[solved] Variable number of dictionnaries as argument in def()

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[solved] Variable number of dictionnaries as argument in def()

ndc85430
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Posts: 1,823

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Joined: Apr 2017

Reputation: 85

Apr-19-2021, 04:52 PM

What are you actually trying to achieve? I'm concerned about all this dynamic code creation you're doing - here using locals. Do you really need to be doing that? Code like that tends to be quite difficult to maintain.

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[solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-19-2021, 02:49 PM

RE: Variable number of dictionnaries as argument in def() - by ndc85430 - Apr-19-2021, 04:52 PM

RE: Variable number of dictionnaries as argument in def() - by buran - Apr-19-2021, 05:03 PM

RE: Variable number of dictionnaries as argument in def() - by paul18fr - Apr-19-2021, 05:04 PM

RE: Variable number of dictionnaries as argument in def() - by ndc85430 - Apr-20-2021, 05:28 AM

RE: Variable number of dictionnaries as argument in def() - by paul18fr - Apr-20-2021, 07:29 AM

RE: Variable number of dictionnaries as argument in def() - by buran - Apr-19-2021, 05:18 PM

RE: [solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-19-2021, 08:17 PM

RE: [solved] Variable number of dictionnaries as argument in def() - by buran - Apr-20-2021, 09:03 AM

RE: [solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-20-2021, 10:07 AM

RE: [solved] Variable number of dictionnaries as argument in def() - by buran - Apr-20-2021, 10:47 AM

RE: [solved] Variable number of dictionnaries as argument in def() - by paul18fr - Apr-20-2021, 11:15 AM

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[solved] Variable number of dictionnaries as argument in def()

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