isinstance() bug

[Michal Vitecek]

hello,

please consider the following situation:

under the current directory there's a subdirectory 'package' with two
files: __init__.py and module.py

./package:
__init__.py
module.py

module.py contains:

class A(object):
pass

aModule = A()
now, let's do:

current directory:

import package.module
type(package.module.aModule) <class 'package.module.A'> isinstance(package.module.aModule, package.module.A) # so far good 1 a = package.module.A()
isinstance(a, package.module.A) # so far good
import sys
sys.path.append('package')
import module
a = module.A()
isinstance(a, package.module.A) # will return 0 !!! 0 isinstance(package.module.aModule, module.A) # will return 0 !!! 0

how is it possible that it IS important how you imported a class
definition for isinstance() to work? it's insane!

--
fuf (fu*@mageo.cz)

[Sidharth Kuruvila]

"Michal Vitecek" <fu*@mageo.cz> wrote in message
news:ma**************************************@pyth on.org...

hello,

please consider the following situation:

under the current directory there's a subdirectory 'package' with two
files: __init__.py and module.py

./package:
__init__.py
module.py

module.py contains:

class A(object):
pass

aModule = A()
now, let's do:

current directory:

>>> import package.module
>>> type(package.module.aModule) <class 'package.module.A'> >>> isinstance(package.module.aModule, package.module.A) # so far good 1 >>> a = package.module.A()
>>> isinstance(a, package.module.A) # so far good
>>> import sys
>>> sys.path.append('package')
>>> import module
>>> a = module.A()
>>> isinstance(a, package.module.A) # will return 0 !!! 0 >>> isinstance(package.module.aModule, module.A) # will return 0 !!! 0 >>>

how is it possible that it IS important how you imported a class
definition for isinstance() to work? it's insane!

--
fuf (fu*@mageo.cz)

its got to do with how python imports modules.
both imports are treated as seperate modules because the relative paths are
diferent.
python is a dynamic language it figures out where modules are located at run
time, it would be expensive to check all paths to see if they ended up at
the same file.

[Skip Montanaro]

Sidharth> python is a dynamic language it figures out where modules are
Sidharth> located at run time, it would be expensive to check all paths
Sidharth> to see if they ended up at the same file.

Maybe not:

Help on function abspath in module posixpath:

abspath(path)
Return an absolute path.

Considering all the other stuff import has to do, this seems like only a
small extra bit of work.

Skip

[Sidharth Kuruvila]

i dont know unix.
how does abs path work with sym-links.

[Sidharth Kuruvila]

Stupid post must be ignored at all costs

"Sidharth Kuruvila" <si*******@hotmail.com> wrote in message
news:bv************@ID-223265.news.uni-berlin.de...

i dont know unix.
how does abs path work with sym-links.

[Skip Montanaro]

Sidharth> i dont know unix.
Sidharth> how does abs path work with sym-links.

Not well. Seems that os.path.realpath does the right thing though:

% ls -l /etc/rc2.d
lrwxrwxrwx 1 root root 10 May 21 2002 /etc/rc2.d -> rc.d/rc2.d/
% python
Python 2.2.3 (#1, Jun 21 2003, 08:08:22)
[GCC 3.0.1] on linux2
Type "help", "copyright", "credits" or "license" for more information.

import os
os.path.abspath("/etc/rc2.d") '/etc/rc2.d' os.path.realpath("/etc/rc2.d") '/etc/rc.d/rc2.d'

As someone else pointed out, the issue of multiple hard links to the same
file is a bit tougher nut to crack.

% cd ~/tmp
% cat > a.py
class A: pass
% ln a.py b.py
% python
Python 2.2.3 (#1, Jun 21 2003, 08:08:22)
[GCC 3.0.1] on linux2
Type "help", "copyright", "credits" or "license" for more information. import a
import b
a.A == b.A 0

You need to compare inode numbers to see if you have the same or different
files.
import stat
os.stat(a.__file__)[stat.ST_INO] 37041L os.stat(b.__file__)[stat.ST_INO]

37041L

Skip

[Jeff Epler]

On Wed, Jan 28, 2004 at 10:37:39PM +0530, Sidharth Kuruvila wrote:

i dont know unix.
how does abs path work with sym-links.

os.abspath does not check whether any components are symlinks. In the
presence of symlinks,
/a/../b
and
/b
can refer to separate files if /a is a symlink to a directory. As far as
I know, posixpath.abspath normalizes the first into the second anyway.

Similarly, if you have a symlink
/a -> c
then posixpath.abspath("a") doesn't return /c, it still returns /a

You can use os.readlink() on posix systems to get the thing a symlink
points to:

os.readlink("/proc/self") '14717' os.readlink("/")

OSError: [Errno 22] Invalid argument: '/'

Jeff