Re: How can a function know what module it's in?

[Joe Strout]

Some corrections, to highlight the depth of my confusion...

On Nov 11, 2008, at 9:10 PM, Joe Strout wrote:

doctest.testmod (mymodule)

This actually works fine if I'm importing the module (with the
standard name) somewhere else

Actually, it does not.

I noticed that a function object has a __module__ attribute, that is
a reference to the module the function is in.

And no, it isn't; it's the NAME of the module the function is in. I'm
not sure what good that does me. docstring.testm od does take an
optional "name" parameter, but the documentation (at least in 2.5.2)
does not explain what this parameter is used for. I tried using it
thusly:

doctest.testmod (name=_test.__m odule__)

but that didn't work; it appears to still be testing the __main__
module. (Perhaps the name parameter is only used to describe the
module in the output, in which case, all I've accomplished here is
getting doctest to lie.)

I'm sure there is a magic identifier somewhere that lets a code get
a reference to its own module, but I haven't been able to find it.
Can someone share a clue?

This question remains open. :)

Thanks,
- Joe

[Rafe]

On Nov 12, 11:17*am, Joe Strout <[e-mail] twrote:

Some corrections, to highlight the depth of my confusion...

On Nov 11, 2008, at 9:10 PM, Joe Strout wrote:

* *doctest.testmo d(mymodule)

This actually works fine if I'm importing the module (with the *
standard name) somewhere else

Actually, it does not.

I noticed that a function object has a __module__ attribute, that is *
a reference to the module the function is in.

And no, it isn't; it's the NAME of the module the function is in. *I'm *
not sure what good that does me. *docstring.test mod does take an *
optional "name" parameter, but the documentation (at least in 2.5.2) *
does not explain what this parameter is used for. *I tried using it *
thusly:

* * * * doctest.testmod (name=_test.__m odule__)

but that didn't work; it appears to still be testing the __main__ *
module. *(Perhaps the name parameter is only used to describe the *
module in the output, in which case, all I've accomplished here is *
getting doctest to lie.)

I'm sure there is a magic identifier somewhere that lets a code get *
a reference to its own module, but I haven't been able to find it. *
Can someone share a clue?

This question remains open. *:)

Thanks,
- Joe

import sys
this_module = sys.modules[__name__]

sys.modules is a dictionary of all modules which have been imported
during the current session. Since the module had to be imported to
access it, it will be in there. '__name__' is available inside
functions because it is in the module scope.

Classes are a little more tricky because doing something like:
this_module = sys.modules[self.__class__. __module__]
will return a different module if the class is inherited in a
different module (since the base class is __class__ now). However,
putting a function at the module level (in the super-class module)
should anchor the results (untested though).

I'm not sure if this is the answer you need with regards to doctest,
but it I think it answers the question in the subject of this thread.

- Rafe

[Joe Strout]

On Nov 11, 2008, at 9:49 PM, Rafe wrote:

>>I'm sure there is a magic identifier somewhere that lets a code get
a reference to its own module, but I haven't been able to find it.

import sys
this_module = sys.modules[__name__]

Beautiful! Thanks very much. For the archives, here is my standard
module-testing idiom now:

def _test():
import doctest, sys
doctest.testmod (sys.modules[__name__])

if __name__ == "__main__":
_test()

Now, when I execute the module directly, it will test itself; and if I
need to test it from the outside (for example, under pdb) I can import
the module and then run themodule._test ().

To whom should I make the suggestion that this go into doctest docs?

Cheers,
- Joe