python class attribute

Question

i have a question about class attribute in python.

class base :
    def __init__ (self):
        pass
    derived_val = 1

t1 = base()
t2 = base()

t2.derived_val +=1
t2.__class__.derived_val +=2
print t2.derived_val             # its value is 2
print t2.__class__.derived_val   # its value is 3

The results are different. I also use id() function to find t2.derived_val and t2.__class__.derived_val have different memory address. My problem is derived_val is class attribute. Why it is different in above example? Is it because the instance of class copy its own derived_val beside the class attribute?

unutbu · Accepted Answer · 2013-01-07 11:24:41Z

47

There are class attributes, and instance attributes. When you say

class base :
    derived_val = 1

You are defining a class attribute. derived_val becomes a key in base.__dict__.

t2=base()
print(base.__dict__)
# {'derived_val': 1, '__module__': '__main__', '__doc__': None}
print(t2.__dict__)
# {}

When you say t2.derived_val Python tries to find 'derived_val' in t2.__dict__. Since it is not there, it looks if there is a 'derived_val' key in any of t2's base classes.

print(t2.derived_val)
print(t2.__dict__)
# 1
# {}

But when you assign a value to t2.derived_val, you are now adding an instance attribute to t2. A derived_val key is added to t2.__dict__.

t2.derived_val = t2.derived_val+1
print(t2.derived_val)
print(t2.__dict__)
# 2
# {'derived_val': 2}

Note that at this point, there are two derived_val attributes, but only the instance attribute is easily accessible. The class attribute becomes accessible only through referencing base.derived_val or direct access to the class dict base.__dict__.

edited Jan 7, 2013 at 11:24

answered May 27, 2010 at 18:11

unutbu

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1 Comment

user10217630 Over a year ago

Is this a defect? I don't remember c++ but isn't it supposed to hold on to its ID (I mean to the same object). Why did the derived_val got added to t2.__dict__? You didn't do something like t2['derived_val'] = value.

Community · Accepted Answer · 2017-05-23 12:00:25Z

4

Check it out here and here.

The __class__ attribute is the class that the object belongs to. So in your example, the situation is similar to static variables. The t2.__class__.derived_val is referring to the variable that belongs to the class, not the variable that belongs to t2.

edited May 23, 2017 at 12:00

CommunityBot

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answered May 27, 2010 at 18:08

jakecar

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Comments

Greppur · Accepted Answer · 2015-10-21 13:33:06Z

4

Another way (perhaps a more concise one) to demonstrate this:

class A():
   a1 = []
x = A()
y = A()
x.a1.append("test")
x.a1, y.a1
(['test'], ['test'])

class B():
   b1 = None
   def __init__(self):
      self.b1 = list()
r = B()
s = B()
r.b1.append("test")
r.b1, s.b1
(["test"], [])

edited Oct 21, 2015 at 13:33

answered Oct 21, 2015 at 13:26

Greppur

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