Just wanted some help understanding these lines of code:
class Parent:
def __init__(self):
print("instance created")
parent1=Parent()
parent2=Parent.__init__(parent1)
output
instance created
instance created
I am trying to understand how a constructor is called in OOP for python.
In the first line the the method __init__ is called by default and the self argument that is passed is somehow parent1?
The second line is the more traditional way I would've thought methods would be called. Since __init__ takes an instance of the parent class as an argument I passed parent1 and it works. I get what is happening in the second line, just wanted to ask what the computer is doing to create the instance parent1 in the first line.
__init__ is not a constructor, it's an initializer. When Python creates an object, it's actually created in __new__ (usually left as the default, which just makes an empty object of the right class), which receives a reference to the class, and returns an instance (typically empty; no attributes set). The resulting instance is passed implicitly as the self in __init__, which then establishes the instance attributes.
Typically, you don't call special methods like __init__ directly (aside from cases involving super() with cooperative inheritance), you just let Python do it for you. The only way to avoid calling __init__ would be to explicitly invoke the class's __new__ (which is also extremely unusual).
__init__ is the equivalent to a constructor in Python. Think of an object oriented language as one that has a mandatory argument for functions that represent object methods, so you always have access to that object in that function. Most languages don't make you type out the way you pass in this. Python uses self, and makes you type it out for every method. It's the same thing, it's just not doing extra work for you.
So when Python instantiates a class, it passes the class to the class's __new__ function, generates an object, and then passes that object to the class's __init__ function as the first argument.
You are correct that __init__() work like a constructor, is automatically runs when an object is instantiated (as would happen with Java constructor, if that helps). Although you can call __init__, you shouldn't call functions/methods starting with _ or __, they are meant to be called from with the class/object.
When self appears as a parameter in a class method you won't have to supply the object's name, Python will figure it out. So the second line above (Parent2 = ...) is not recommended.
See the documentation:
object.__init__(self[, ...])
Called after the instance has been created (by __new__()), but before it is returned to the caller. The arguments are those passed to the class constructor expression.
object.__new__(cls[, ...])
Called to create a new instance of class cls. __new__() is a static method (special-cased so you need not declare it as such) that takes the class of which an instance was requested as its first argument. The remaining arguments are those passed to the object constructor expression
So under the hood in parent1 = Parent(), Python's basically doing this:
_temp_new_parent = Parent.__new__(Parent) # Inherited from "object.__new__"
Parent.__init__(_temp_new_parent)
parent1 = _temp_new_parent
(_temp_new_parent doesn't really exist, I'm just using it as an abstraction.)
Note that __init__() doesn't return anything, so in your code, parent2 is None. And if __init__() had set instance attributes, it would have set them on parent1 since that's what you passed in.
Related
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.
does objects have its methods or invoke them from superclass in python?
for example:
class tst():
def __init__(self,name,family):
self.name=name
self.family=family
def fun(self,a,b):
print(a+b)
newtst=tst("myname","my family")
tst.fun(newtst,3,5)
newtst.fun(3,5)
in the code above does newtst object invoke fun function from tst class or it has own method and run it directly
and if the latter is true why we need to self parameter in definition class's functions
and in know that id(newtst.fun) is different from id(tst.fun) but i think that differences is because of creating new method's object in memory.
Instance methods are implemented as descriptors in python.
This implies, the separate instances are not keeping individual copies of method, but getting the same attribute of a different instances may produce individual results for each of them.
Despite we have the same function object beyond the attributes of all the instances we have, it still doesn't means that
id(tst.fun) == id(newtst.fun)
because method descriptor gives us different things when calling on a class and a class instance.
For a class we'll get an unbound method from tst, but for newtst it will be bound to an instance.
Can anyone simply explain the difference between constructors and methods in Python
When a class is instantiated, its __init__ method is called to initialize the class instance. Memory is allocated for the class instance, __init__ is called, and the new class is returned. __init__ is the constructor for the class. For example:
c = MyClass(123)
When __init__ is called, the first argument, self, is bound to the new class instance, and the second argument is 123.
An ordinary method operates on an existing class instance:
c.myMethod(456)
In this case, the first argument, self, is bound to c, which is an existing class instance, and the second argument is 456.
In most ways __init__ is like any other method of the class, except it is implicitly called when a new class instance is created.
This question already has answers here:
How does `super` interacts with a class's `__mro__` attribute in multiple inheritance?
(2 answers)
Closed 4 years ago.
From Python3's documentation super() "returns a proxy object that delegates method calls to a parent or sibling class of type." What does that mean?
Suppose I have the following code:
class SuperClass():
def __init__(self):
print("__init__ from SuperClass.")
print("self object id from SuperClass: " + str(id(self)))
class SubClass(SuperClass):
def __init__(self):
print("__init__ from SubClass.")
print("self object id from SubClass: " + str(id(self)))
super().__init__()
sc = SubClass()
The output I get from this is:
__init__ from SubClass.
self object id from SubClass: 140690611849200
__init__ from SuperClass.
self object id from SuperClass: 140690611849200
This means that in the line super().__init__(), super() is returning the current object which is then implicitly passed to the superclass' __init__() method. Is this accurate or am I missing something here?
To put it simply, I want to understand the following:
When super().__init__() is run,
What exactly is being passed to __init__() and how? We are calling it on super() so whatever this is returning should be getting passed to the __init__() method from what I understand about Python so far.
Why don't we have to pass in self to super().__init__()?
returns a proxy object that delegates method calls to a parent or
sibling class of type.
This proxy is an object that acts as the method-calling portion of the parent class. It is not the class itself; rather, it's just enough information so that you can use it to call the parent class methods.
If you call __init__(), you get your own, local, sub-class __init__ function. When you call super(), you get that proxy object, which will redirect you to the parent-class methods. Thus, when you call super().__init__(), that proxy redirects the call to the parent-class __init__ method.
Similarly, if you were to call super().foo, you would get the foo method from the parent class -- again, re-routed by that proxy.
Is that clear to you?
Responses to OP comments
But that must mean that this proxy object is being passed to
__init__() when running super().__init__() right?
Wrong. The proxy object is like a package name, such as calling math.sqrt(). You're not passing math to sqrt, you're using it to denote which sqrt you're using. If you wanted to pass the proxy to __init__, the call would be __init__(super()). That call would be semantically ridiculous, of course.
When we have to actually pass in self which is the sc object in my example.
No, you are not passing in sc; that is the result of the object creation call (internal method __new__), which includes an invocation of init. For __init__, the self object is a new item created for you by the Python run-time system. For most class methods, that first argument (called self out of convention, this in other languages) is the object that invoked the method.
This means that in the line super().__init__(), super() is returning the current object which is then implicitly passed to the superclass' __init__() method. Is this accurate or am I missing something here?
>>> help(super)
super() -> same as super(__class__, <first argument>)
super call returns a proxy/wrapper object which remembers:
The instance invoking super()
The class of the calling object
The class that's invoking super()
This is perfectly sound. super always fetches the attribute of the next class in the hierarchy ( really the MRO) that has the attribute that you're looking for. So it's not returning the current object, but rather and more accurately, it returns an object that remembers enough information to search for attributes higher in the class hierarchy.
What exactly is being passed to __init__() and how? We are calling it on super() so whatever this is returning should be getting passed to the __init__() method from what I understand about Python so far.
You're almost right. But super loves to play tricks on us. super class defines __getattribute__, this method is responsible for attribute search. When you do something like: super().y(), super.__getattribute__ gets called searching for y. Once it finds y it passes the instance that's invoking the super call to y. Also, super has __get__ method, which makes it a descriptor, I'll omit the details of descriptors here, refer to the documentation to know more. This answers your second question as well, as to why self isn't passed explicitly.
*Note: super is a little bit different and relies on some magic. Almost for all other classes, the behavior is the same. That is:
a = A() # A is a class
a.y() # same as A.y(a), self is a
But super is different:
class A:
def y(self):
return self
class B(A):
def y(self)
return super().y() # equivalent to: A.y(self)
b = B()
b.y() is b # True: returns b not super(), self is b not super()
I wrote a simple test to investigate what CPython does for super:
class A:
pass
class B(A):
def f(self):
return super()
#classmethod
def g(cls):
return super()
def h(selfish):
selfish = B()
return super()
class C(B):
pass
c = C()
for method in 'fgh':
super_object = getattr(c, method)()
print(super_object, super_object.__self__, super_object.__self_class__, super_object.__thisclass__) # (These methods were found using dir.)
The zero-argument super call returns an object that stores three things:
__self__ stores the object whose name matches the first parameter of the method—even if that name has been reassigned.
__self_class__ stores its type, or itself in the case of a class method.
__thisclass__ stores the class in which the method is defined.
(It is unfortunate that __thisclass__ was implemented this way rather than fetching an attribute on the method because it makes it impossible to use the zero-argument form of super with meta-programming.)
The object returned by super implements getattribute, which forwards method calls to the type found in the __mro__ of __self_class__ one step after __thisclass__.
As might be familiar to most of you, this is from Mark Pilgrim's book DIP, chapter 5
class FileInfo(UserDict):
"store file metadata"
def __init__(self, filename=None):
UserDict.__init__(self)
self["name"] = filename
Well I am new to python, coming from basic C background and having confusion understanding it. Stating what I understand, before what I don't understand.
Statement 0: FileInfo is inheriting from class UserDict
Statement 1: __init__ is not a constructor, however after the class instantiates, this is the first method that is defined.
Statement2: self is almost like this
Now the trouble:
as per St1 init is defined as the first function.
UserDict.__init__(self)
Now within the same function __init__ why is the function being referenced, there is no inherent recursion I guess. Or is it trying to override the __init__ method of the class UserDict which the class FileInfo has inherited and put an extra parameter(key value pair) of filename and reference it to the filename being passed to __init__ method.
I am partly sure, I have answered my question, however as you can sense there is confusion, would be great if someone can explain me how to rule this confusion out with some more advanced use case and detailed example of how generally code is written.
You're correct, the __init__ method is not a constructor, it's an initializer called after the object is instantiated.
In the code you've presented, the __init__ method on the FileInfo class is extending the functionality of the __init__ method of the base class, UserDict. By calling the base __init__ method, it executes any code in the base class's initialization, and then adds its own. Without a call to the base class's __init__ method, only the code explicitly added to FileInfo's __init__ method would be called.
The conventional way to do this is by using the super method.
class FileInfo(UserDict):
"store file metadata"
def __init__(self, filename=None):
super(UserDict, self).__init__()
self["name"] = filename
A common use case is returning extra values or adding additional functionality. In Django's class based views, the method get_context_data is used to get the data dictionary for rendering templates. So in an extended method, you'd get whatever values are returned from the base method, and then add your own.
class MyView(TemplateView):
def get_context_data(self, **kwargs):
context = super(MyClass, self).get_context_data(**kwargs)
context['new_key'] = self.some_custom_method()
return kwargs
This way you do not need to reimplement the functionality of the base method when you want to extend it.
Creating an object in Python is a two-step process:
__new__(self, ...) # constructor
__init__(self, ...) # initializer
__new__ has the responsibility of creating the object, and is used primarily when the object is supposed to be immutable.
__init__ is called after __new__, and does any further configuration needed. Since most objects in Python are mutable, __new__ is usually skipped.
self refers to the object in question. For example, if you have d = dict(); d.keys() then in the keys method self would refer to d, not to dict.
When a subclass has a method of the same name as its parent class, Python calls the subclass' method and ignores the parent's; so if the parent's method needs to be called, the subclass method must call it.
"Or is it trying to override the init method of the class UserDict which the class FileInfo has inherited and put an extra parameter(key value pair) of filename and reference it to the filename being passed to init method."
It's exactly that. UserDict.__init__(self) calls the superclass init method.
Since you come from C, maybe you're not well experienced with OOP, so you could read this article : http://en.wikipedia.org/wiki/Inheritance_(object-oriented_programming) to understand the inheritance principle better (and the "superclass" term I used).
.. the self variable represents the instance of the object itself. In python this is not a hidden parameter as in other languages. You have to declare it explicitly. When you create an instance of the FileInfo class and call its methods, it will be passed automatically,
The __init__ method is roughly what represents a constructor in Python.
The __init__ method of FileInfo is overriding the __init__ method of UserDict.
Then FileInfo.__init__ calls UserDict.__init__ on the newly created FileInfo instance (self). This way all properties and magic available to UserDict are now available to that FileInfo instance (ie. they are inherited from UserDict).
The last line is the reason for overriding UserDict.__init__ : UserDict does not create the wanted property self.filename.
When you call __init__ method for a class that is inheriting from a base class, you generally modify the ancestor class and as a part of customization, you extend the ancestor's init method with proper arguements.
__init__ is not a constructor, however after the class instantiates, this is the first method that is defined.
This method is called when an instance is being initialized, after __new__ (i.e. when you call ClassName()). I'm not sure what difference there is as opposed to a constructor.
Statement2: self is almost like this
Yes but it is not a language construct. The name self is just convention. The first parameter passed to an instance method is always a reference to the class instance itself, so writing self there is just to name it (assign it to variable).
UserDict.__init__(self)
Here you are calling the UserDict's __init__ method and passing it a reference to the new instance (because you are not calling it with self.method_name, it is not passed automatically. You cannot call an inherited class's constructor without referencing its name, or using super). So what you are doing is initializing your object the same way any UserDict object would be initialized.