How can I tell if variable is defined

[Dave Angel]

Mel wrote:
> <snip>
>> This is an artifact of the interactive interpreter,
>>     
>
> True.  You can avoid the artifact by wrapping the test in a function:
>
> Python 2.6.2 (release26-maint, Apr 19 2009, 01:56:41) 
> [GCC 4.3.3] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
>   
>>>> class A():
>>>>         
> ...   def __repr__ (self):
> ...     return b
> ... 
>   
>>>> def is_defined (obj):
>>>>         
> ...   try:
> ...     obj
> ...   except NameError:
> ...     return False
> ...   return True
> ... 
>   
>>>> a = A()
>>>> if is_defined (a):
>>>>         
> ...   print "`a` is defined"
> ... else:
> ...   print "`a` is not defined"
> ... 
> `a` is defined
>   
>>>> if is_defined (b):
>>>>         
> ...   print "`b` is defined"
> ... else:
> ...   print "`b` is not defined"
> ... 
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
> NameError: name 'b' is not defined
>   
>
> At the cost of it not quite working when the function is called with an 
> undefined name.  I suppose the print statements could be crafted to make it 
> look better.
>
>
> 	Mel.
>
>   
No clue what you're trying to accomplish here.  The exception inside the 
function is_defined() will never fire;  the method always returns True.

You're getting an exception during the attempt to call such a function, 
and so of course there's no point in putting the if test around that either.

This may be the last time I try to read such complexity in an 
interactive transcript.  It took a couple of tries before I got the 
indentation understood.  I thought that is_defined() was intended to be 
a method inside the class.

DaveA