> In a function like this:
>
> def f(x):
> return eval("x")
>
> , eval uses the local function namespace, and the above works.
> This is according to chapter 2.3 of the Python library ref.
>
> Now on my problem: When eval() is used with map, the same
> mechanism takes place:
>
> def f(x):
> return map(eval,["x"])
>
> It works the same as the above, because map is a builtin function
> that does not modify the frame chain, so eval finds the local
> namespace.
> Not so with Stackless Python (at the moment), since Stackless map
> assigns an own frame to map without passing the correct namespaces
> to it. (Reported by Bernd Rinn)
>
> Question: Is this by chance, or is eval() *meant* to function with
> the local namespace, even if it is executed in the context of
> a function like map() ?
Map, being a built-in, is transparent to namespaces.
> The description of map() does not state whether it has to pass
> its surrounding namespace to the mapped function, and if one
> simulates map() by writing one's own python implementation,
> it will fail exactly like Stackless does today. The same
> applies to apply().
So you can't simulate a built-in.
> I think I should fix Stackless here, anyway?
Yes.
Note: beware of Jeremy's nested scopes. That adds a whole slew of
namespaces! (But eval() is more crippled there.)
--Guido van Rossum (home page: http://www.python.org/~guido/)
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