[Python-ideas] *var()*

[Guido van Rossum]

Right, that's why I said "won't access intermediate scopes"...

On Thursday, January 2, 2014, Steven D'Aprano wrote:

> On Thu, Jan 02, 2014 at 02:16:39PM -1000, Guido van Rossum wrote:
> > On Thu, Jan 2, 2014 at 3:35 AM, Steven D'Aprano <steve at pearwood.info<javascript:;>>
> wrote:
> > > I would have guessed that you could get this working with eval, but if
> > > there is such a way, I can't work it out.
> >
> > It's trivial if you directly invoke eval():
>
> That's what I thought too, but I get surprising results with
> nonlocals.
>
>
> a = b = "global"
>
> def test1():
>     b = c = "nonlocal"
>     def inner():
>         d = "local"
>         return (a, b, c, d)
>     return inner()
>
> def test2():
>     b = c = "nonlocal"
>     def inner():
>         d = "local"
>         c  # Need this or the function fails with NameError.
>         return (eval('a'), eval('b'), eval('c'), eval('d'))
>     return inner()
>
> assert test1() == test2()  # Fails.
>
>
> test1() returns ('global', 'nonlocal', 'nonlocal', 'local'), which is
> what I expect. But test2() returns ('global', 'global', 'nonlocal',
> 'local'),
> which surprises me.
>
> If I understand what is going on in test2's inner function, eval('b')
> doesn't see the nonlocal b so it picks up the global b. (If there is no
> global b, you get NameError.) But eval('c') sees the nonlocal c because
> we have a closure, due to the reference to c in the previous line.
>
> If there's a way to get eval('b') to return "nonlocal" without having a
> closure, I don't know it. This suggests to me that you can't reliably
> look-up a nonlocal from an inner function using eval.
>
>
> --
> Steven
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-- 
--Guido van Rossum (on iPad)
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