All I want to do is timeout a function if it does not return before that
It all started because urllib2 supports timeout for urlopen, but not for reading part
and my program hangs. Changing defaulttimeout for sockets does not work. Using signal.sigalrm
does not work. I can't switch to requests because then I will have to rewrite and test a lot more.
I DON'T want to make a thread run the function and then timeout the thread, I want to timeout the function. Any ideas how?
I like to use David's class here in my projects. I find it's very effective and I like that it provides a simple way to implement in existing code via a decorator. For example:
# Timeout after 30 seconds
#timeout(30)
def your_function():
...
CAUTION: This is not thread-safe! If you're using multithreading, the signal will get caught by a random thread. For single-threaded programs however, this is the easiest solution.
Yes, it can be done in windows without signal and it will also work in other os as well. This is using thread but not to run the function but to raise a signal for timeout. The logic is to create a new thread and wait for a given time and raise an exception using _thread(in python3 and thread in python2). This exception will be thrown in the main thread and the with block will get exit if any exception occurs.
import threading
import _thread # import thread in python2
class timeout():
def __init__(self, time):
self.time= time
self.exit=False
def __enter__(self):
threading.Thread(target=self.callme).start()
def callme(self):
time.sleep(self.time)
if self.exit==False:
_thread.interrupt_main() # use thread instead of _thread in python2
def __exit__(self, a, b, c):
self.exit=True
Usuage Example :-
with timeout(2):
func()
The program in the with block should exit within 2 seconds otherise it will be exited after 2 seconds.
Related
I looked online and found some SO discussing and ActiveState recipes for running some code with a timeout. It looks there are some common approaches:
Use thread that run the code, and join it with timeout. If timeout elapsed - kill the thread. This is not directly supported in Python (used private _Thread__stop function) so it is bad practice
Use signal.SIGALRM - but this approach not working on Windows!
Use subprocess with timeout - but this is too heavy - what if I want to start interruptible task often, I don't want fire process for each!
So, what is the right way? I'm not asking about workarounds (eg use Twisted and async IO), but actual way to solve actual problem - I have some function and I want to run it only with some timeout. If timeout elapsed, I want control back. And I want it to work on Linux and Windows.
A completely general solution to this really, honestly does not exist. You have to use the right solution for a given domain.
If you want timeouts for code you fully control, you have to write it to cooperate. Such code has to be able to break up into little chunks in some way, as in an event-driven system. You can also do this by threading if you can ensure nothing will hold a lock too long, but handling locks right is actually pretty hard.
If you want timeouts because you're afraid code is out of control (for example, if you're afraid the user will ask your calculator to compute 9**(9**9)), you need to run it in another process. This is the only easy way to sufficiently isolate it. Running it in your event system or even a different thread will not be enough. It is also possible to break things up into little chunks similar to the other solution, but requires very careful handling and usually isn't worth it; in any event, that doesn't allow you to do the same exact thing as just running the Python code.
What you might be looking for is the multiprocessing module. If subprocess is too heavy, then this may not suit your needs either.
import time
import multiprocessing
def do_this_other_thing_that_may_take_too_long(duration):
time.sleep(duration)
return 'done after sleeping {0} seconds.'.format(duration)
pool = multiprocessing.Pool(1)
print 'starting....'
res = pool.apply_async(do_this_other_thing_that_may_take_too_long, [8])
for timeout in range(1, 10):
try:
print '{0}: {1}'.format(duration, res.get(timeout))
except multiprocessing.TimeoutError:
print '{0}: timed out'.format(duration)
print 'end'
If it's network related you could try:
import socket
socket.setdefaulttimeout(number)
I found this with eventlet library:
http://eventlet.net/doc/modules/timeout.html
from eventlet.timeout import Timeout
timeout = Timeout(seconds, exception)
try:
... # execution here is limited by timeout
finally:
timeout.cancel()
For "normal" Python code, that doesn't linger prolongued times in C extensions or I/O waits, you can achieve your goal by setting a trace function with sys.settrace() that aborts the running code when the timeout is reached.
Whether that is sufficient or not depends on how co-operating or malicious the code you run is. If it's well-behaved, a tracing function is sufficient.
An other way is to use faulthandler:
import time
import faulthandler
faulthandler.enable()
try:
faulthandler.dump_tracebacks_later(3)
time.sleep(10)
finally:
faulthandler.cancel_dump_tracebacks_later()
N.B: The faulthandler module is part of stdlib in python3.3.
If you're running code that you expect to die after a set time, then you should write it properly so that there aren't any negative effects on shutdown, no matter if its a thread or a subprocess. A command pattern with undo would be useful here.
So, it really depends on what the thread is doing when you kill it. If its just crunching numbers who cares if you kill it. If its interacting with the filesystem and you kill it , then maybe you should really rethink your strategy.
What is supported in Python when it comes to threads? Daemon threads and joins. Why does python let the main thread exit if you've joined a daemon while its still active? Because its understood that someone using daemon threads will (hopefully) write the code in a way that it wont matter when that thread dies. Giving a timeout to a join and then letting main die, and thus taking any daemon threads with it, is perfectly acceptable in this context.
I've solved that in that way:
For me is worked great (in windows and not heavy at all) I'am hope it was useful for someone)
import threading
import time
class LongFunctionInside(object):
lock_state = threading.Lock()
working = False
def long_function(self, timeout):
self.working = True
timeout_work = threading.Thread(name="thread_name", target=self.work_time, args=(timeout,))
timeout_work.setDaemon(True)
timeout_work.start()
while True: # endless/long work
time.sleep(0.1) # in this rate the CPU is almost not used
if not self.working: # if state is working == true still working
break
self.set_state(True)
def work_time(self, sleep_time): # thread function that just sleeping specified time,
# in wake up it asking if function still working if it does set the secured variable work to false
time.sleep(sleep_time)
if self.working:
self.set_state(False)
def set_state(self, state): # secured state change
while True:
self.lock_state.acquire()
try:
self.working = state
break
finally:
self.lock_state.release()
lw = LongFunctionInside()
lw.long_function(10)
The main idea is to create a thread that will just sleep in parallel to "long work" and in wake up (after timeout) change the secured variable state, the long function checking the secured variable during its work.
I'm pretty new in Python programming, so if that solution has a fundamental errors, like resources, timing, deadlocks problems , please response)).
solving with the 'with' construct and merging solution from -
Timeout function if it takes too long to finish
this thread which work better.
import threading, time
class Exception_TIMEOUT(Exception):
pass
class linwintimeout:
def __init__(self, f, seconds=1.0, error_message='Timeout'):
self.seconds = seconds
self.thread = threading.Thread(target=f)
self.thread.daemon = True
self.error_message = error_message
def handle_timeout(self):
raise Exception_TIMEOUT(self.error_message)
def __enter__(self):
try:
self.thread.start()
self.thread.join(self.seconds)
except Exception, te:
raise te
def __exit__(self, type, value, traceback):
if self.thread.is_alive():
return self.handle_timeout()
def function():
while True:
print "keep printing ...", time.sleep(1)
try:
with linwintimeout(function, seconds=5.0, error_message='exceeded timeout of %s seconds' % 5.0):
pass
except Exception_TIMEOUT, e:
print " attention !! execeeded timeout, giving up ... %s " % e
I am writing an queue processing application which uses threads for waiting on and responding to queue messages to be delivered to the app. For the main part of the application, it just needs to stay active. For a code example like:
while True:
pass
or
while True:
time.sleep(1)
Which one will have the least impact on a system? What is the preferred way to do nothing, but keep a python app running?
I would imagine time.sleep() will have less overhead on the system. Using pass will cause the loop to immediately re-evaluate and peg the CPU, whereas using time.sleep will allow the execution to be temporarily suspended.
EDIT: just to prove the point, if you launch the python interpreter and run this:
>>> while True:
... pass
...
You can watch Python start eating up 90-100% CPU instantly, versus:
>>> import time
>>> while True:
... time.sleep(1)
...
Which barely even registers on the Activity Monitor (using OS X here but it should be the same for every platform).
Why sleep? You don't want to sleep, you want to wait for the threads to finish.
So
# store the threads you start in a your_threads list, then
for a_thread in your_threads:
a_thread.join()
See: thread.join
If you are looking for a short, zero-cpu way to loop forever until a KeyboardInterrupt, you can use:
from threading import Event
Event().wait()
Note: Due to a bug, this only works on Python 3.2+. In addition, it appears to not work on Windows. For this reason, while True: sleep(1) might be the better option.
For some background, Event objects are normally used for waiting for long running background tasks to complete:
def do_task():
sleep(10)
print('Task complete.')
event.set()
event = Event()
Thread(do_task).start()
event.wait()
print('Continuing...')
Which prints:
Task complete.
Continuing...
signal.pause() is another solution, see https://docs.python.org/3/library/signal.html#signal.pause
Cause the process to sleep until a signal is received; the appropriate handler will then be called. Returns nothing. Not on Windows. (See the Unix man page signal(2).)
I've always seen/heard that using sleep is the better way to do it. Using sleep will keep your Python interpreter's CPU usage from going wild.
You don't give much context to what you are really doing, but maybe Queue could be used instead of an explicit busy-wait loop? If not, I would assume sleep would be preferable, as I believe it will consume less CPU (as others have already noted).
[Edited according to additional information in comment below.]
Maybe this is obvious, but anyway, what you could do in a case where you are reading information from blocking sockets is to have one thread read from the socket and post suitably formatted messages into a Queue, and then have the rest of your "worker" threads reading from that queue; the workers will then block on reading from the queue without the need for neither pass, nor sleep.
Running a method as a background thread with sleep in Python:
import threading
import time
class ThreadingExample(object):
""" Threading example class
The run() method will be started and it will run in the background
until the application exits.
"""
def __init__(self, interval=1):
""" Constructor
:type interval: int
:param interval: Check interval, in seconds
"""
self.interval = interval
thread = threading.Thread(target=self.run, args=())
thread.daemon = True # Daemonize thread
thread.start() # Start the execution
def run(self):
""" Method that runs forever """
while True:
# Do something
print('Doing something imporant in the background')
time.sleep(self.interval)
example = ThreadingExample()
time.sleep(3)
print('Checkpoint')
time.sleep(2)
print('Bye')
I've slightly modified the signal example from the official docs (bottom of page).
I'm calling sleep 10 but I would like an alarm to be raised after 1 second. When I run the following snippet it takes way more than 1 second to trigger the exception (I think it runs the full 10 seconds).
import signal, os
def handler(signum, frame):
print 'Interrupted', signum
raise IOError("Should after 1 second")
signal.signal(signal.SIGALRM, handler)
signal.alarm(1)
os.system('sleep 10')
signal.alarm(0)
How can I be sure to terminate a function after a timeout in a single-threaded application?
From the docs:
A Python signal handler does not get executed inside the low-level (C)
signal handler. Instead, the low-level signal handler sets a flag
which tells the virtual machine to execute the corresponding Python
signal handler at a later point(for example at the next bytecode
instruction).
Therefore, a signal such as that generated by signal.alarm() can't terminate a function after a timeout in some cases. Either the function should cooperate by allowing other Python code to run (e.g., by calling PyErr_CheckSignals() periodically in C code) or you should use a separate process, to terminate the function in time.
Your case can be fixed if you use subprocess.check_call('sleep 10'.split()) instead of os.system('sleep 10').
I have a simple app that listens to a socket connection. Whenever certain chunks of data come in a callback handler is called with that data. In that callback I want to send my data to another process or thread as it could take a long time to deal with. I was originally running the code in the callback function, but it blocks!!
What's the proper way to spin off a new task?
threading is the threading library usually used for resource-based multithreading. The multiprocessing library is another library, but designed more for running intensive parallel computing tasks; threading is generally the recommended library in your case.
Example
import threading, time
def my_threaded_func(arg, arg2):
print "Running thread! Args:", (arg, arg2)
time.sleep(10)
print "Done!"
thread = threading.Thread(target=my_threaded_func, args=("I'ma", "thread"))
thread.start()
print "Spun off thread"
The multiprocessing module has worker pools. If you don't need a pool of workers, you can use Process to run something in parallel with your main program.
import threading
from time import sleep
import sys
# assume function defs ...
class myThread (threading.Thread):
def __init__(self, threadID):
threading.Thread.__init__(self)
self.threadID = threadID
def run(self):
if self.threadID == "run_exe":
run_exe()
def main():
itemList = getItems()
for item in itemList:
thread = myThread("run_exe")
thread.start()
sleep(.1)
listenToSocket(item)
while (thread.isAlive()):
pass # a way to wait for thread to finish before looping
main()
sys.exit(0)
The sleep between thread.start() and listenToSocket(item) ensures that the thread is established before you begin to listen. I implemented this code in a unit test framework were I had to launch multiple non-blacking processes (len(itemList) number of times) because my other testing framework (listenToSocket(item)) was dependent on the processes.
un_exe() can trigger a subprocess call that can be blocking (i.e. invoking pipe.communicate()) so that output data from the execution will still be printed in time with the python script output. But the nature of threading makes this ok.
So this code solves two problems - print data of a subprocess without blocking script execution AND dynamically create and start multiple threads sequentially (makes maintenance of the script better if I ever add more items to my itemList later).
I'm working with the Gnuradio framework. I handle flowgraphs I generate to send/receive signals. These flowgraphs initialize and start, but they don't return the control flow to my application:
I imported time
while time.time() < endtime:
# invoke GRC flowgraph for 1st sequence
if not seq1_sent:
tb = send_seq_2.top_block()
tb.Run(True)
seq1_sent = True
if time.time() < endtime:
break
# invoke GRC flowgraph for 2nd sequence
if not seq2_sent:
tb = send_seq_2.top_block()
tb.Run(True)
seq2_sent = True
if time.time() < endtime:
break
The problem is: only the first if statement invokes the flow-graph (that interacts with the hardware). I'm stuck in this. I could use a Thread, but I'm unexperienced how to timeout threads in Python. I doubt that this is possible, because it seems killing threads isn't within the APIs. This script only has to work on Linux...
How do you handle blocking functions with Python properly - without killing the whole program.
Another more concrete example for this problem is:
import signal, os
def handler(signum, frame):
# print 'Signal handler called with signal', signum
#raise IOError("Couldn't open device!")
import time
print "wait"
time.sleep(3)
def foo():
# Set the signal handler and a 5-second alarm
signal.signal(signal.SIGALRM, handler)
signal.alarm(3)
# This open() may hang indefinitely
fd = os.open('/dev/ttys0', os.O_RDWR)
signal.alarm(0) # Disable the alarm
foo()
print "hallo"
How do I still get print "hallo". ;)
Thanks,
Marius
First of all - the use of signals should be avoided at all cost:
1) It may lead to a deadlock. SIGALRM may reach the process BEFORE the blocking syscall (imagine super-high load in the system!) and the syscall will not be interrupted. Deadlock.
2) Playing with signals may have some nasty non-local consequences. For example, syscalls in other threads may be interrupted which usually is not what you want. Normally syscalls are restarted when (not a deadly) signal is received. When you set up a signal handler it automatically turns off this behavior for the whole process, or thread group so to say. Check 'man siginterrupt' on that.
Believe me - I met two problems before and they are not fun at all.
In some cases the blocking can be avoided explicitely - I strongly recommend using select() and friends (check select module in Python) to handle blocking writes and reads. This will not solve blocking open() call, though.
For that I've tested this solution and it works well for named pipes. It opens in a non-blocking way, then turns it off and uses select() call to eventually timeout if nothing is available.
import sys, os, select, fcntl
f = os.open(sys.argv[1], os.O_RDONLY | os.O_NONBLOCK)
flags = fcntl.fcntl(f, fcntl.F_GETFL, 0)
fcntl.fcntl(f, fcntl.F_SETFL, flags & ~os.O_NONBLOCK)
r, w, e = select.select([f], [], [], 2.0)
if r == [f]:
print 'ready'
print os.read(f, 100)
else:
print 'unready'
os.close(f)
Test this with:
mkfifo /tmp/fifo
python <code_above.py> /tmp/fifo (1st terminal)
echo abcd > /tmp/fifo (2nd terminal)
With some additional effort select() call can be used as a main loop of the whole program, aggregating all events - you can use libev or libevent, or some Python wrappers around them.
When you can't explicitely force non-blocking behavior, say you just use an external library, then it's going to be much harder. Threads may do, but obviously it is not a state-of-the-art solution, usually being just wrong.
I'm afraid that in general you can't solve this in a robust way - it really depends on WHAT you block.
IIUC, each top_block has a stop method. So you actually can run the top_block in a thread, and issue a stop if the timeout has arrived. It would be better if the top_block's wait() also had a timeout, but alas, it doesn't.
In the main thread, you then need to wait for two cases: a) the top_block completes, and b) the timeout expires. Busy-waits are evil :-), so you should use the thread's join-with-timeout to wait for the thread. If the thread is still alive after the join, you need to stop the top_run.
You can set a signal alarm that will interrupt your call with a timeout:
http://docs.python.org/library/signal.html
signal.alarm(1) # 1 second
my_blocking_call()
signal.alarm(0)
You can also set a signal handler if you want to make sure it won't destroy your application:
def my_handler(signum, frame):
pass
signal.signal(signal.SIGALRM, my_handler)
EDIT:
What's wrong with this piece of code ? This should not abort your application:
import signal, time
def handler(signum, frame):
print "Timed-out"
def foo():
# Set the signal handler and a 5-second alarm
signal.signal(signal.SIGALRM, handler)
signal.alarm(3)
# This open() may hang indefinitely
time.sleep(5)
signal.alarm(0) # Disable the alarm
foo()
print "hallo"
The thing is:
The default handler for SIGALRM is to abort the application, if you set your handler then it should no longer stop the application.
Receiving a signal usually interrupts system calls (then unblocks your application)
The easy part of your question relates to the signal handling. From the perspective of the Python runtime a signal which has been received while the interpreter was making a system call is presented to your Python code as an OSError exception with an errno attributed corresponding to errno.EINTR
So this probably works roughly as you intended:
#!/usr/bin/env python
import signal, os, errno, time
def handler(signum, frame):
# print 'Signal handler called with signal', signum
#raise IOError("Couldn't open device!")
print "timed out"
time.sleep(3)
def foo():
# Set the signal handler and a 5-second alarm
signal.signal(signal.SIGALRM, handler)
try:
signal.alarm(3)
# This open() may hang indefinitely
fd = os.open('/dev/ttys0', os.O_RDWR)
except OSError, e:
if e.errno != errno.EINTR:
raise e
signal.alarm(0) # Disable the alarm
foo()
print "hallo"
Note I've moved the import of time out of the function definition as it seems to be poor form to hide imports in that way. It's not at all clear to me why you're sleeping in your signal handler and, in fact, it seems like a rather bad idea.
The key point I'm trying to make is that any (non-ignored) signal will interrupt your main line of Python code execution. Your handler will be invoked with arguments indicating which signal number triggered the execution (allowing for one Python function to be used for handling many different signals) and a frame object (which could be used for debugging or instrumentation of some sort).
Because the main flow through the code is interrupted it's necessary for you to wrap that code in some exception handling in order to regain control after such events have occurred. (Incidentally if you're writing code in C you'd have the same concern; you have to be prepared for any of your library functions with underlying system calls to return errors and handle -EINTR in the system errno by looping back to retry or branching to some alternative in your main line (such as proceeding to some other file, or without any file/input, etc).
As others have indicated in their responses to your question, basing your approach on SIGALARM is likely to be fraught with portability and reliability issues. Worse, some of these issues may be race conditions that you'll never encounter in your testing environment and may only occur under conditions that are extremely hard to reproduce. The ugly details tend to be in cases of re-entrancy --- what happens if signals are dispatched during execution of your signal handler?
I've used SIGALARM in some scripts and it hasn't been an issue for me, under Linux. The code I was working on was suitable to the task. It might be adequate for your needs.
Your primary question is difficult to answer without knowing more about how this Gnuradio code behaves, what sorts of objects you instantiate from it, and what sorts of objects they return.
Glancing at the docs to which you've linked, I see that they don't seem to offer any sort of "timeout" argument or setting that could be used to limit blocking behavior directly. In the table under "Controlling Flow Graphs" I see that they specifically say that .run() can execute indefinitely or until SIGINT is received. I also note that .start() can start threads in your application and, it seems, returns control to your Python code line while those are running. (That seems to depend on the nature of your flow graphs, which I don't understand sufficiently).
It sounds like you could create your flow graphs, .start() them, and then (after some time processing or sleeping in your main line of Python code) call the .lock() method on your controlling object (tb?). This, I'm guessing, puts the Python representation of the state ... the Python object ... into a quiescent mode to allow you to query the state or, as they say, reconfigure your flow graph. If you call .run() it will call .wait() after it calls .start(); and .wait() will apparently run until either all blocks "indicate they are done" or until you call the object's .stop() method.
So it sounds like you want to use .start() and neither .run() nor .wait(); then call .stop() after doing any other processing (including time.sleep()).
Perhaps something as simple as:
tb = send_seq_2.top_block()
tb.start()
time.sleep(endtime - time.time())
tb.stop()
seq1_sent = True
tb = send_seq_2.top_block()
tb.start()
seq2_sent = True
.. though I'm suspicious of my time.sleep() there. Perhaps you want to do something else where you query the tb object's state (perhaps entailing sleeping for smaller intervals, calling its .lock() method, and accessing attributes that I know nothing about and then calling its .unlock() before sleeping again.
if not seq1_sent:
tb = send_seq_2.top_block()
tb.Run(True)
seq1_sent = True
if time.time() < endtime:
break
If the 'if time.time() < endtime:' then you will break out of the loop and the seq2_sent stuff will never be hit, maybe you mean 'time.time() > endtime' in that test?
you could try using Deferred execution... Twisted framework uses them alot
http://www6.uniovi.es/python/pycon/papers/deferex/
You mention killing threads in Python - this is partialy possible although you can kill/interrupt another thread only when Python code runs, not in C code, so this may not help you as you want.
see this answer to another question:
python: how to send packets in multi thread and then the thread kill itself
or google for killable python threads for more details like this:
http://code.activestate.com/recipes/496960-thread2-killable-threads/
If you want to set a timeout on a blocking function, threading.Thread as the method join(timeout) which blocks until the timeout.
Basically, something like that should do what you want :
import threading
my_thread = threading.Thread(target=send_seq_2.top_block)
my_thread.start()
my_thread.join(TIMEOUT)