Question

I have some class:

class RSA:
 CONST_MOD=2
 def __init__(self):
  print "created"

 def fast_powering(self,number,power,mod):
  print "powering"

I want to instantiate it and call method fast_powering:

 def main():
  obj=RSA() # here instant of class is created 
  val=obj.fast_powering(10,2,obj.CONST_MOD)  #  and we call method of object
  print val

And it works fine!

But I found that I can do it a little bit different way too, like:

def main():
 obj=RSA #do we create a link to the class without creating of object , or what?
 val=obj().fast_powering(10,2,obj().CONST_MOD) # and here we do something like
          # calling of static method of class in C++ without class instantiation,
          #  or ?
 print val

Sorry, I think a little bit in C++ way, but anyway
to my great astonishment it works too!
What’s actually happening here? Which wayn is more preffered? It’s some misterious for me.

Thanks in advance for any replies!

Answer 1

In your example, you’re doing:

obj = RSA

which is just binding the name obj to whatever was bound to RSA, which is the class RSA in your instance. Then you’re doing:

obj().fast_powering(…)

Which is equivalent to creating an instance of RSA and calling the method fast_powering on it. Note that this way, you’ll get a new instance of RSA on each call, which is probably not what you want. You’ll also have noticed, that the __init__ method has been called in the line cited above. Also consider:

>>> class RSA:
...   def __init__(self):
...     print("foo")
... 
>>> obj = RSA
>>> obj() is obj()
foo
foo
False

Here we see that the statement obj() is obj() in fact creates two objects, which are of course not identical. This is opposed to your first example, as demonstrated using:

>>> class RSA:
...   def __init__(self):
...     print("foo")
... 
>>> obj = RSA()
foo
>>> obj is obj
True