I am reading python 3.7 documentation. And I am very confused about the following sentences:
"Derived classes may override methods of their base classes. Because methods have no special privileges when calling other methods of the same object, a method of a base class that calls another method defined in the same base class may end up calling a method of a derived class that overrides it. (For C++ programmers: all methods in Python are effectively virtual.)"
Can you show me an example code that "a base class that calls another method defined in the same base class may end up calling a method of a derived class that overrides it."?
And here is my understanding:
class A:
def me(self):
print("This is A")
def idet(self):
self.me()
class B(A):
def me(self):
print("this is B")
a = A()
b = B()
b.me()
b.idet()
the result is
this is B
this is B
I am not sure if it is the case.
And the last question is what does "all methods in Python are effectively virtual" mean? (I am familiar with Java but not C++)
The example shows this principle exactly. b, which is an instance of B, calls the method ident defined in class A, which in turn calls me. Since B overrides me, B's method is called, and you get this is me printed out.
In C++, methods can't be overridden by default - you have to declare them as virtual to get this behavior. In Python (and in Java, which you mentioned you are familiar with) this is is the default behavior. In Java you can modify a method so it can't be overrided, by defining it as final.
When you derive a class, all the methods of the super class are copied into the base class. When you redefine a method that already exists in the super class while defining methods in the base class (or super class), it is called as overriding. When you override a method from the base class (or super class) in the sub class, it will kinda cut the connections between the method in the base class and the sub class.
In your program, you're calling b.me() first, which is an overrided method; So, it will execute the me() from the class B. Then, you've got b.idet() which is a copied method from the base class A; So, its code will not change. But, when you look closely in the body of the idet() method, what it does is that it will call the me() method of the class which it is being called from. In this case, as the class calling that method is the class B, it will execute the me() method from the class B.
idet() has got self.me(); self keyword references to the attributes/methods class within which it is written.
I think the Python3 documentation could have also mentioned that when an instance of class B, 'b', refers to a method that is pulled from class A, it passes itself into that method as the first argument (self). Therefore any reference to "self" inside any method called by 'b' (including its inherited methods) will first search class B functions before class A functions, even if the method called by 'b' was derived from class A.
b.idet() is equivalent to A.idet(b) which calls b.me()
"a method of a base class that calls another method defined in the
same base class may end up calling a method of a derived class that overrides it"
I can see how this seems misleading, and I think its because the method idet(self) of base class A only calls another method defined in the same base class (A) if "self" refers to an instance of A, but in the above scenario when "self" refers to an instance of subclass B, idet(self) doesn't really call another method defined in A, it calls a method defined in B that overrides A.
Related
I am trying to gain a better understanding of class variables and the #classmethod decorator in python. I've done a lot of googling but I am having difficulty grasping basic OOP concepts. Take the following class:
class Repository:
repositories = []
repository_count = 0
def __init__(self):
self.update_repositories()
Repository.repository_count += 1
#classmethod
def update_repositories(cls):
if not cls.repositories:
print('appending repository')
cls.repositories.append('twenty')
else:
print('list is full')
a = Repository()
b = Repository()
print(Repository.repository_count)
Output:
appending repository
list is full
2
In the __init__ method, why does self.update_repositories() successfully call the update_repositories class method? I thought that self in this case refers to the instantiated object, not the class?
The code works without using the #classmethod decorator. Why?
In the __init__ method why do I need to use the keyword Repository in Repository.repository_count += 1? Am I doing this correctly or is there a better practice?
Class methods can be called from an instance. Look at the documentation here.
A class method can be called either on the class (such as C.f()) or on an instance (such as C().f()). The instance is ignored except for its class. If a class method is called for a derived class, the derived class object is passed as the implied first argument.
The function works without the decorator, but it is not a class method. The cls and self parameter names are simply convention. You can put anything in the place of cls or self. For example:
class Demo:
def __init__(self):
pass
def instance_method(test):
print(test)
#classmethod
def class_method(test):
print(test)
demo = Demo()
This results in:
demo.instance_method()
>>> <__main__.Demo object at 0x7facd8e34510>
demo.class_method()
>>> <class '__main__.Demo'>
So all non decorated methods in a class are a considered instance
methods and all methods decorated with #classmethod are
class methods. Naming your parameters cls, self or
anything else for that matter does not effect the functionality, but I
would strongly advice sticking with convention.
In your case specifcally removing the #classmethod decorator turns the method into an instance method and cls is now actually what self would normally be, a reference to the class's instance. Since class methods and attributes can be called from an instance cls.update_repositories still points to the class variable.
Depends on what you are trying to do. Generally if you want to access a class variable or method inside a class, but outside a class method, your approach is correct.
In Python, I can create a class method using the #classmethod decorator:
>>> class C:
... #classmethod
... def f(cls):
... print(f'f called with cls={cls}')
...
>>> C.f()
f called with cls=<class '__main__.C'>
Alternatively, I can use a normal (instance) method on a metaclass:
>>> class M(type):
... def f(cls):
... print(f'f called with cls={cls}')
...
>>> class C(metaclass=M):
... pass
...
>>> C.f()
f called with cls=<class '__main__.C'>
As shown by the output of C.f(), these two approaches provide similar functionality.
What are the differences between using #classmethod and using a normal method on a metaclass?
As classes are instances of a metaclass, it is not unexpected that an "instance method" on the metaclass will behave like a classmethod.
However, yes, there are differences - and some of them are more than semantic:
The most important difference is that a method in the metaclass is not "visible" from a class instance. That happens because the attribute lookup in Python (in a simplified way - descriptors may take precedence) search for an attribute in the instance - if it is not present in the instance, Python then looks in that instance's class, and then the search continues on the superclasses of the class, but not on the classes of the class. The Python stdlib make use of this feature in the abc.ABCMeta.register method.
That feature can be used for good, as methods related with the class themselves are free to be re-used as instance attributes without any conflict (but a method would still conflict).
Another difference, though obvious, is that a method declared in the metaclass can be available in several classes, not otherwise related - if you have different class hierarchies, not related at all in what they deal with, but want some common functionality for all classes, you'd have to come up with a mixin class, that would have to be included as base in both hierarchies (say for including all classes in an application registry). (NB. the mixin may sometimes be a better call than a metaclass)
A classmethod is a specialized "classmethod" object, while a method in the metaclass is an ordinary function.
So, it happens that the mechanism that classmethods use is the "descriptor protocol". While normal functions feature a __get__ method that will insert the self argument when they are retrieved from an instance, and leave that argument empty when retrieved from a class, a classmethod object have a different __get__, that will insert the class itself (the "owner") as the first parameter in both situations.
This makes no practical differences most of the time, but if you want access to the method as a function, for purposes of adding dynamically adding decorator to it, or any other, for a method in the metaclass meta.method retrieves the function, ready to be used, while you have to use cls.my_classmethod.__func__ to retrieve it from a classmethod (and then you have to create another classmethod object and assign it back, if you do some wrapping).
Basically, these are the 2 examples:
class M1(type):
def clsmethod1(cls):
pass
class CLS1(metaclass=M1):
pass
def runtime_wrap(cls, method_name, wrapper):
mcls = type(cls)
setattr(mcls, method_name, wrapper(getatttr(mcls, method_name)))
def wrapper(classmethod):
def new_method(cls):
print("wrapper called")
return classmethod(cls)
return new_method
runtime_wrap(cls1, "clsmethod1", wrapper)
class CLS2:
#classmethod
def classmethod2(cls):
pass
def runtime_wrap2(cls, method_name, wrapper):
setattr(cls, method_name, classmethod(
wrapper(getatttr(cls, method_name).__func__)
)
)
runtime_wrap2(cls1, "clsmethod1", wrapper)
In other words: apart from the important difference that a method defined in the metaclass is visible from the instance and a classmethod object do not, the other differences, at runtime will seem obscure and meaningless - but that happens because the language does not need to go out of its way with special rules for classmethods: Both ways of declaring a classmethod are possible, as a consequence from the language design - one, for the fact that a class is itself an object, and another, as a possibility among many, of the use of the descriptor protocol which allows one to specialize attribute access in an instance and in a class:
The classmethod builtin is defined in native code, but it could just be coded in pure python and would work in the exact same way. The 5 line class bellow can be used as a classmethod decorator with no runtime differences to the built-in #classmethod" at all (though distinguishable through introspection such as calls toisinstance, and evenrepr` of course):
class myclassmethod:
def __init__(self, func):
self.__func__ = func
def __get__(self, instance, owner):
return lambda *args, **kw: self.__func__(owner, *args, **kw)
And, beyond methods, it is interesting to keep in mind that specialized attributes such as a #property on the metaclass will work as specialized class attributes, just the same, with no surprising behavior at all.
When you phrase it like you did in the question, the #classmethod and metaclasses may look similar but they have rather different purposes. The class that is injected in the #classmethod's argument is usually used for constructing an instance (i.e. an alternative constructor). On the other hand, the metaclasses are usually used to modify the class itself (e.g. like what Django does with its models DSL).
That is not to say that you can't modify the class inside a classmethod. But then the question becomes why didn't you define the class in the way you want to modify it in the first place? If not, it might suggest a refactor to use multiple classes.
Let's expand the first example a bit.
class C:
#classmethod
def f(cls):
print(f'f called with cls={cls}')
Borrowing from the Python docs, the above will expand to something like the following:
class ClassMethod(object):
"Emulate PyClassMethod_Type() in Objects/funcobject.c"
def __init__(self, f):
self.f = f
def __get__(self, obj, klass=None):
if klass is None:
klass = type(obj)
def newfunc(*args):
return self.f(klass, *args)
return newfunc
class C:
def f(cls):
print(f'f called with cls={cls}')
f = ClassMethod(f)
Note how __get__ can take either an instance or the class (or both), and thus you can do both C.f and C().f. This is unlike the metaclass example you give which will throw an AttributeError for C().f.
Moreover, in the metaclass example, f does not exist in C.__dict__. When looking up the attribute f with C.f, the interpreter looks at C.__dict__ and then after failing to find, looks at type(C).__dict__ (which is M.__dict__). This may matter if you want the flexibility to override f in C, although I doubt this will ever be of practical use.
In your example, the difference would be in some other classes that will have M set as their metaclass.
class M(type):
def f(cls):
pass
class C(metaclass=M):
pass
class C2(metaclass=M):
pass
C.f()
C2.f()
class M(type):
pass
class C(metaclass=M):
#classmethod
def f(cls):
pass
class C2(metaclass=M):
pass
C.f()
# C2 does not have 'f'
Here is more on metaclasses
What are some (concrete) use-cases for metaclasses?
Both #classmethod and Metaclass are different.
Everything in python is an object. Every thing means every thing.
What is Metaclass ?
As said every thing is an object. Classes are also objects in fact classes are instances of other mysterious objects formally called as meta-classes. Default metaclass in python is "type" if not specified
By default all classes defined are instances of type.
Classes are instances of Meta-Classes
Few important points are to understand metioned behaviour
As classes are instances of meta classes.
Like every instantiated object, like objects(instances) get their attributes from class. Class will get it's attributes from Meta-Class
Consider Following Code
class Meta(type):
def foo(self):
print(f'foo is called self={self}')
print('{} is instance of {}: {}'.format(self, Meta, isinstance(self, Meta)))
class C(metaclass=Meta):
pass
C.foo()
Where,
class C is instance of class Meta
"class C" is class object which is instance of "class Meta"
Like any other object(instance) "class C" has access it's attributes/methods defined in it's class "class Meta"
So, decoding "C.foo()" . "C" is instance of "Meta" and "foo" is method calling through instance of "Meta" which is "C".
First argument of method "foo" is reference to instance not class unlike "classmethod"
We can verify as if "class C" is instance of "Class Meta
isinstance(C, Meta)
What is classmethod?
Python methods are said to be bound. As python imposes the restriction that method has to be invoked with instance only.
Sometimes we might want to invoke methods directly through class without any instance (much like static members in java) with out having to create any instance.By default instance is required to call method. As a workaround python provides built-in function classmethod to bind given method to class instead of instance.
As class methods are bound to class. It takes at least one argument which is reference to class itself instead of instance (self)
if built-in function/decorator classmethod is used. First argument
will be reference to class instead of instance
class ClassMethodDemo:
#classmethod
def foo(cls):
print(f'cls is ClassMethodDemo: {cls is ClassMethodDemo}')
As we have used "classmethod" we call method "foo" without creating any instance as follows
ClassMethodDemo.foo()
Above method call will return True. Since first argument cls is indeed reference to "ClassMethodDemo"
Summary:
Classmethod's receive first argument which is "a reference to class(traditionally referred as cls) itself"
Methods of meta-classes are not classmethods. Methods of Meta-classes receive first argument which is "a reference to instance(traditionally referred as self) not class"
What is the purpose of subclassing an instance of a class in Python? Here's an example:
class A:
def __init__(*args): print(args)
base = A()
class Test(base): pass
This code works properly under Python, but base is an instance of class A (1) Why do we need to subclass an instance of a class? Is it related to metaclasses?
From this question:
What happens when you inherent from a module instead of a class in Python?
I understand that Test(base) will become type(base).__init__, (2) does this happen at definition time, when the class is defined? (3) How does Python know/decide that base is an instance of a class? Is it becuase type(base) doesn't return type?
Python actually uses type(base)(classname, bases, body) to produce the class object. This is the normal metaclass invocation (unless the class specifies a specific metaclass directly).
For an instance (or a module, which is basically an instance too), that means the class __new__ static method is called to produce a new instance, and on that new instance __init__ is called. So for your class Test(base): pass syntax, you are essentially creating a new instance of A, when Python executes the class statement.
This isn't really 'definition time'; there is no such phase in Python code. Python loads the bytecode for a module and executes it when it is imported for the first time. Class statements at the top of the module (so not inside functions) are executed at that time. It is at that time then that the type(base)(...) call is executed, yes.
Python doesn't 'know' or 'decide' anything about the base class. Python's philosophy is to trust the developer that writes code in it. The assumption is that you know what you are doing, and the base classes are just treated as if they'll respond correctly. Since type(base)(....) didn't raise an exception, Python just continued.
You can use anything that's callable as a metaclass, really.
This question is in relation to posts at What does 'super' do in Python? , How do I initialize the base (super) class? , and Python: How do I make a subclass from a superclass? which describes two ways to initialize a SuperClass from within a SubClass as
class SuperClass:
def __init__(self):
return
def superMethod(self):
return
## One version of Initiation
class SubClass(SuperClass):
def __init__(self):
SuperClass.__init__(self)
def subMethod(self):
return
or
class SuperClass:
def __init__(self):
return
def superMethod(self):
return
## Another version of Initiation
class SubClass(SuperClass):
def __init__(self):
super(SubClass, self).__init__()
def subMethod(self):
return
So I'm a little confused about needing to explicitly pass self as a parameter in
SuperClass.__init__(self)
and
super(SubClass, self).__init__().
(In fact if I call SuperClass.__init__() I get the error
TypeError: __init__() missing 1 required positional argument: 'self'
). But when calling constructors or any other class method (ie :
## Calling class constructor / initiation
c = SuperClass()
k = SubClass()
## Calling class methods
c.superMethod()
k.superMethod()
k.subMethod()
), The self parameter is passed implicitly .
My understanding of the self keyword is it is not unlike the this pointer in C++, whereas it provides a reference to the class instance. Is this correct?
If there would always be a current instance (in this case SubClass), then why does self need to be explicitly included in the call to SuperClass.__init__(self)?
Thanks
This is simply method binding, and has very little to do with super. When you can x.method(*args), Python checks the type of x for a method named method. If it finds one, it "binds" the function to x, so that when you call it, x will be passed as the first parameter, before the rest of the arguments.
When you call a (normal) method via its class, no such binding occurs. If the method expects its first argument to be an instance (e.g. self), you need to pass it in yourself.
The actual implementation of this binding behavior is pretty neat. Python objects are "descriptors" if they have a __get__ method (and/or __set__ or __delete__ methods, but those don't matter for methods). When you look up an attribute like a.b, Python checks the class of a to see if it has a attribute b that is a descriptor. If it does, it translates a.b into type(a).b.__get__(a, type(a)). If b is a function, it will have a __get__ method that implements the binding behavior I described above. Other kinds of descriptors can have different behaviors. For instance, the classmethod decorator replaces a method with a special descriptor that binds the function the class, rather than the instance.
Python's super creates special objects that handle attribute lookups differently than normal objects, but the details don't matter too much for this issue. The binding behavior of methods called through super is just like what I described in the first paragraph, so self gets passed automatically to the bound method when it is called. The only thing special about super is that it may bind a different function than you'd get lookup up the same method name on self (that's the whole point of using it).
The following example might elucidate things:
class Example:
def method(self):
pass
>>> print(Example.method)
<unbound method Example.method>
>>> print(Example().method)
<bound method Example.method of <__main__.Example instance at 0x01EDCDF0>>
When a method is bound, the instance is passed implicitly. When a method is unbound, the instance needs to be passed explicitly.
The other answers will definitely offer some more detail on the binding process, but I think it's worth showing the above snippet.
The answer is non-trivial and would probably warrant a good article. A very good explanation of how super() works is brilliantly given by Raymond Hettinger in a Pycon 2015 talk, available here and a related article.
I will attempt a short answer and if it is not sufficient I (and hopefully the community) will expand on it.
The answer has two key pieces:
Python's super() needs to have an object on which the method being overridden is called, so it is explicitly passed with self. This is not the only possible implementation and in fact, in Python 3, it is no longer required that you pass the self instance.
Python super() is not like Java, or other compiled languages, super. Python's implementation is designed to support the multiple collaborative inheritance paradigm, as explained in Hettinger's talk.
This has an interesting consequence in Python: the method resolution in super() depends not only on the parent class, but on the children classes as well (consequence of multiple inheritance). Note that Hettinger is using Python 3.
The official Python 2.7 documentation on super is also a good source of information (better understood after watching the talk, in my opinion).
Because in SuperClass.__init__(self), you're calling the method on the class, not the instance, so it cannot be passed implicitly. Similarly you cannot just call SubClass.subMethod(), but you can call SubClass.subMethod(k) and it'll be equivalent to k.subMethod(). Similarly if self refers to a SubClass then self.__init__() means SubClass.__init__(self), so if you want to call SuperClass.__init you have to call it directly.
This question already has an answer here:
Why not constructor of super class invoked when we declare the object of sub class?
(1 answer)
Closed 9 years ago.
class A:
def __init__(self):
print 'A'
class B(A):
def __init__(self):
print 'B'
b = B()
B
In C++, I would have expected to see A B output, but in Python I am getting only B. I know that I can do super(B, self).__init__() to achieve the same in Python, but as this is apparently not the default (or is it - I am new to the syntax as well), I am worried that the paradigms for instatinating objects are completely different.
So what are objects in Python, what is their relation with classes and what is the standard way to initialize all data in all parent classes in Python?
Python rarely does anything automatically. As you say, if you want to invoke the superclass __init__, then you need to do it yourself, usually by calling super:
class B(A):
def __init__(self):
print 'B'
super(B, self).__init__()
The point to note is that instance attributes, like everything else in Python, are dynamic. __init__ is not the constructor, that's __new__ which you rarely need to meddle with. The object is fully constructed by the time __init__ is called, but since instance attributes are dynamic they are usually added by that method, which is only special in that it's called first once the object is created.
You can of course create instance attributes in any other method, or even from outside the class itself by simply doing something like myBobj.foo = 'bar'.
So what are objects in Python
Well, objects in python are like dictionaries of members and methods. It's no more sophisticated than that. You don't have visibility handling (if you want to hide a member, just do not talk about it in the public documentation, only with a comment).
what is their relation with classes
A class defines the method/member skeleton that will instantiate that dict/object. So you got the constructor __init__(), which is only a handler used when you create that object.
what is the standard way to initialize all data in all parent classes in Python?
Either you do not redefine the constructor, and then all parent classes will have their constructor initiated (like the default C++ behavior) or you do redefine the constructor, and then have to make an explicit call to your parent class's constructor.
Remember the zen of python: "Explicit is better than implicit". It totally applies here.
You need to invoke the base constructor in your inherited class constructor:
class B(A):
def __init__(self):
A.__init__(self)
# super(B, self).__init__() you can use this line as well
print 'B'