obj.__dict__ expected behavior or bug?

[Sean Ross]

Hi.
It's expected behaviour.

Let's go through your code with a few additional print statements to see if
we can demonstrate what's happening;

class Config:
    a = 1
    b = 2
    c = 3
    d = None
    e = None
    h = {'d' : 22, 'e' : 33}

    def __init__(self, factor):
        for attr in self.h.keys():
            self.__dict__[attr] = self.h[attr] * factor

    def moda(self):
        self.a *= 5


c = Config(2)

# Here's what to pay attention to ...........
print "c.__dict__: ", c.__dict__                                        #
this is the instance dictionary
print "c.__class__.__dict__: ", c.__class__.__dict__  # this is the class
dictionary


print c.a, c.b, c.c, c.d, c.e
for attr in c.__dict__:
    print 'c.%s = %s' % (attr, c.__dict__[attr])
print

c.moda()

print "c.moda() --------------"
print "c.__dict__: ", c.__dict__
print "c.__class__.__dict__: ", c.__class__.__dict__

print c.a, c.b, c.c, c.d, c.e
for attr in c.__dict__:
    print 'c.%s = %s' % (attr, c.__dict__[attr])
print



Now, here's the output with #annotations:


c.__dict__:  {'e': 66, 'd': 44}
c.__class__.__dict__:  {'a': 1, 'moda': <function moda at 0x015209B0>,
'__module__': '__main__', 'b': 2, 'e': None, 'd': None, 'h': {'e': 33, 'd':
22}, 'c': 3, '__init__': <function __init__ at 0x01520B30>, '__doc__': None}
1 2 3 44 66
c.e = 66
c.d = 44

# Okay. We can see that the values for 'a', 'b', 'c' were all found
# in the class dictionary of instance c, while 'd', and 'e' were
# found in the instance dictionary of c. More on this later....


# Now we're about to call moda() ....
c.moda() --------------

# What's changed?
c.__dict__:  {'a': 5, 'e': 66, 'd': 44}
c.__class__.__dict__:  {'a': 1, 'moda': <function moda at 0x015209B0>,
'__module__': '__main__', 'b': 2, 'e': None, 'd': None, 'h': {'e': 33, 'd':
22}, 'c': 3, '__init__': <function __init__ at 0x01520B30>, '__doc__': None}
5 2 3 44 66
c.a = 5
c.e = 66
c.d = 44

# This time only 'b' and 'c''s values were pulled from instance c's class'
dictionary.
# What about 'a'? 'a' was pulled from c's instance dictionary. Nothing's
changed
# for 'd' and 'e'.


Okay then. What's going on?

class Config:
    a = 1
    b = 2
    c = 3
    d = None
    e = None
    h = {'d' : 22, 'e' : 33}

The code above adds class variables a - h to the class Config. So, if you
have an instance c of Config,
variables a-h are stored in c's class dictionary (c.__class__.__dict__) and
NOT c's instance dictionary
(c.__dict__). Moving on...

    def __init__(self, factor):
        for attr in self.h.keys():
            self.__dict__[attr] = self.h[attr] * factor

Inside the constructor, you call self.h.keys(). To find self.h, Python looks
first in self.__dict__. But 'h' isn't there.
Next it looks in self.__class__.__dict__. That's were 'h' is!   Now this:

            self.__dict__[attr] = self.h[attr] * factor

Here, you're assigning NEW attributes 'd' and 'e' to self's __dict__. What
you are not doing is assigning new values to class variables 'd' and 'e' in
self.__class__.__dict__ .


    def moda(self):
        self.a *= 5

Something similar is happening in here. This one is a bit more complicated.

self.a *= 5

is the same as

self.a = self.a * 5

What does this really mean? Well,

self.a =  ....

is equivalent to

self.__dict__['a']  =   ....

But

self.a = self.a ....

is not necessarily equivalent to

self.__dict__['a']  =   self.__dict__['a']


because the self.a on the right hand side of the assignment has to be looked
up by Python. And, as we showed earlier,
look up starts with self.__dict__. But 'a' is not yet a key in that
dictionary, so we move up to self.__class__.__dict__.
That's where 'a' is! It's value is '1', so we get

self.__dict__['a'] = 1*5
                                 ^
                                 self.__class__.__dict__['a']

We finish the evaluation, and assign 5 to self.__dict__['a'], creating a new
instance variable.
The class variable 'a' is unchanged. If you call c.moda() again later then,
that time, Python's lookup
would find 'a' in self.__dict__, and the expression self.a *= 5 would be
equivalent to

 self.__dict__['a']  = 5*5
                                  ^
                                  self.__dict__['a']

So, the thing is, yes the behaviour is expected, if you know what behaviour
to expect ...


Okay, then. Hopefully that was helpful.
Sean