Develop Reference

How to create an abstract class attribute (potentially read-only)

I have spent a lot of time researching this, but none of the answers seem to work how I would like.
I have an abstract class with a class attribute I want each subclass to be forced to implement
class AbstractFoo():
forceThis = 0
So that when I do this
class RealFoo(AbstractFoo):
pass
it throws an error telling me it can't create the class until I implement forceThis.
How can I do that?
(I don't want the attribute to be read-only, but if that's the only solution, I'll accept it.)
For a class method, I've discovered I can do
from abc import ABCMeta, abstractmethod
class AbstractFoo(metaclass=ABCMeta):
#classmethod
#abstractmethod
def forceThis():
"""This must be implemented"""
so that
class RealFoo(AbstractFoo):
pass
at least throws the error TypeError: Can't instantiate abstract class EZ with abstract methods forceThis
(Although it doesn't force forceThis to be a class method.)
How can I get a similar error to pop up for the class attribute?

You can do this by defining your own metaclass. Something like:
class ForceMeta(type):
required = ['foo', 'bar']
def __new__(mcls, name, bases, namespace):
cls = super().__new__(mcls, name, bases, namespace)
for prop in mcls.required:
if not hasattr(cls, prop):
raise NotImplementedError('must define {}'.format(prop))
return cls
Now you can use this as the metaclass of your own classes:
class RequiredClass(metaclass=ForceMeta):
foo = 1
which will raise the error 'must define bar'.

I came up with a solution based on those posted earlier. (Thank you #Daniel Roseman and #martineau)
I created a metaclass called ABCAMeta (the last 'A' stands for 'Attributes').
The class has two ways of working.
A class which just uses ABCAMeta as a metaclass must have a property called required_attributes which should contain a list of the names of all the attributes you want to require on future subclasses of that class
A class whose parent's metaclass is ABCAMeta must have all the required attributes specified by its parent class(es).
For example:
class AbstractFoo(metaclass=ABCAMeta):
required_attributes = ['force_this']
class RealFoo(AbstractFoo):
pass
will throw an error:
NameError: Class 'RealFoo' has not implemented the following attributes: 'force_this'
Exactly how I wanted.
from abc import ABCMeta
class NoRequirements(RuntimeError):
def __init__(self, message):
RuntimeError.__init__(self, message)
class ABCAMeta(ABCMeta):
def __init__(mcls, name, bases, namespace):
ABCMeta.__init__(mcls, name, bases, namespace)
def __new__(mcls, name, bases, namespace):
def get_requirements(c):
"""c is a class that should have a 'required_attributes' attribute
this function will get that list of required attributes or
raise a NoRequirements error if it doesn't find one.
"""
if hasattr(c, 'required_attributes'):
return c.required_attributes
else:
raise NoRequirements(f"Class '{c.__name__}' has no 'required_attributes' property")
cls = super().__new__(mcls, name, bases, namespace)
# true if no parents of the class being created have ABCAMeta as their metaclass
basic_metaclass = True
# list of attributes the class being created must implement
# should stay empty if basic_metaclass stays True
reqs = []
for parent in bases:
parent_meta = type(parent)
if parent_meta==ABCAMeta:
# the class being created has a parent whose metaclass is ABCAMeta
# the class being created must contain the requirements of the parent class
basic_metaclass=False
try:
reqs.extend(get_requirements(parent))
except NoRequirements:
raise
# will force subclasses of the created class to define
# the attributes listed in the required_attributes attribute of the created class
if basic_metaclass:
get_requirements(cls) # just want it to raise an error if it doesn't have the attributes
else:
missingreqs = []
for req in reqs:
if not hasattr(cls, req):
missingreqs.append(req)
if len(missingreqs)!=0:
raise NameError(f"Class '{cls.__name__}' has not implemented the following attributes: {str(missingreqs)[1:-1]}")
return cls
Any suggestions for improvement are welcome in the comments.

Although you can do something very similar with a metaclass, as illustrated in #Daniel Roseman's answer, it can also be done with a class decorator. A couple of advantages they have are that errors will occur when the class is defined, instead of when an instance of one is created, and the syntax for specifying them is the same in both Python 2 and 3. Some folks also find them simpler and easier to understand.
def check_reqs(cls):
requirements = 'must_have',
missing = [req for req in requirements if not hasattr(cls, req)]
if missing:
raise NotImplementedError(
'class {} did not define required attribute{} named {}'.format(
cls.__name__, 's' if len(missing) > 1 else '',
', '.join('"{}"'.format(name) for name in missing)))
return cls
#check_reqs
class Foo(object): # OK
must_have = 42
#check_reqs
class Bar(object): # raises a NotImplementedError
pass

Python __super black magic failed

I want to add an attribute for every class created by a metaclass. For example, when a class named C is created, I want add an attribute C._C__sup whose value is the descriptor super(C).
Here is what I've tried:
class Meta(type):
def __init__(cls, name, bases, dict): # Not overriding type.__new__
cls.__dict__['_' + name + '__sup'] = super(cls)
# Not calling type.__init__; do I need it?
class C(object):
__metaclass__ = Meta
c = C()
print c._C__sup
This gives me:
TypeError: Error when calling the metaclass bases
'dictproxy' object does not support item assignment
Some background information:
(You don't have to read this part)
Inspired by this article, what I'm doing is trying to avoid "hardcoding" the class name when using super:
The idea there is to use the unbound super objects as private
attributes. For instance, in our example, we could define the private
attribute __sup in the class C as the unbound super object
super(C):
>>> C._C__sup = super(C)
With this definition inside the methods the syntax self.__sup.meth
can be used as an alternative to super(C, self).meth. The advantage
is that you avoid to repeat the name of the class in the calling
syntax, since that name is hidden in the mangling mechanism of private
names. The creation of the __sup attributes can be hidden in a
metaclass and made automatic. So, all this seems to work: but
actually this not the case.

Use setattr instead of assignment to cls.__dict__:
class Meta(type):
def __init__(cls, name, bases, clsdict): # Not overriding type.__new__
setattr(cls, '_' + name + '__sup', super(cls))
super(Meta, cls).__init__(name, bases, clsdict)
class C(object):
__metaclass__ = Meta
def say(self):
return 'wow'
class D(C):
def say(self):
return 'bow' + self.__sup.say()
c = C()
print(c._C__sup)
# <super: <class 'C'>, <C object>>
d = D()
print(d.say())
prints
bowwow
By the way, it is a good idea to call
super(Meta, cls).__init__(name, bases, clsdict)
inside Meta.__init__ to allow Meta to participate in class hierarchies which
might need super to properly call a chain of __init__s. This seems
particularly appropriate since you are building a metaclass to assist with the
use of super.

Python metaclasses: Why isn't __setattr__ called for attributes set during class definition?

I have the following python code:
class FooMeta(type):
def __setattr__(self, name, value):
print name, value
return super(FooMeta, self).__setattr__(name, value)
class Foo(object):
__metaclass__ = FooMeta
FOO = 123
def a(self):
pass
I would have expected __setattr__ of the meta class being called for both FOO and a. However, it is not called at all. When I assign something to Foo.whatever after the class has been defined the method is called.
What's the reason for this behaviour and is there a way to intercept the assignments that happen during the creation of the class? Using attrs in __new__ won't work since I'd like to check if a method is being redefined.

A class block is roughly syntactic sugar for building a dictionary, and then invoking a metaclass to build the class object.
This:
class Foo(object):
__metaclass__ = FooMeta
FOO = 123
def a(self):
pass
Comes out pretty much as if you'd written:
d = {}
d['__metaclass__'] = FooMeta
d['FOO'] = 123
def a(self):
pass
d['a'] = a
Foo = d.get('__metaclass__', type)('Foo', (object,), d)
Only without the namespace pollution (and in reality there's also a search through all the bases to determine the metaclass, or whether there's a metaclass conflict, but I'm ignoring that here).
The metaclass' __setattr__ can control what happens when you try to set an attribute on one of its instances (the class object), but inside the class block you're not doing that, you're inserting into a dictionary object, so the dict class controls what's going on, not your metaclass. So you're out of luck.
Unless you're using Python 3.x! In Python 3.x you can define a __prepare__ classmethod (or staticmethod) on a metaclass, which controls what object is used to accumulate attributes set within a class block before they're passed to the metaclass constructor. The default __prepare__ simply returns a normal dictionary, but you could build a custom dict-like class that doesn't allow keys to be redefined, and use that to accumulate your attributes:
from collections import MutableMapping
class SingleAssignDict(MutableMapping):
def __init__(self, *args, **kwargs):
self._d = dict(*args, **kwargs)
def __getitem__(self, key):
return self._d[key]
def __setitem__(self, key, value):
if key in self._d:
raise ValueError(
'Key {!r} already exists in SingleAssignDict'.format(key)
)
else:
self._d[key] = value
def __delitem__(self, key):
del self._d[key]
def __iter__(self):
return iter(self._d)
def __len__(self):
return len(self._d)
def __contains__(self, key):
return key in self._d
def __repr__(self):
return '{}({!r})'.format(type(self).__name__, self._d)
class RedefBlocker(type):
#classmethod
def __prepare__(metacls, name, bases, **kwargs):
return SingleAssignDict()
def __new__(metacls, name, bases, sad):
return super().__new__(metacls, name, bases, dict(sad))
class Okay(metaclass=RedefBlocker):
a = 1
b = 2
class Boom(metaclass=RedefBlocker):
a = 1
b = 2
a = 3
Running this gives me:
Traceback (most recent call last):
File "/tmp/redef.py", line 50, in <module>
class Boom(metaclass=RedefBlocker):
File "/tmp/redef.py", line 53, in Boom
a = 3
File "/tmp/redef.py", line 15, in __setitem__
'Key {!r} already exists in SingleAssignDict'.format(key)
ValueError: Key 'a' already exists in SingleAssignDict
Some notes:
__prepare__ has to be a classmethod or staticmethod, because it's being called before the metaclass' instance (your class) exists.
type still needs its third parameter to be a real dict, so you have to have a __new__ method that converts the SingleAssignDict to a normal one
I could have subclassed dict, which would probably have avoided (2), but I really dislike doing that because of how the non-basic methods like update don't respect your overrides of the basic methods like __setitem__. So I prefer to subclass collections.MutableMapping and wrap a dictionary.
The actual Okay.__dict__ object is a normal dictionary, because it was set by type and type is finicky about the kind of dictionary it wants. This means that overwriting class attributes after class creation does not raise an exception. You can overwrite the __dict__ attribute after the superclass call in __new__ if you want to maintain the no-overwriting forced by the class object's dictionary.
Sadly this technique is unavailable in Python 2.x (I checked). The __prepare__ method isn't invoked, which makes sense as in Python 2.x the metaclass is determined by the __metaclass__ magic attribute rather than a special keyword in the classblock; which means the dict object used to accumulate attributes for the class block already exists by the time the metaclass is known.
Compare Python 2:
class Foo(object):
__metaclass__ = FooMeta
FOO = 123
def a(self):
pass
Being roughly equivalent to:
d = {}
d['__metaclass__'] = FooMeta
d['FOO'] = 123
def a(self):
pass
d['a'] = a
Foo = d.get('__metaclass__', type)('Foo', (object,), d)
Where the metaclass to invoke is determined from the dictionary, versus Python 3:
class Foo(metaclass=FooMeta):
FOO = 123
def a(self):
pass
Being roughly equivalent to:
d = FooMeta.__prepare__('Foo', ())
d['Foo'] = 123
def a(self):
pass
d['a'] = a
Foo = FooMeta('Foo', (), d)
Where the dictionary to use is determined from the metaclass.

There are no assignments happening during the creation of the class. Or: they are happening, but not in the context you think they are. All class attributes are collected from class body scope and passed to metaclass' __new__, as the last argument:
class FooMeta(type):
def __new__(self, name, bases, attrs):
print attrs
return type.__new__(self, name, bases, attrs)
class Foo(object):
__metaclass__ = FooMeta
FOO = 123
Reason: when the code in the class body executes, there's no class yet. Which means there's no opportunity for metaclass to intercept anything yet.

Class attributes are passed to the metaclass as a single dictionary and my hypothesis is that this is used to update the __dict__ attribute of the class all at once, e.g. something like cls.__dict__.update(dct) rather than doing setattr() on each item. More to the point, it's all handled in C-land and simply wasn't written to call a custom __setattr__().
It's easy enough to do whatever you want to the attributes of the class in your metaclass's __init__() method, since you're passed the class namespace as a dict, so just do that.

During the class creation, your namespace is evaluated to a dict and passed as an argument to the metaclass, together with the class name and base classes. Because of that, assigning a class attribute inside the class definition wouldn't work the way you expect. It doesn't create an empty class and assign everything. You also can't have duplicated keys in a dict, so during class creation attributes are already deduplicated. Only by setting an attribute after the class definition you can trigger your custom __setattr__.
Because the namespace is a dict, there's no way for you to check duplicated methods, as suggested by your other question. The only practical way to do that is parsing the source code.

class __init__ (not instance __init__)

Here's a very simple example of what I'm trying to get around:
class Test(object):
some_dict = {Test: True}
The problem is that I cannot refer to Test while it's still being defined
Normally, I'd just do this:
class Test(object):
some_dict = {}
def __init__(self):
if self.__class__.some_dict == {}:
self.__class__.some_dict = {Test: True}
But I never create an instance of this class. It's really just a container to hold a group of related functions and data (I have several of these classes, and I pass around references to them, so it is necessary for Test to be it's own class)
So my question is, how could I refer to Test while it's being defined, or is there something similar to __init__ that get's called as soon as the class is defined? If possible, I want self.some_dict = {Test: True} to remain inside the class definition. This is the only way I know how to do this so far:
class Test(object):
#classmethod
def class_init(cls):
cls.some_dict = {Test: True}
Test.class_init()

The class does in fact not exist while it is being defined. The way the class statement works is that the body of the statement is executed, as a block of code, in a separate namespace. At the end of the execution, that namespace is passed to the metaclass (such as type) and the metaclass creates the class using the namespace as the attributespace.
From your description, it does not sound necessary for Test to be a class. It sounds like it should be a module instead. some_dict is a global -- even if it's a class attribute, there's only one such attribute in your program, so it's not any better than having a global -- and any classmethods you have in the class can just be functions.
If you really want it to be a class, you have three options: set the dict after defining the class:
class Test:
some_dict = {}
Test.some_dict[Test] = True
Use a class decorator (in Python 2.6 or later):
def set_some_dict(cls):
cls.some_dict[cls] = True
#set_some_dict
class Test:
some_dict = {}
Or use a metaclass:
class SomeDictSetterType(type):
def __init__(self, name, bases, attrs):
self.some_dict[self] = True
super(SomeDictSetterType, self).__init__(name, bases, attrs)
class Test(object):
__metaclass__ = SomeDictSetterType
some_dict = {}

You could add the some_dict attribute after the main class definition.
class Test(object):
pass
Test.some_dict = {Test: True}

I've tried to use classes in this way in the past, and it gets ugly pretty quickly (for example, all the methods will need to be class methods or static methods, and you will probably realise eventually that you want to define certain special methods, for which you will have to start using metaclasses). It could make things a lot easier if you just use class instances instead - there aren't really any downsides.
A (weird-looking) alternative to what others have suggested: you could use __new__:
class Test(object):
def __new__(cls):
cls.some_dict = {cls: True}
Test()
You could even have __new__ return a reference to the class and use a decorator to call it:
def instantiate(cls):
return cls()
#instantiate
class Test(object):
def __new__(cls):
cls.some_dict = {cls: True}
return cls

You can also use a metaclass (a function here but there are other ways):
def Meta(name, bases, ns):
klass = type(name, bases, ns)
setattr(klass, 'some_dict', { klass: True })
return klass
class Test(object):
__metaclass__ = Meta
print Test.some_dict

Thomas's first example is very good, but here's a more Pythonic way of doing the same thing.
class Test:
x = {}
#classmethod
def init(cls):
# do whatever setup you need here
cls.x[cls] = True
Test.init()

Accessing the class that owns a decorated method from the decorator

I'm writing a decorator for methods that must inspect the parent methods (the methods of the same name in the parents of the class in which I'm decorating).
Example (from the fourth example of PEP 318):
def returns(rtype):
def check_returns(f):
def new_f(*args, **kwds):
result = f(*args, **kwds)
assert isinstance(result, rtype), \
"return value %r does not match %s" % (result,rtype)
return result
new_f.func_name = f.func_name
# here I want to reach the class owning the decorated method f,
# it should give me the class A
return new_f
return check_returns
class A(object):
#returns(int)
def compute(self, value):
return value * 3
So I'm looking for the code to type in place of # here I want...
Thanks.

As bobince said it, you can't access the surrounding class, because at the time the decorator is invoked, the class does not exist yet. If you need access to the full dictionary of the class and the bases, you should consider a metaclass:
__metaclass__
This variable can be any callable accepting arguments for name, bases, and dict. Upon class creation, the callable is used instead of the built-in type().
Basically, we convert the returns decorator into something that just tells the metaclass to do some magic on class construction:
class CheckedReturnType(object):
def __init__(self, meth, rtype):
self.meth = meth
self.rtype = rtype
def returns(rtype):
def _inner(f):
return CheckedReturnType(f, rtype)
return _inner
class BaseInspector(type):
def __new__(mcs, name, bases, dct):
for obj_name, obj in dct.iteritems():
if isinstance(obj, CheckedReturnType):
# do your wrapping & checking here, base classes are in bases
# reassign to dct
return type.__new__(mcs, name, bases, dct)
class A(object):
__metaclass__ = BaseInspector
#returns(int)
def compute(self, value):
return value * 3
Mind that I have not tested this code, please leave comments if I should update this.
There are some articles on metaclasses by the highly recommendable David Mertz, which you might find interesting in this context.

here I want to reach the class owning the decorated method f
You can't because at the point of decoration, no class owns the method f.
class A(object):
#returns(int)
def compute(self, value):
return value * 3
Is the same as saying:
class A(object):
pass
#returns(int)
def compute(self, value):
return value*3
A.compute= compute
Clearly, the returns() decorator is built before the function is assigned to an owner class.
Now when you write a function to a class (either inline, or explicitly like this) it becomes an unbound method object. Now it has a reference to its owner class, which you can get by saying:
>>> A.compute.im_class
<class '__main__.A'>
So you can read f.im_class inside ‘new_f’, which is executed after the assignment, but not in the decorator itself.
(And even then it's a bit ugly relying on a CPython implementation detail if you don't need to. I'm not quite sure what you're trying to do, but things involving “get the owner class” are often doable using metaclasses.)