Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Accessing a decorator in a parent class from the child in Python

how does one go about accessing a decorator from a base class in a child?

I assumed (wrongly) that the ffg. would work:

class baseclass(object):
    def __init__(self):
        print 'hey this is the base'

    def _deco(func):
        def wrapper(*arg):
            res = func(*arg)
            print 'I\'m a decorator. This is fabulous, but that colour, so last season sweetiedarling'
            return res
        return wrapper

    @_deco
    def basefunc(self):
        print 'I\'m a base function'

This class works fine, but then I create a child class inheriting from this:

class otherclass(baseclass):
    def __init__(self):
        super(otherclass, self).__init__()
        print 'other class'


    @_deco
    def meh(self):
        print 'I\'m a function'

This won't even import properly, let alone run. @_deco is undefined. Trying baseclass._deco throws an unbound method _deco() error, which isn't really surprising.

Any idea how to do this, I'd really like to encapsulate the decorator in the class, but I'm not married to the idea and I'd need to call it in the base & the child class.

like image 892
dochead Avatar asked Aug 06 '10 05:08

dochead


2 Answers

class baseclass(object):
    def __init__(self):
        print 'hey this is the base'

    def _deco(func):
        def wrapper(*arg):
            res = func(*arg)
            print 'I\'m a decorator. This is fabulous, but that colour, so last season sweetiedarling'
            return res
        return wrapper

    @_deco
    def basefunc(self):
        print 'I\'m a base function'

    @_deco
    def basefunc2(self):
        print "I'm another base function"

   #no more uses of _deco in this class
    _deco = staticmethod(_deco) 
   # this is the key. it must be executed after all of the uses of _deco in 
   # the base class. this way _deco is some sort weird internal function that 
   # can be called from within the class namespace while said namespace is being 
   # created and a proper static method for subclasses or external callers.


class otherclass(baseclass):
    def __init__(self):
        super(otherclass, self).__init__()
        print 'other class'


    @baseclass._deco
    def meh(self):
        print 'I\'m a function'
like image 116
aaronasterling Avatar answered Nov 15 '22 15:11

aaronasterling


There is also python3-specific way to use that decorator in child class without mentioning parent, exactly as OP suggested. It requires decorator to be implemented in parent's metaclass (nice explanation of metaclases can be found here), using its __prepare__() method.

aaronasterling's answer is valid and preferred way how to solve that, I am posting this only as an interesting example to help others understand the basics of language. Use metaclasses only when there is no other way to achive what you need!

class metaclass(type):
    @classmethod
    def __prepare__(metacls, name, bases):
        def _deco(func):
            def wrapper(*arg):
                res = func(*arg)
                print('I\'m a decorator. This is fabulous, but that colour, so last season sweetiedarling')
                return res
            return wrapper
        return {"_deco": _deco}

class baseclass(metaclass=metaclass):
    def __init__(self):
        print('hey this is the base')

    @_deco
    def basefunc(self):
        print('I\'m a base function')
        
class otherclass(baseclass):
    def __init__(self):
        super(otherclass, self).__init__()
        print('other class')

    @_deco
    def meh(self):
        print('I\'m a function')

The sample code works well in python3:

>>> obj = otherclass()
hey this is the base
other class
>>> obj.meh()
I'm a function
I'm a decorator. This is fabulous, but that colour, so last season sweetiedarling

Important notes about __prepare__() method:

  • If present, it runs before the object body is executed
  • Its return value is used as local namespace for the class body at the begining of its evaluation (this way, decorator can be availabe from child's body without using parent's namespace)
  • It should be implemented as classmethod() and should return mapping object (i.e. dict)
  • If not present, empty mapping is used as initial local namespace.
like image 40
Michal Avatar answered Nov 15 '22 13:11

Michal