Why is this code using class variables not working as it would in Java? [duplicate]

Question

I am moving from Java to Python. In Java, Static variables will return the same value to all objects of a class even if it is changed by another object of the same class. In the Python book that I'm reading, the author says, a data member is entered into class namespace if it's assignment takes place within the immediate scope of the class definition.

I'm using IDLE, the Python Interpreter, and below is the Python code:

class ABC:
    OVRLIMIT_FEE=5

>>> a=ABC()
>>> b=ABC()

>>> a.OVRLIMIT_FEE+=1
>>> a.OVRLIMIT_FEE
6
>>> b.OVRLIMIT_FEE
5

But if it enters the class namespace, should it not return same value when it is accessed with both a and b ?

Even though you mentioned java, this question is totally unrelated to it — SomeJavaGuy
– SomeJavaGuy, Commented Aug 11, 2016 at 11:05
you are changing an objects attribute not ABC's attribute, do ABC.OVRLIMIT_FEE+=1 then you will get 6 for both — dnit13
– dnit13, Commented Aug 11, 2016 at 11:06
If you assign to a.OVRLIMIT_FEE (as you have done with a.OVRLIMIT_FEE+=1), you are creating an instance variable on a. If you want to access a class variable, use ABC.OVRLIMIT_FEE. — khelwood
– khelwood, Commented Aug 11, 2016 at 11:06
Why is this marked as duplicate? The referenced question asks about existence of "static" variables, this one about a specific behaviour of them. Admittedly the answers cover this problem as well. — Torben Klein
– Torben Klein, Commented Aug 11, 2016 at 15:35

Torben Klein · Accepted Answer · 2016-08-11 15:30:18Z

0

What happens is that you created a so-called Oldstyle class, which does not support your (correctly) expected behaviour. Inherit from object to create a new-style class. Then it will work as you expect [...]

Using a.foo += 1 is implicitly unterstood as a.foo = a.foo + 1, which

takes the class property foo
adds 1
puts the result into instance property foo.

Code:

>>> class B(object):
...   x=5
>>> b1=B()
>>> b2=B()
>>> b1.x += 1
>>> b1.__dict__
{'x': 6}
>>> b2.__dict__
{}
>>> B.__dict__
{'x': 5  ... and other stuff ... }

edited Aug 11, 2016 at 15:30

answered Aug 11, 2016 at 11:16

Torben Klein

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2 Comments

Kaushik Over a year ago

b2.x prints 6 because x is 6 for class B. If it is to work as I expect, should it not print 7 ?

Torben Klein Over a year ago

oh my, I should not answer python questions in the lunch break.. I will correct the answer right away.

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