Re: __del__ pattern?
On Tue, 16 Aug 2005 08:03:58 -0400, Peter Hansen [EMAIL PROTECTED] wrote: Tom Anderson wrote: On Mon, 15 Aug 2005, Peter Hansen wrote: Using '' instead of 'localhost' means bind to *all* interfaces, not just the loopback one. Doesn't '' mean 'bind to the *default* interface'? What does default mean, and is that definition in conflict with what I said? The docs say it means INADDR_ANY. They don't say what that means, so you'd have to read up on the C socket calls to learn more. Or some helpful soul will clarify for the class... :-) INADDR_ANY means every network interface you can find. This includes the local loopback and all physical and logical network interfaces. /Jorgen -- // Jorgen Grahn jgrahn@ Ph'nglui mglw'nafh Cthulhu \X/algonet.se R'lyeh wgah'nagl fhtagn! -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
BranoZ wrote: [EMAIL PROTECTED] wrote: For a reasonably portable solution, leave the lock file open. On most systems, you cannot delete an open file,.. On most UNIXes, you can delete an open file. Even flock-ed. This is BTW also an hack around flock. Yes, sorry; my bad. Use file that is writeable by A and B in a directory that is writeable only by root. Is that portable? What's the sequence the program should try? -- --Bryan -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
Bryan Olson wrote:
Use file that is writeable by A and B in a directory that is
writeable only by root.
Is that portable?
I have the feeling that you are asking if it works on Windows.
No idea! I have only user experience with Windows.
On UNIX it is as portable as 'flock', which means all modern
Unices (be careful about NFS).
What's the sequence the program should try?
1.
open a file, which name was previously agreed on
(like /var/tmp/prog-name-user-name)
If it fails, report error and exit. System error or
somebody has created unaccessible file by the same name.
2.
Try to aquire a flock on the descriptor from step 1.
If it fails, some running process already has the lock, exit
3.
lock will be released and lockfile closed automaticaly by OS
on process exit.
import sys, fcntl
try:
lockfile=open('/var/tmp/test1', 'w')
fcntl.flock(lockfile.fileno(),
fcntl.LOCK_EX | fcntl.LOCK_NB)
except IOError:
print sys.exc_info()[1]
sys.exit(-1)
You can flock any open file, no matter if it is read/write/append.
BranoZ
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Re: __del__ pattern?
On Mon, 15 Aug 2005, Peter Hansen wrote:
Tom Anderson wrote:
Only one socket can be bound to a given port at any time, so the second
instance of SpecialClass will get an exception from the bind call, and
will be stillborn. This is a bit of a crufty hack, though - you end up
with an open port on your machine for no good reason. If
If you bind with self.sock.bind(('localhost', 4242)) instead, at least
you don't have much of a security risk since the port won't be available
for connections from outside the same machine.
Excellent suggestion, thanks!
Using '' instead of 'localhost' means bind to *all* interfaces, not just
the loopback one.
Doesn't '' mean 'bind to the *default* interface'?
tom
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Re: __del__ pattern?
Tom Anderson wrote: On Mon, 15 Aug 2005, Peter Hansen wrote: Using '' instead of 'localhost' means bind to *all* interfaces, not just the loopback one. Doesn't '' mean 'bind to the *default* interface'? What does default mean, and is that definition in conflict with what I said? The docs say it means INADDR_ANY. They don't say what that means, so you'd have to read up on the C socket calls to learn more. Or some helpful soul will clarify for the class... :-) -Peter -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
What does default mean, and is that definition in conflict with what
I
said? The docs say it means INADDR_ANY.
someSocket.bind(('', somePort)) means accept connections from any
machine. (We use INADDR_ANY instead of '' in C.)
someSocket.bind(('localhost', somePort)) means accept only connections
from the local machine.
Obviously, the second form is much more secure. Even if you don't plan
on accepting any of the connections, it's always more secure to let the
operating system turn everyone else away.
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problem at his university. He always wore a red baseball,
and so was known as the guy in the red hat. Later, when he
started his own open-source company, he wondered what to call it...
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Re: __del__ pattern?
Tom Anderson wrote: On Mon, 15 Aug 2005, Chris Curvey wrote: Is there a better pattern to follow than using a __del__ method? I just need to be absolutely, positively sure of two things: An old hack i've seen before is to create a server socket - ie, make a socket and bind it to a port: import socket class SpecialClass: def __init__(self): self.sock = socket.socket() self.sock.bind((, 4242)) def __del__(self): self.sock.close() Something like that, anyway. Only one socket can be bound to a given port at any time, so the second instance of SpecialClass will get an exception from the bind call, and will be stillborn. This is a bit of a crufty hack, though - you end up with an open port on your machine for no good reason. Much worse, it's a bug. That pattern is for programs that need to respond at a well-known port. In this case it doesn't work; the fact that *someone* has a certain socket open does not mean that this particular program is running. -- --Bryan -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
Dan wrote:
someSocket.bind(('localhost', somePort)) means accept only connections
from the local machine.
Almost: accept only attempts to connect *to* localhost, from the local
machine. Attempting to connect -- even locally -- using one of the IP
addresses bound to an external interface will fail.
-Peter
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Re: __del__ pattern?
Chris Curvey wrote: I need to ensure that there is only one instance of my python class on my machine at a given time. (Not within an interpreter -- that would just be a singleton -- but on the machine.) These instances are created and destroyed, but there can be only one at a time. So when my class is instantiated, I create a little lock file, and I have a __del__ method that deletes the lock file. Unfortunately, there seem to be some circumstances where my lock file is not getting deleted. Then all the jobs that need that special class start queueing up requests, and I get phone calls in the middle of the night. For a reasonably portable solution, leave the lock file open. On most systems, you cannot delete an open file, and if the program terminates, normally or abnormally, the file will be closed. When the program starts, it looks for the lock file, and if it's there, tries to delete it; if the delete fails, another instance is probably running. It then tries to create the lock file, leaving it open; if the create fails, you probably lost a race with another instance. When exiting cleanly, the program closes the file and deletes it. If the program crashes without cleaning up, the file will still be there, but a new instance can delete it, assuming permissions are right. There are neater solutions that are Unix-only or Windows-only. See BranzoZ's post for a Unix method. -- --Bryan -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
[EMAIL PROTECTED] wrote: For a reasonably portable solution, leave the lock file open. On most systems, you cannot delete an open file,.. On most UNIXes, you can delete an open file. Even flock-ed. This is BTW also an hack around flock. 1. Process A opens file /var/tmp/test1, and flocks descriptor. 2. Process H unlinks /var/tmp/test1 3. Process B opens file /var/tmp/test1, and flocks _another_ descriptor 4. Processes A and B are running simultaneously Do you need protection agains H ? Use file that is writeable by A and B in a directory that is writeable only by root. BranoZ -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
[EMAIL PROTECTED] writes: Chris Curvey wrote: I need to ensure that there is only one instance of my python class on my machine at a given time. (Not within an interpreter -- that would just be a singleton -- but on the machine.) These instances are created and destroyed, but there can be only one at a time. So when my class is instantiated, I create a little lock file, and I have a __del__ method that deletes the lock file. Unfortunately, there seem to be some circumstances where my lock file is not getting deleted. Then all the jobs that need that special class start queueing up requests, and I get phone calls in the middle of the night. For a reasonably portable solution, leave the lock file open. On most systems, you cannot delete an open file, Uh, you can on unix -- what else did you have in mind for most systems? Cheers, mwh -- Well, yes. I don't think I'd put something like penchant for anal play and able to wield a buttplug in a CV unless it was relevant to the gig being applied for... -- Matt McLeod, asr -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
So when my class is instantiated, I create a little lock file, and I
have a __del__ method that deletes the lock file. Unfortunately, there
seem to be some circumstances where my lock file is not getting
deleted.
Maybe the interpreter died by the signal.. in that case the __del__
is not called.
You can try 'flock', instead of lock files.
import fcntl
class Test1(object):
def __init__(self):
self.lock=open('/var/tmp/test1', 'w')
fcntl.flock(self.lock.fileno(), fcntl.LOCK_EX)
print 'Lock aquired!'
def __del__(self):
fcntl.flock(self.lock.fileno(), fcntl.LOCK_UN)
self.lock.close()
In this case, if interpreter dies, the lock is released by OS.
If you try to create another instance in the same interpreter
or another, the call will block in __init__. You can change it to
raise an exception instead.
BranoZ
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Re: __del__ pattern?
Chris Curvey [EMAIL PROTECTED] writes: I need to ensure that there is only one instance of my python class on my machine at a given time. I recommend modifying your requirements such that you ensure that there is only one active instance of your class at any one time (or something like that), and then use try:finally: blocks to ensure your locks get removed. Is there a better pattern to follow than using a __del__ method? I just need to be absolutely, positively sure of two things: 1) There is only one instance of my special class on the machine at a time. 2) If my special class is destroyed for any reason, I need to be able to create another instance of the class. As another poster mentioned, you also need to work out what you're going to do if your process gets killed in a way that doesn't allow finally blocks to run (this doesn't have much to do with Python). Cheers, mwh -- The above comment may be extremely inflamatory. For your protection, it has been rot13'd twice. -- the signature of JWhitlock on slashdot -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
On Mon, 15 Aug 2005, Chris Curvey wrote: Is there a better pattern to follow than using a __del__ method? I just need to be absolutely, positively sure of two things: An old hack i've seen before is to create a server socket - ie, make a socket and bind it to a port: import socket class SpecialClass: def __init__(self): self.sock = socket.socket() self.sock.bind((, 4242)) def __del__(self): self.sock.close() Something like that, anyway. Only one socket can be bound to a given port at any time, so the second instance of SpecialClass will get an exception from the bind call, and will be stillborn. This is a bit of a crufty hack, though - you end up with an open port on your machine for no good reason. If you're running on unix, you could try using a unix-domain socket instead; i'm not sure what the binding semantics of those are, though. I think Brano's suggestion of using flock is a better solution. tom -- Gin makes a man mean; let's booze up and riot! -- http://mail.python.org/mailman/listinfo/python-list
Re: __del__ pattern?
Tom Anderson wrote:
Only one socket can be bound to a given port at any time, so the second
instance of SpecialClass will get an exception from the bind call, and
will be stillborn. This is a bit of a crufty hack, though - you end up
with an open port on your machine for no good reason. If
If you bind with self.sock.bind(('localhost', 4242)) instead, at least
you don't have much of a security risk since the port won't be available
for connections from outside the same machine. Using '' instead of
'localhost' means bind to *all* interfaces, not just the loopback one.
-Peter
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