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.deriv         


        

Answer 1

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"], [])

Answer 2

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.

Answer 3

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__.