exec and globals and locals ...

[Eko palypse]

Am Donnerstag, 19. September 2019 12:45:35 UTC+2 schrieb Richard Damon:
> On 9/19/19 6:16 AM, Eko palypse wrote:
> >> In all cases, if the optional parts are omitted, the code is executed in the current scope. ...
> >>
> >>
> >> You can see from it that "globals" is optional.
> >> And that, if "globals" is missing, then
> >> "exec" is executed in the current scope ("f1" in your case).
> > Thank you for your answer, and that is exactly what confuses me?
> > Where does x come from? If I only would read x then I would understand why
> > it can be found/read but I alter it and as such I either have to provide the
> > info that this is a global variable, declare it inside of f1 or provide 
> > the globals dict to exec. But I don't do any of it. Why is exec able to use
> > the global x?
> >
> > Eren
> 
> I think the issue is that x += 1 isn't exactly like x = x + 1, and this
> is one case that shows it. x = x + 1 is an assignment to the symbol x,
> which makes x a local, and thus the read becomes an undefined symbol. x
> += 1 is different, it isn't a plain assignment so doesn't create the
> local. The read of x is inherently tied to the writing of x so x stays
> referring to the global.
> 
> -- 
> Richard Damon

Thank you that would never have come to my mind.
I thought +=1 is just syntactic sugar which clearly isn't.
If I do the regular x = x + 1 then I do get the expected exception.

Thank you
Eren