[Tutor] Assessing local variable outside function

[nils wagenaar]

Thank you all! It is clear now:)

Verstuurd vanaf mijn iPhone

> Op 28 okt. 2016 om 19:31 heeft Ben Finney <ben+python at benfinney.id.au> het volgende geschreven:
> 
> Alan Gauld via Tutor <tutor at python.org> writes:
> 
>>> On 28/10/16 02:38, nils wagenaar wrote:
>>> Could i use a variable defined in a function in another function?
> 
> My answer would be: You can't because Python variables don't exist
> outside their namespace.
> 
> You can make the object available in various ways, but not the variable.
> 
>> By returning it to the caller.
> 
> That's somewhat misleading. Returning the *object* would not grant
> access to the local *variable*.
> 
> Nils, it's important to realise that a variable in Python is not
> tractable: you can't hand them around, you can't access the name itself.
> A Python variable exists only in its namespace, and can't move.
> 
> The variable is (at any given point) bound to an object; you can get
> *other* variables bound to the same object by explicitly doing that.
> Alan suggests one way.
> 
> Whether that meets your request to “use a variable defined in a function
> in another function” will have to wait for you to check how the Python
> data model actually works. Does that answer it, or do you need something
> different?
> 
> -- 
> \     “Dare to be naïve.” —Richard Buckminster Fuller, personal motto |
>  `\                                                                   |
> _o__)                                                                  |
> Ben Finney
> 
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