At 06:55 AM 12/23/2008 -0500, Rocky Bernstein wrote:
Now that there is a package mechanism (are package mechanisms?) like
zipimporter that bundle source code into a single file, should the
notion of a "file" location should be adjusted to include the package
and/or importer?
Is there a standard API or routine which can extract this information
given a code object?
The inspect module (in 2.5 and up) supports retrieving the source
lines for any object that has module globals. So you could do it for
a class, a function, a method, module-level code, or even a frame,
but not for a standalone code object.
I believe there are also certain inspect module APIs that will return
a pseudo-filename, i.e. the zipfile name followed by the path within
the zipfile.
Also I'm not sure there *is* a standard print string way to show
member inside a package. zipimporter may insert co_filename strings
like:
/usr/lib/python2.5/site-packages/tracer-0.1.0-py2.5.egg/tracer.py
AFAIK, it'll only do this if the zipfile doesn't contain a usable
.pyc or .pyo. Ordinarily, co_filename will be the name of the
original source file before the zipfile was created.
_______________________________________________
Python-Dev mailing list
Python-Dev@python.org
http://mail.python.org/mailman/listinfo/python-dev
Unsubscribe:
http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
