How not to ignore KeyboardInterrupt in try and except

Question

How to break only when KeyboardInterrupt is raised?

I have a something function that is not fully debugged that it can have a lot of random exceptions that I don't know about. But I want to ignore them except when I interrupt the process. Then I use a try...except... for it. For example

counter = 0
data = []
while True : 
    try : 
       data.append(something())
       counter+=1
    except : 
       continue

    if counter % 100 == 0 : 
       yield data
       data = []
yield data

How do I break it when I KeyboardInterrupt it?

"How do I break it when I KeyboardInterrupt it?" - it automatically stops, doesn't it ? — TheEagle, Feb 02 '21 at 03:58
Hi! Can you please better the question and the example. The ` something` functions is not there, and please clarify what you use the `try...except...` for it would be great. Do you want the function to yield when the KeyboardInterrupt occurs or do you want the whole program to end - which would be happening already? What is it? — Rohan Asokan, Feb 02 '21 at 03:58
@RohanAsokan I think he doesn't like the look of a stack trace when hitting ctrl-c. — Kris, Feb 02 '21 at 04:03
@IKnowHowBitcoinWorks I updated my answer, is that what you had in mind? — Kris, Feb 02 '21 at 04:23
@Programmer it doesn't seems to be the case. it just keeps throwing errors untils the iteration is stopped. I added another try and except inside the something() and have the inner exception to print something then iterate 1000 times. It printed a lot of lines when I interrupt it. — IKnowHowBitcoinWorks, Feb 02 '21 at 04:36

Kris · Accepted Answer · 2021-02-02T04:22:22.510

You could collect exceptions and only show them to the user if a KeyboardInterrupt is raised, e.g.

from time import sleep
import random


def something():
    """ flip a coin to decide whether to throw an error """
    if random.choice((True, False)):
        raise ValueError('bla bla')
        

exceptions = []

for i in range(20):
    print('i:', i)
    
    try:
        something()
        sleep(1)
        
    except KeyboardInterrupt:
        print('interrupted')
        raise KeyboardInterrupt(repr(exceptions))
        
    except Exception as e:
        exceptions.append(e)

This will give a stack trace like:

KeyboardInterrupt: [ValueError('bla bla'), ValueError('bla bla'), ValueError('bla bla'), ValueError('bla bla'), ValueError('bla bla')]

Vansh Sachdeva · Answer 2 · 2021-02-02T04:34:33.173

0

You can do this if you only want to break on keyboardInterupt:-

counter = 0
data = []
while True : 
    try : 
       data.append(something())
       counter+=1
    except KeyboardInterrupt:
        break
    except:
        continue

    if counter % 100 == 0 : 
       yield data
       data = []
yield data

This will do the trick.

edited Feb 02 '21 at 04:34

answered Feb 02 '21 at 04:18

Vansh Sachdeva

148
7

How not to ignore KeyboardInterrupt in try and except

2 Answers2