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Getting private attribute in parent class using super(), outside of a method

I have a class with a private constant _BAR = object().
In a child class, outside of a method (no access to self), I want to refer to _BAR.
Here is a contrived example:
class Foo:
_BAR = object()
def __init__(self, bar: object = _BAR):
...
class DFoo(Foo):
"""Child class where I want to access private class variable from parent."""
def __init__(self, baz: object = super()._BAR):
super().__init__(baz)
Unfortunately, this doesn't work. One gets an error: RuntimeError: super(): no arguments
Is there a way to use super outside of a method to get a parent class attribute?
The workaround is to use Foo._BAR, I am wondering though if one can use super to solve this problem.

Inside of DFoo, you cannot refer to Foo._BAR without referring to Foo. Python variables are searched in the local, enclosing, global and built-in scopes (and in this order, it is the so called LEGB rule) and _BAR is not present in any of them.
Let's ignore an explicit Foo._BAR.
Further, it gets inherited: DFoo._BAR will be looked up first in DFoo, and when not found, in Foo.
What other means are there to get the Foo reference? Foo is a base class of DFoo. Can we use this relationship? Yes and no. Yes at execution time and no at definition time.
The problem is when the DFoo is being defined, it does not exist yet. We have no start point to start following the inheritance chain. This rules out an indirect reference (DFoo -> Foo) in a def method(self, ....): line and in a class attribute _DBAR = _BAR.
It is possible to work around this limitation using a class decorator. Define the class and then modify it:
def deco(cls):
cls._BAR = cls.__mro__[1]._BAR * 2 # __mro__[0] is the class itself
return cls
class Foo:
_BAR = 10
#deco
class DFoo(Foo):
pass
print(Foo._BAR, DFoo._BAR) # 10 20
Similar effect can be achieved with a metaclass.
The last option to get a reference to Foo is at execution time. We have the object self, its type is DFoo, and its parent type is Foo and there exists the _BAR. The well known super() is a shortcut to get the parent.
I have assumed only one base class for simplicity. If there were several base classes, super() returns only one of them. The example class decorator does the same. To understand how several bases are sorted to a sequence, see how the MRO works (Method Resolution Order).
My final thought is that I could not think up a use-case where such access as in the question would be required.

Short answer: you can't !
I'm not going into much details about super class itself here. (I've written a pure Python implementation in this gist if you like to read.)
But now let's see how we can call super:
1- Without arguments:
From PEP 3135:
This PEP proposes syntactic sugar for use of the super type to
automatically construct instances of the super type binding to the
class that a method was defined in, and the instance (or class object
for classmethods) that the method is currently acting upon.
The new syntax:
super()
is equivalent to:
super(__class__, <firstarg>)
...and <firstarg> is the first parameter of the method
So this is not an option because you don't have access to the "instance".
(Body of the function/methods is not executed unless it gets called, so no problem if DFoo doesn't exist yet inside the method definition)
2- super(type, instance)
From documentation:
The zero argument form only works inside a class definition, as the
compiler fills in the necessary details to correctly retrieve the
class being defined, as well as accessing the current instance for
ordinary methods.
What were those necessary details mentioned above? A "type" and A "instance":
We can't pass neither "instance" nor "type" which is DFoo here. The first one is because it's not inside the method so we don't have access to instance(self). Second one is DFoo itself. By the time the body of the DFoo class is being executed there is no reference to DFoo, it doesn't exist yet. The body of the class is executed inside a namespace which is a dictionary. After that a new instance of type type which is here named DFoo is created using that populated dictionary and added to the global namespaces. That's what class keyword roughly does in its simple form.
3- super(type, type):
If the second argument is a type, issubclass(type2, type) must be
true
Same reason mentioned in above about accessing the DFoo.
4- super(type):
If the second argument is omitted, the super object returned is
unbound.
If you have an unbound super object you can't do lookup(unless for the super object's attributes itself). Remember super() object is a descriptor. You can turn an unbound object to a bound object by calling __get__ and passing the instance:
class A:
a = 1
class B(A):
pass
class C(B):
sup = super(B)
try:
sup.a
except AttributeError as e:
print(e) # 'super' object has no attribute 'a'
obj = C()
print(obj.sup.a) # 1
obj.sup automatically calls the __get__.
And again same reason about accessing DFoo type mentioned above, nothing changed. Just added for records. These are the ways how we can call super.

Why must instance variables be defined inside of methods?

Why must instance variables be defined inside of methods? In other words why must self only be used to define new variables inside of methods in a class. Why can't you define variables using self as part of the class, but outside of methods.
"Instance variables are those variables for which each class object has it's own copy of it" - this definition doesn't say anything about methods. So, given that the definition doesn't mention methods why can't I define an instance variable (in other words use self to define a new variable) inside of a class, but outside of a method?

Python requires the object reference (implicit or explicit this in Java, for example) to be explicit. Inside methods -- bound functions -- the first param in the function definition is the instance. (This is conventionally called self but you can use any name.)
If you define
class C:
x = 1
there is no self reference, unlike, e.g. Java, where this is implicit.

Because the mechanism which Python uses to deal with OOP are very simple. There's no special syntax to define classes really, the class keyword is a very thin layer over what amounts to creating a dict. Everything you define inside a class Foo: block basically ends up as the contents of Foo.__dict__. So there's no syntax to define attributes of the instance resulting from calling Foo(). You add instance attributes simply by attaching them to the object you get from calling Foo(), which is self in __init__ or other instance methods.

For that to answer you need to know a little bit how the Python interpreter works.
In general every class and method definition are separate objects.
What you do when calling a method is that you pass the class instance as first parameter to the method. With that the method knows on what instance it is running on (and therefore where to allocate instance variables to).
This however only counts for instance methods.
Of course you can also create classmethods with #classmethod these take the class type as argument instead of an instance and can therefore not be used to create variables on the self context.

Why must instance variables be defined inside of methods?
They don't. You can define them from anywhere, as long as you have an instance (of a mutable type):
class Foo(object):
pass
f = Foo()
f.bar = 42
print(f.bar)
In other words why must self only be used to define new variables inside of methods in a class. Why can't you define variables using self as part of the class, but outside of methods.
self (which is only a naming convention, there's absolutely nothing magical here) is used to represent the current instance. How could you use it at the class block's top-level where you don't have any instance at all (and not even the class itself FWIW) ?
Defining the class "members" at the class top-level is mostly a static languages thing, where "objects" are mainly (technically) structs (C style structs, or Pascal style records if you prefer) with a statically defined memory structure.
Python is a dynamic language, which instead uses dicts as supporting data structure, so someobj.attribute is usually (minus computed attributes etc) resolved as someobj.__dict__["attribute"] (and someobj.attribute = value as someobj.__dict__["attribute"] = value).
So 1/ it doesn't NEED to have a fixed, explicitely defined data structure, and 2/ yet it DOES need to have an instance at end to set an attribute on it.
Note that you can force a class to use a fixed memory structure (instead of a plain dict) using slots, but you will still need to set the values from within a method (canonically the __init__, which exists for this very reason: initializing the instance's attributes).

Object that has neither attributes nor methods in python

'Every thing in python is an object'
So, should all objects have to have attributes and methods ?
I read below statemts in tutorial site, could you give example of pre-defined object in python that has neither attributes nor methods ?
Some objects have neither attributes nor methods

Everything in Python is indeed an object, this is true. Even classes themselves are considered to be objects and they're indeed the product of the builtin class typewhich is not surprisingly an object too. But objects will almost certainly inherit attributes including methods or data in most circumstances.
So, should all objects have to have attributes and methods ?
Not necessarily all objects have their own attributes. For instance, an object can inherit the attribute from its class and its class's superclasses, but that attribute or method doesn't necessarily live within the instance's namespace dictionary. In fact, instances' namespaces could be empty like the following:
class Foo:
pass
a = A()
print(a.__dict__)
a here doesn't have any attributes aside from those inherited from its class so if you check its namespace through the builtin attribute __dict__ you'll find the namespace to be an empty dictionary. But you might wonder isn't a.__dict__ an attribute of a? Make a distinction between class-level attributes--attributes inherited from the class or its superclasses and instance-attributes--attributes that belong to the instance and usually live in its namespace __dict__.
Could you give example of pre-defined object in python that has neither attributes nor methods ?
If you meant by predefined object, a builtin object, I couldn't imagine such scenario. Again, even if there are no attributes at the object itself, there would be attributes inherited from its class or the class's superclasses if there's any superclass in most cases. Probably and I'm guessing here, the tutorial is asking you to create class that assigns no attributes to its objects, just like the code I included above.
And this already answers your question better: Is everything an object in python like ruby?

There's a hackish way to emulate a Python object with no attributes.
class NoAttr(object):
def __getattribute__(self, attr):
raise AttributeError("no attribute: %s" % attr)
def __setattr__(self, attr, value):
raise AttributeError("can't set attribute: %s" % attr)
def __delattr__(self, attr):
raise AttributeError("no attribute: %s" % attr)
a = NoAttr()
This instance a, for all intents and purposes, in pure Python, behaves like an object with no attributes (you can try hasattr on it).
There may be a low-level way to do this in a C extension by implementing a type in C that pathologically stops Python's object implementation from working. Anyway the margin here is too small for writing one.
A pre-defined object with no attributes would defeat the purpose of pre-defining it.

python metaclasses at module level

I read What is a metaclass in Python?
and I tried to replicate the upper metaclass from the example and found that this doesn't work in all cases:
def upper(cls_name, cls_parents, cls_attr):
""" Make all class attributes uppper case """
attrs = ((name, value) for name, value in cls_attr.items()
if not name.startswith('__'))
upper_atts = dict((name.upper(), value) for name, value in attrs)
return type(cls_name, cls_parents, upper_atts)
__metaclass__ = upper #Module level
class Foo:
bar = 1
f = Foo()
print(f.BAR) #works in python2.6
The above fails (with an attribute error) in python3 which I think is natural because all classes in python3 already have object as their parent and metaclass resolution goes into the object class.
The question:
How do I make a module level metaclass in python3?

The module level metaclass isn't really "module level", it has to do with how class initialization worked. The class creation would look for the variable "__metaclass__" when creating the class, and if it wasn't in the local environment it would look in the global. Hence, if you had a "module level" __metaclass__ that would be used for every class afterwards, unless they had explicit metaclasses.
In Python 3, you instead specify the metaclass with a metaclass= in the class definition. Hence there is no module level metaclasses.
So what do you do? Easy: You specify it explicitly for each class.
It's really not much extra work, and you can even do it with a nice regexp search and replace if you really have hundreds of classes and don't want to do it manually.

If you want to change all the attributes to upper case, you should probably use the __init__ method to do so, than use a metaclass.
Metaclasses are deeper magic than 99% of users should ever worry about. If you wonder whether you need them, you don't (the people who actually need them know with certainty that they need them, and don't need an explanation about why).
-- Python Guru Tim Peters
If you need something deeper, you should also evaluate using Class Decorators.
Using MetaClasses and understanding how the classes are created is so unnecessary as long as you want to do something that you can do using class decorators or initialization.
That said, if you really want to use a Metaclass tho' pass that as a keyword argument to the class.
class Foo(object, metaclass=UpperCaseMetaClass)
where UpperCaseMetaClass is a class that extends type and not a method.
class UpperCaseMetaClass(type):
def __new__():
#Do your Magic here.

Intercepting changes of attributes in classes within a class - Python

I have been messing around with pygame and python and I want to be able to call a function when an attribute of my class has changed. My current solution being:
class ExampleClass(parentClass):
def __init__(self):
self.rect = pygame.rect.Rect(0,0,100,100)
def __setattr__(self, name, value):
parentClass.__setattr__(self,name,value)
dofancystuff()
Firstclass = ExampleClass()
This works fine and dofancystuff is called when I change the rect value with Firsclass.rect = pygame.rect.Rect(0,0,100,100). However if I say Firstclass.rect.bottom = 3. __setattr__ and there for dofancystuff is not called.
So my question I guess is how can I intercept any change to an attribute of a subclass?
edit: Also If I am going about this the wrong way please do tell I'm not very knowledgable when it comes to python.

Well, the simple answer is you can't. In the case of Firstclass.rect = <...> your __setattr__ is called. But in the case of Firstclass.rect.bottom = 3 the __setattr__ method of the Rect instance is called. The only solution I see, is to create a derived class of pygame.rect.Rect where you overwrite the __setattr__ method. You can also monkey patch the Rect class, but this is discouraged for good reasons.

You could try __getattr__, which should be called on Firstclass.rect.
But do this instead: Create a separate class (subclass of pygame.rect?) for ExampleClass.rect. Implement __setattr__ in that class. Now you will be told about anything that gets set in your rect member for ExampleClass. You can still implement __setattr__ in ExampleClass (and should), only now make sure you instantiate a version of your own rect class...
BTW: Don't call your objects Firstclass, as then it looks like a class as opposed to an object...

This isn't answering the question but it is relevant:
self.__dict__[name] = value
is probably better than
parentClass.__setattr__(self, name, value)
This is recommended by the Python documentation (setattr">http://docs.python.org/2/reference/datamodel.html?highlight=setattr#object.setattr) and makes more sense anyway in the general case since it does not assume anything about the behaviour of parentClass setattr.
Yay for unsolicited advice four years too late!

I think the reason why you have this difficulty deserves a little more information than is provided by the other answers.
The problem is, when you do:
myObject.attribute.thing = value
You're not assigning a value to attribute. The code is equivalent to this:
anAttribute = myObject.attribute
anAttribute.thing = value
As it's seen by myObject, all you're doing it getting the attribute; you're not setting the attribute.
Making subclasses of your attributes that you control, and can define __setattr__ for is one solution.
An alternative solution, that may make sense if you have lots of attributes of different types and don't want to make lots of individual subclasses for all of them, is to override __getattribute__ or __getattr__ to return a facade to the attribute that performs the relevant operations in its __setattr__ method. I've not attempted to do this myself, but I imagine that you should be able to make a simple facade class that will act as a facade for any object.
Care would need to be taken in the choice of __getattribute__ and __getattr__. See the documentation linked in the previous sentence for information, but basically if __getattr__ is used, the actual attributes will have top be encapsulated/obfuscated somehow so that __getattr__ handles requests for them, and if __getattribute__ is used, it'll have to retrieve attributes via calls to a base class.
If all you're trying to do is determine if some rects have been updated, then this is overkill.