Instancing a class - difference between with and without brackets

Question

Say I have a very simple class like the following:

class myClass:
    def __init__(self): 
        self.myProp = 2

If I instantiate using the brackets, everything works as I expect:

>>> a = myClass()
>>> a.myProp
2

However, if I don't use the brackets on the two lines above, i.e.:

>>> a = myClass

I get the following error:

>>> a.myProp
Traceback (most recent call last):
File "<pyshell#45>", line 1, in <module>
a.myProp
AttributeError: class myClass has no attribute 'myProp'

If I print the object,

>>> a = myClass
>>> a

I get

<class __main__.myClass at 0x0275C538>

It seems that a is an instance of the class, but somehow is not initialized. In other languages, I would expect a compile error if trying to cast a class instance into an object without initalizing it (e.g. in C#, myClass a = new myClass(); would work fine but myClass a = new myClass; would return a compile error).

So my question is: what is, technically speaking, the object a = myClass without brackets?

Answer 1

a is the class itself -- In python, classes are first class objects¹. You can pass them around as parameters, alias them to different names (as you've done in your example) and then you can instances from any reference that you have in the current namespace.

a = myClass  # a and myClass identical at this point.  The interpretter won't care which you use.
a_instance = a()  # instance of myClass

def make_instance(cls):
    return cls()

another_instance = make_instance(a)
yet_another_instance = make_instance(myClass)

You see, python doesn't have any "compile time checking" because really -- there is no compile time. Python code gets interpreted at runtime. True, you can have SyntaxErrors pop up at when you import something, but that is still during runtime.

^{1No pun intended}