Capture keyboardinterrupt in Python without try-except

Is there some way in Python to capture KeyboardInterrupt event without putting all the code inside a try-except statement?

I want to cleanly exit without trace if user presses Ctrl+C.

Yes, you can install an interrupt handler using the module signal, and wait forever using a threading.Event:

import signal
import sys
import time
import threading

def signal_handler(signal, frame):
    print('You pressed Ctrl+C!')
    sys.exit(0)

signal.signal(signal.SIGINT, signal_handler)
print('Press Ctrl+C')
forever = threading.Event()
forever.wait()

If all you want is to not show the traceback, make your code like this:

## all your app logic here
def main():
   ## whatever your app does.


if __name__ == "__main__":
   try:
      main()
   except KeyboardInterrupt:
      # do nothing here
      pass

(Yes, I know that this doesn't directly answer the question, but it's not really clear why needing a try/except block is objectionable -- maybe this makes it less annoying to the OP)

An alternative to setting your own signal handler is to use a context-manager to catch the exception and ignore it:

>>> class CleanExit(object):
...     def __enter__(self):
...             return self
...     def __exit__(self, exc_type, exc_value, exc_tb):
...             if exc_type is KeyboardInterrupt:
...                     return True
...             return exc_type is None
... 
>>> with CleanExit():
...     input()    #just to test it
... 
>>>

This removes the try-except block while preserving some explicit mention of what is going on.

This also allows you to ignore the interrupt only in some portions of your code without having to set and reset again the signal handlers everytime.