how to close python process on program end - python-3.x

Question

I fail to close a serial connection that runs in a process properly at the end of the program. (on windows/VSCode and Ctrl-C)
I get an error message and most of the time the port is already opened at the next start of the program.
Do I have to finish the run process first?
class serialOne(Process):
def __init__(self, serial_port, debug, baudrate=57600, timeout=1):
...
def terminate(self):
print("close ports")
self.active = False
self.ser.close()
def run(self):
while self.active:
self.initCom()
self.readCom()
time.sleep(0.005)
def main():
global processList
global debug
while True:
if debug == True:
print("main")
time.sleep(1.0)
for process in processList:
process.terminate()
and my main:
def main():
global processList
global debug
while True:
if debug == True:
print("main") # actually doing nothing
time.sleep(1.0)
for process in processList:
process.terminate()
that's the error message:
Process serialOne-1:
Traceback (most recent call last):
File "C:\Users\dgapp\AppData\Local\Programs\Python\Python37-32\lib\multiprocessing\process.py", line 297, in _bootstrap
self.run()
Traceback (most recent call last):
File "e:\_python\rfid_jacky\simple_multiprocess_rfid_02.py", line 129, in run
time.sleep(0.005)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\ptvsd_launcher.py", line 45, in <module>
KeyboardInterrupt
main(ptvsdArgs)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\__main__.py", line 265, in main
wait=args.wait)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\__main__.py", line 258, in handle_args
debug_main(addr, name, kind, *extra, **kwargs)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\_local.py", line 45, in debug_main
run_file(address, name, *extra, **kwargs)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\_local.py", line 79, in run_file
run(argv, addr, **kwargs)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\_local.py", line 140, in _run
_pydevd.main()
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\_vendored\pydevd\pydevd.py", line 1925, in main
debugger.connect(host, port)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\_vendored\pydevd\pydevd.py", line 1283, in run
return self._exec(is_module, entry_point_fn, module_name, file, globals, locals)
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\_vendored\pydevd\pydevd.py", line 1290, in _exec
pydev_imports.execfile(file, globals, locals) # execute the script
File "c:\Users\dgapp\.vscode\extensions\ms-python.python-2018.12.1\pythonFiles\lib\python\ptvsd\_vendored\pydevd\_pydev_imps\_pydev_execfile.py", line 25, in execfile
exec(compile(contents+"\n", file, 'exec'), glob, loc)
File "e:\_python\rfid_jacky\simple_multiprocess_rfid_02.py", line 161, in <module>
main()
File "e:\_python\rfid_jacky\simple_multiprocess_rfid_02.py", line 140, in main
time.sleep(1.0)
KeyboardInterrupt

Answer 1

When you press Ctrl+C, a KeyboardInterrupt exception is thrown, and interrupts your infinite sleep loop. But since you don't catch this exception, the code after this loop (with process.terminate()) is never called, which probably causes your issue.
So you have several options:
catch KeyboardInterrupt and use that to exit the inifite loop:
def main():
global processList
global debug
try:
while True:
if debug == True:
print("main") # actually doing nothing
time.sleep(1.0)
except KeyboardInterrupt:
pass
for process in processList:
process.terminate()
Which is simple and very readable.
register an exit handler that will be run when your program exits:
import atexit
#atexit.register
def shutdown():
global processList
for process in processList:
process.terminate()
def main():
global debug
while True:
if debug == True:
print("main") # actually doing nothing
time.sleep(1.0)
Which is more reliable since it will work even if your process is terminated by another signal.

Answer 2

The KeyboardInterrupt happens when the user hits the interrupt key.
A simple solution would be to catch the exception:
while True:
if debug == True:
print("main") # actually doing nothing
try:
# do your things
except KeyboardInterrupt:
print("program was interrupted by user")
break
You could also use the finally keyword to properly end your program:
try:
while True:
# do your things
except KeyboardInterrupt:
print("program was interrupted by user")
break
finally:
close() # this will always happen, even if an exception was raised