Pythonista 3.4 error: KeyboardInterrupt

DavinE

Hello all,

I have the following error in the new Pythonista version which worked in the previous version:

Traceback (most recent call last):
File
"/private/var/mobile/Containers/
Shared/AppGroup/
7502F935-14D3-4DA1-9D3B-1D998BE8A1E2/
Pythonista3/Documents/BarcodeScanner/
Barcode Scanner.py"
line 1202, in
checkCloseButton
ACTION Button
sys.exit(1)
File
"/var/containers/Bundle/
Application/872E2289-CF4B-43A6-BA49-
E163FB9FA4F0/Pythonista3.app/Frameworks/
Py3Kit. framework/pykit_startup.py"
line
33, in exit
raise KeyboardInterrupt(f'Exit:
{arg}')
KeyboardInterrupt: Exit: 1

BSoD

@DavinE
Me too.
sys.exit() always raise the key interrupt error.

DavinE

@BSoD said in Pythonista 3.4 error: KeyboardInterrupt:

@DavinE
Me too.
sys.exit() always raise the key interrupt error.

How did you get around that ?
Is there an alternative ?

BSoD

@DavinE
No.
But this error only occurs at the end of the program and has no actual impact.
So, until the official fixes this issue, you can temporarily ignore this error.

DavinE

@BSoD said in Pythonista 3.4 error: KeyboardInterrupt:

@DavinE
No.
But this error only occurs at the end of the program and has no actual impact.
So, until the official fixes this issue, you can temporarily ignore this error.

Not in my case...

It crash at the middle why ever... 😤

quer87

@DavinE

hi there
i also got the same error
since i updated pythonista3 latest v3.4

iphone 13 mini / iOS v15.6.1

any py program with input function has same results

in my case
main process is ok but the last sys.exit() got NG

Traceback (most recent call last):

line 647, in <module>
sys.exit()

line 33, in exit
raise KeyboardInterrupt(f'Exit: {arg}')

KeyboardInterrupt: Exit: 0

after that console is unstable

i wish fix that ...

JonB

@BSoD one workaround would be to simply monkeypatch sys so that

def exit ():
pass

sys.exit = exit

BSoD

@DavinE
I found a way to make the sys.exit work like old version.

In pythonista 3.4, there is a new script named pykit_startup.py that override the sys.exit method to avoid app really exit, and backup it to sys._exit .
So, you can just restore sys.exit to make it work as before.
Add these codes before your main entry:

import sys
sys.exit = sys._exit

ccc

try:
    x()
    y = z
except KeyboardInterrupt:
    pass

BSoD

@ccc
Neither 'pass' nor 'try/except:pass' can truly exit the executing script.

print('This should be shown.')
try:
    sys.exit()
except:
    pass
print('This should NOT be shown')