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exec and globals and locals ...


Am Donnerstag, 19. September 2019 18:31:43 UTC+2 schrieb Peter Otten:
> Eko palypse wrote:
> 
> > No, I have to correct myself
> > 
> > x = 5
> > def f1():
> >     exec("x = x + 1; print('f1 in:', x)")
> >     return x
> > print('f1 out', f1())
> > 
> > results in the same, for me confusing, results.
> > 
> > f1 in: 6
> > f1 out 5
> 
> Inside a function exec assignments go to a *copy* of the local namespace.
> Also LOAD_NAME is used to look up names. Therefore you can read and then 
> shade a global name with its local namesake.
> 
> Inside a function the namespace is determined statically. As f1() has no 
> assignment to x (code inside exec(...) is not considered) x is looked up in 
> directly the global namespace using LOAD_GLOBAL.
> 
> If you want to access the local namespace used by exec() you have to provide 
> one explicitly:
> 
> >>> x = 5
> >>> def f():
> ...     ns = {}
> ...     exec("x += 1", globals(), ns)
> ...     return ns["x"]
> ... 
> >>> f()
> 6
> >>> x
> 5
> 
> By the way, in Python 2 where exec was a statement the local namespace is 
> shared:
> 
> >>> x = 5
> >>> def f():
> ...     exec "x += 1"
> ...     return x
> ... 
> >>> f()
> 6
> >>> x
> 5

Sorry, missed that. 
Thank you, may I ask you how I could have come myself
to that explanation? What do I have to read to get that understanding?
Hopefully you don't say read the C code, because that is something
I tried but failed miserably.

Thank you
Eren