__name__). Please tell where to do that.
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Look at how base_events.py logs stuff.
tests/test_unix_events.py
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| with self.assertRaisesRegex(OSError, | ||
| 'Address.*is already in use'): | ||
| self.loop.run_until_complete(coro) | ||
| self.loop.run_until_complete(coro) |
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How does this work? We have a live socket, bound to path. Then we create a server. Does this mean that the first socket is in an invalid state? Or they are both bound to path now?
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Seems, potential client will connect to second socket. First one will be hidden.
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I did not check. Maybe it is implementation specific - i.e. Different on bsd and linux.
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Please try to figure out what exactly is going on in this unittest.
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I have checked. If socket-inode on FS is deleted, no one can connect. If it overrided with another UNIX-socket, all clients will be connected to that new one, as I expect eariler.
import socket
import os
if os.path.exists('/tmp/qwe'):
os.unlink('/tmp/qwe')
sock1 = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
sock1.bind('/tmp/qwe')
sock1.listen()
os.unlink('/tmp/qwe')
sock2 = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
sock2.bind('/tmp/qwe')
sock2.listen()
#os.unlink('/tmp/qwe')
sock3 = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
sock3.connect('/tmp/qwe')
sock3.send(b'asd')
(sk, addr) = sock2.accept()
print(sk.recv(6))There was a problem hiding this comment.
I can not find POSIX standard, describinf this...reading man 7 unix say that many things are implementation specific.
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I think, that unittests are fine. Unittests must check that all assertions are not failed on target OS.
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What assertions? Clearly, there is some system-depending behaviour here: some systems allow Unix sockets to bind to existing in-use Unix sockets paths. Why should we have a test for that?
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Well. I will remove that test.
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I transform that test: It now checks that binding to same path as CLOSED socket works. This will improve code-coverage. And also I have found a bug with logging :).
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@socketpair FWIW we're just a couple days from a complete feature freeze of asyncio. Will you have time to finish this, or should I submit a new PR? |
asyncio/unix_events.py
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| 'path was not specified, and no sock specified') | ||
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| if sock.family != socket.AF_UNIX: | ||
| if sock.family != socket.AF_UNIX or sock.type != socket.SOCK_STREAM: |
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BTW, I've just committed this check to asyncio, please rebase your PR. Regardless of this PR, this is something that needed to be fixed (I wanted to fix this before 3.5 came out, but completely forgot about it). /cc @gvanrossum
Actually (as I think), removing before binding and removing after closing should be done at Python's |
Maybe. But let's fix this in asyncio first. |
@1st1 Can you advice best way to add |
After giving this some thought - let's not add this. I now remember an issue with uvloop: originally I was removing the path, but then someone's program broke because it was cleaning up the path manually and assumed it's always there. Trying to remove the path before connecting to it makes total sense, removing it after is nice, but not strictly necessary. |
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Thank you, Mark! |
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