Getting globals of the caller, not the defining module

[Terry Reedy]

On 11/11/2013 7:02 AM, sg552 at hotmail.co.uk wrote:
> (Sorry for posting through GG, I'm at work.)
>
> On Monday, November 11, 2013 11:25:42 AM UTC, Steven D'Aprano wrote:
>> Suppose I have a function that needs access to globals:
>>
>> # module A.py
>> def spam():
>>      g = globals()  # this gets globals from A
>>      introspect(g)
>>
>> As written, spam() only sees its own globals, i.e. those of the module in
>> which spam is defined. But I want spam to see the globals of the caller.
>>
>> # module B
>> import A
>> A.spam()  # I want spam to see globals from B
>>
>> I can have the caller explicitly pass the globals itself:
>>
>> def spam(globs=None):
>>      if globs is None:
>>          globs = globals()
>>      introspect(globs)
>>
>> But since spam is supposed to introspect as much information as possible,
>> I don't really want to do that. What (if anything) are my other options?
>
> How about this?
>
> # module A.py
> import inspect
> def spam():
>      return inspect.stack()[1][0].f_globals

In Python 3, the attribute is __globals__. In either case, it is only 
defined on Python coded functions, so one should be prepared for it to 
not exist. That possibility is real because there *are* builtins like 
map and filter that take function args and call them.

Inspect has been modified in Py 3, but stack is still there.

> # module B.py
> import A
> print(A.spam() is globals()) # prints True
> def f():
>      return A.spam()
>
> # module C.py
> import B
> print(B.f() is vars(B)) # prints True
>
> I don't really know what I'm doing but I guess it won't work in alternative implementations of Python.
>


-- 
Terry Jan Reedy