Create decorator that can see current class method

早过忘川 提交于 2019-12-05 11:53:58

You can have it be a class but you need to use the descriptor protocol

 import types

 class canRun(object):
    def __init__(self, f):
        self.f = f
        self.o = object  # <-- What the hell is this about? 

    def __call__(self, *args):
        if self.longcondition():
            self.f(*args)

    def __get__(self, instance, owner):
         return types.MethodType(self, instance)

You always need to use a descriptor when you want to decorate class methods with a class instance using the __call__ method. The reason for this is that there will only be one self passed in which refers to the instance of the decorating class and not the instance of the decorated method.

There's no real need to implement this decorator as a class, and there's no need to implement it inside the definition of the Foo class. The following will suffice:

def canRun(meth):
    def decorated_meth(self, *args, **kwargs):
        if self.longcondition():
            print 'Can run'
            return meth(self, *args, **kwargs)
        else:
            print 'Cannot run'
            return None
    return decorated_meth

Using that decorator seems to work:

>>> Foo('hello').run(5)
Can run
0. run... hello
1. run... hello
2. run... hello
3. run... hello
4. run... hello
>>> Foo(123).run(5)
Cannot run

My previous answer was made in haste. If you're wanting to write a decorator you should really use wraps from the functools module. It takes care of the hard stuff for you.

A proper way to define the canRun decorator is:

from functools import wraps
def canRun(f):
  @wraps(f)
  def wrapper(instance, *args):
    if instance.longcondition():
      return f(instance, *args)
  return wrapper

The canRun function should be defined outside of the class.

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