micropython unexpected result

Reply to micropython unexpected result on Tue, 17 Dec 2019 02:35:00 GMT

kjm — Tue, 17 Dec 2019 02:35:00 GMT

OK programming gurus I expected

old_error=0
def _logerrorin(error):
    global old_error
    print('old_error =', old_error)
    print('new_error =', error)
    if error!=old_error: print('a new error')
    old_error=error

for i in range(2):
  try:
    1/0
  except Exception as e: _logerrorin(e)

to print 'a new error' just once, but I actually get

old_error = 0
new_error = divide by zero
a new error
old_error = divide by zero
new_error = divide by zero
a new error

Can someone tell me what I'm doing wrong?

Reply to micropython unexpected result on Tue, 17 Dec 2019 05:16:37 GMT

johand — Tue, 17 Dec 2019 05:16:37 GMT

I am guessing a bit now... Looks like you are compairing two different Exception-objects, they are not the same. You most likley want to compare the description string of the exceptions.

Not sure if this is the correct way, but you should get an idea of what to do:

if str(error)!=str(old_error): print('a new error')

Johan

Reply to micropython unexpected result on Tue, 17 Dec 2019 16:15:38 GMT

mnovella — Tue, 17 Dec 2019 16:15:38 GMT

@johand Indeed you are correct.

As new_error and old_error are Objects, comparing them directly will result in that they are different all the time.

Thus, as you suggested one may want to compare the string or something else that Exception throws.

Reply to micropython unexpected result on Tue, 17 Dec 2019 16:43:55 GMT

robert-hh — Tue, 17 Dec 2019 16:43:55 GMT

@kjm e is an instance of the class . With every new Exception raised, you create a new instance. Both instances have a different id. In MicroPython as well a Python, class instances are equal if their id is the same, which is not the case here.
If that logic is important for you, you can for instance compare the types in line 6:
if type(error) != type(old_error): print('a new error')