CODEWITHSUNDEEP
pythonmultithreadingpyqtpyqt5
Six
Six

Reputation: 5512

How to kill hung threads in Python

I've created a function that uses PyQt5 to "render" HTML and return the result. It's as follows:

def render(source_html):
    """Fully render HTML, JavaScript and all."""

    import sys
    from PyQt5.QtWidgets import QApplication
    from PyQt5.QtWebKitWidgets import QWebPage

    class Render(QWebPage):
        def __init__(self, html):
            self.html = None
            self.app = QApplication(sys.argv)
            QWebPage.__init__(self)
            self.loadFinished.connect(self._loadFinished)
            self.mainFrame().setHtml(html)
            self.app.exec_()

        def _loadFinished(self, result):
            self.html = self.mainFrame().toHtml()
            self.app.quit()

    return Render(source_html).html

Occasionally it's threads will hang indefinitely and I'll have to kill the whole program. Unfortunately PyQt5 may as well be a black box as I'm not sure how to kill it when it misbehaves.

Ideally I'd be able to implement a timeout of n seconds. As a workaround, I've put the function in it's own script render.py and am calling it from via subprocess with this monstrosity:

def render(html):
    """Return fully rendered HTML, JavaScript and all."""
    args = ['render.py', '-']
    timeout = 20

    try:
        return subprocess.check_output(args,
                                       input=html,
                                       timeout=timeout,
                                       universal_newlines=True)
    # Python 2's subprocess.check_output doesn't support input or timeout
    except TypeError:
        class SubprocessError(Exception):
            """Base exception from subprocess module."""
            pass

        class TimeoutExpired(SubprocessError):
            """
            This exception is raised when the timeout expires while
            waiting for a child process.
            """

            def __init__(self, cmd, timeout, output=None):
                super(TimeoutExpired, self).__init__()
                self.cmd = cmd
                self.timeout = timeout
                self.output = output

            def __str__(self):
                return ('Command %r timed out after %s seconds' %
                        (self.cmd, self.timeout))

        process = subprocess.Popen(['timeout', str(timeout)] + args,
                                   stderr=subprocess.PIPE,
                                   stdin=subprocess.PIPE,
                                   stdout=subprocess.PIPE)
        # pipe html into render.py's stdin
        output = process.communicate(
            html.encode('utf8'))[0].decode('latin1')
        retcode = process.poll()
        if retcode == 124:
            raise TimeoutExpired(args, timeout)
        return output

The multiprocessing module appears to greatly simplify things:

from multiprocessing import Pool

pool = Pool(1)
rendered_html = pool.apply_async(render, args=(html,)).get(timeout=20)
pool.terminate()

Is there a way to implement a timeout that doesn't necessitate these sort of shenanigans?

Upvotes: 0

Views: 1958

Answers (1)

Locane
Locane

Reputation: 3134

I was looking for a solution too, there apparently isn't one, on purpose.

If you're using Linux and all you want is Python to attempt something for N seconds and then time out and handle an error condition after those N seconds, you can do this:

import time
import signal

# This stuff is so when we get SIGALRM from the timeout functionality we can handle it instead of
# crashing to the ground
class TimeOutError(Exception):
    pass
def raise_timeout(var1, var2):
    raise TimeOutError
signal.signal(signal.SIGALRM, raise_timeout)

# Turn the alarm on
signal.alarm(1)

# Try your thing
try:
    time.sleep(2)
except TimeOutError as e:
    print("  We hit our timeout value and we bailed out of whatever that BS was.")

# Remember to turn the alarm back off if your attempt succeeds!
signal.alarm(0)

The one drawback is that you can't nest signal.alarm() hooks; if, in your try statement, you're calling something else that also then sets a signal.alarm(), it will override the first one and screw your stuff up.

Upvotes: 1

Related Questions