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Static variables [was Re: syntax difference]


  To: Ben Bacarisse
From: "Bart" <bart at 1:261/38.remove-ij1-this>

  To: Ben Bacarisse
From: Bart <bc at freeuk.com>

On 23/06/2018 23:25, Ben Bacarisse wrote:
> Bart <bc at freeuk.com> writes:
>
>> On 23/06/2018 21:13, Chris Angelico wrote:
>>> On Sat, Jun 23, 2018 at 10:41 PM, Bart <bc at freeuk.com> wrote:
>>
>>>> (At what point would that happen anyway; if you do this:
>>
>>> NONE of your examples are taking copies of the function. They all are
>>> making REFERENCES to the same function. That is all.
>>
>> This is about your notion that invocations of the same function via
>> different references, should maintain their own versions of the
>> function's 'static' data.
>>
>> Since these references are created via the return g statement here:
>>
>>      def f():
>>          def g():
>>              ....
>>          return g
>>
>> (say to create function references i and j like this:
>>
>>      i = f()
>>      j = f()
>> )
>>
>> I'm assuming that something special must be happening. Otherwise, how
>> does f() know which reference it's being called via?
>>
>> What is different, what extra bit of information is provided when f()
>> is invoked via i() or j()?
>
> f is not being invoked by either i() or j().  i = f() binds i to the
> function returned by f.  That's a newly minted function.  In languages
> like Python, executing def creates a function.

Do you mean that if the same 'def' block is re-executed, it will create a
different instance of the function? (Same byte-code, but a different set of
everything else the function uses.)

Wow. (Just think of all the times you write a function containing a neat bunch
of local functions, every time it's called it has to create a new function
instances for each of those functions, even if they are not used.)

Anyway just for regular statics, the following appears to work. Not ideal, but
simpler than some alternatives:

def f():
     if not hasattr(f,'x'): f.x=0
     f.x += 1
     return f.x

--
bart.

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