Tim Chase wrote:
> >>> [name for name in dir() if id(eval(name)) == id(banana)]
> ['banana', 'spatula']
>
Please, if you are going to do something like this, then please at least
use the 'is' operator. Using id(expr1)==id(expr2) is just plain stupid: it
will actually work in this case, but as soon as you get into a mindset of
testing for the same object by comparing object ids you are going to shoot
yourself in the foot.
The first of the following tests returns True, which looks sensible at
first glance (even though it shouldn't), but what of the second one?
>>> class C:
def method1(self): pass
def method2(self): pass
>>> inst = C()
>>> id(inst.method1)==id(inst.method1)
True
>>> id(inst.method1)==id(inst.method2)
True
Much better to use 'is' and get consistent results
>>> inst.method1 is inst.method1
False
(In case I didn't make it clear, the problem in general with comparing the
result of calling 'id' is that as soon as the first call to id returns, any
object created when evaluating its parameter can be freed, so the second
call to id can reuse memory and get the same answer even though the objects
are different.)
--
http://mail.python.org/mailman/listinfo/python-list
