[Python-ideas] os.path.here()

[Ryan Gonzalez]

That IS an option...

```python
import inspect

...


module_path = inspect.stack()[1][0].f_globals['__file__']

```



On Wed, Feb 19, 2014 at 11:18 AM, Giampaolo Rodola' <g.rodola at gmail.com>wrote:

>
>
>
> On Wed, Feb 19, 2014 at 6:04 PM, Oscar Benjamin <
> oscar.j.benjamin at gmail.com> wrote:
>
>> On 19 February 2014 16:52, Giampaolo Rodola' <g.rodola at gmail.com> wrote:
>> >
>> > The implementation is pretty straightforward:
>> >
>> > def here(concat=None):
>> >     """Return the absolute path of the parent directory where the
>> >     script is defined.
>> >     """
>> >     here = os.path.abspath(os.path.dirname(__file__))
>> >     if concat is not None:
>> >         here = os.path.abspath(os.path.join(here, concat))
>> >     return here
>>
>> So if I do from os.path import here and get the above function what
>> happens when I call it in another module?
>>
>
> Ouch! You're right, I naively didn't take that into account. =)
> I guess there are ways to inspect the caller's module name but I'm not
> gonna push for that.
> Sorry for the noise.
>
>
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-- 
Ryan
If anybody ever asks me why I prefer C++ to C, my answer will be simple:
"It's becauseslejfp23(@#Q*(E*EIdc-SEGFAULT. Wait, I don't think that was
nul-terminated."
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