'Remove traceback in Python on Ctrl-C

Is there a way to keep tracebacks from coming up when you hit Ctrl+c, i.e. raise KeyboardInterrupt in a Python script?

Solution 1:^[1]

Try this:

import signal
import sys
signal.signal(signal.SIGINT, lambda x, y: sys.exit(0))

This way you don't need to wrap everything in an exception handler.

Solution 2:^[2]

Catch the KeyboardInterrupt:

try:
    # do something
except KeyboardInterrupt:
    pass

Solution 3:^[3]

Catch it with a try/except block:

while True:
   try:
      print "This will go on forever"
   except KeyboardInterrupt:
      pass

Solution 4:^[4]

try:
    your_stuff()
except KeyboardInterrupt:
    print("no traceback")

Solution 5:^[5]

Also note that by default the interpreter exits with the status code 128 + the value of SIGINT on your platform (which is 2 on most systems).

    import sys, signal

    try:
        # code...
    except KeyboardInterrupt: # Suppress tracebacks on SIGINT
        sys.exit(128 + signal.SIGINT) # http://tldp.org/LDP/abs/html/exitcodes.html

Solution 6:^[6]

suppress exception using context manager:

from contextlib import suppress

def output_forever():
    while True:
        print('endless script output. Press ctrl + C to exit')


if __name__ == '__main__':
    with suppress(KeyboardInterrupt):
        output_forever()

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	scrutari
Solution 2	icktoofay
Solution 3	Manny D
Solution 4	?s??o?
Solution 5	toksaitov
Solution 6	Roman Kazakov

'Remove traceback in Python on Ctrl-C

Solution 1:[1]

Solution 2:[2]

Solution 3:[3]

Solution 4:[4]

Solution 5:[5]

Solution 6:[6]