__file__

[John Machin]

On Apr 11, 8:03 pm, "7stud" <bbxx789_0... at yahoo.com> wrote:
> Hi,
>
> Thanks for the response.
>
> On Apr 11, 12:49 am, "Gabriel Genellina" <gagsl-... at yahoo.com.ar>
> wrote:
>
>
>
> > __file__ corresponds to the filename used to locate and load the module,
> > whatever it is. When the module is found on the current directory
> > (corresponding to '' in sys.path), you get just the filename; if sys.path
> > contains a relative path, that's what you get; the same for any absolute
> > path.
> > Whatever path worked to find and load the module, that's stored as
> > __file__.
>
> > If you plan to use it, it's a good idea to make it early into an absolute
> > path (using os.path.abspath(__file__)) just in case the current directory
> > changes.
>
> That last part doesn't seem to fit with your description above.  What
> does the current working directory have to do with the path that was
> used to load a module?  I would think the path that was used to load a
> module is constant.

You are correct, but that is not what GG was talking about. Here is an
example of what he meant:

While your cwd is /bar, you load module foo from a *relative*( path,
e.g. ./foo.py. If you do the os.path.abspath("./foo.py") immediately
as recommended, you get the correct answer: /bar/foo.py. However if
you change your cwd to /zot, then do the os.path.abspath("./foo.py"),
you get /zot/foo.py which is a nonsense.

HTH,
John