How to limit execution time of a function call in Python

import signal

def signal_handler(signum, frame):
    raise Exception("Timed out!")

signal.signal(signal.SIGALRM, signal_handler)
signal.alarm(10)   # Ten seconds
try:
    long_function_call()
catch Exception, msg:
    print "Timed out!"

from __future__ import with_statement # Required in 2.5
import signal
from contextlib import contextmanager

class TimeoutException(Exception): pass

@contextmanager
def time_limit(seconds):
    def signal_handler(signum, frame):
        raise TimeoutException, "Timed out!"
    signal.signal(signal.SIGALRM, signal_handler)
    signal.alarm(seconds)
    try:
        yield
    finally:
        signal.alarm(0)


try:
    with time_limit(10):
        long_function_call()
except TimeoutException, msg:
    print "Timed out!"

def timeout(func, args=(), kwargs={}, timeout_duration=1, default=None):
    '''This function will spwan a thread and run the given function using the args, kwargs and
    return the given default value if the timeout_duration is exceeded
    '''
    import threading
    class InterruptableThread(threading.Thread):
        def __init__(self):
            threading.Thread.__init__(self)
            self.result = default
        def run(self):
            try:
                self.result = func(*args, **kwargs)
            except:
                self.result = default
    it = InterruptableThread()
    it.start()
    it.join(timeout_duration)
    if it.isAlive():
        return it.result
    else:
        return it.result

def function_with_enforced_timeout():
  f = open_temporary_file()
  try:
   ...
  finally:
   here()
   unlink(f.filename)