CODEWITHSUNDEEP
pythonpython-decorators
pinker
pinker

Reputation: 1293

Python Decorators - __call__ in class

I was trying to understand Python decorators and I was trying to write an equivalent program to this one:

class myDecorator(object):
   def __init__(self, f):
      print ("inside myDecorator.__init__()")
      f() # Prove that function definition has completed

   def __call__(self):
      print ("inside myDecorator.__call__()")

@myDecorator
def aFunction():
   print ("inside aFunction()")

print ("Finished decorating aFunction()")

aFunction()

The problem is I am not understanding how the __call__ method of class is being invoked by calling aFunction() in the end.

Is aFunction() being replaced by myDecorator.__call__(aFunction). Can you please help me? How would be an equivalent program without decorators?

Thanks!

Upvotes: 11

Views: 10846

Answers (5)

Soumil Nitin Shah
Soumil Nitin Shah

Reputation: 730

Hello The correct way to do this would be something like this you have a Metaclass > --- Decoratorclass > --> Actual Class

    class Meta(type):
        """ Metaclass  """

        def __call__(cls, *args, **kwargs):
            instance = super(Meta, cls).__call__(*args, **kwargs)
            return instance

        def __init__(cls, name, base, attr):
            super(Meta, cls).__init__(name, base, attr)


    class counted(metaclass=Meta):

        """ counts how often a function is called """

        def __init__(self, func):
            self.func = func

        def __call__(self, *args, **kwargs):
            print("Timer Start ")
            Tem = self.func(*args, **kwargs)
            print("Timer End {} ".format(Tem))
            return Tem


    class Test(object):

        def __init__(self, *args, **kwargs):
            pass

        @counted
        def methodA():
            print("Method A")
            return "1111"


    if __name__ == "__main__":
        obj = Test()
        obj.methodA()

Upvotes: 0

MangoHands
MangoHands

Reputation: 1171

Here's a fix:

class myDecorator(object):
   def __init__(self, f):
      print ("inside myDecorator.__init__()")
      self.f = f #Store the function being wrapped

   def __call__(self):
      print ("inside myDecorator.__call__()")
      self.f() # Call the wrapped function

Upvotes: 0

kindall
kindall

Reputation: 184280

afunction is being replaced by an instance of the class myDecorator. The class has a __call__ method, so the instance can be called like a function. Thus, assuming its signature is compatible (usually a decorator returns something that uses *args and **kwargs), the instance can be used as a replacement for the original function.

Normally the __call__ method would call the wrapped function. Doing it in __init__ is generally incorrect; __init__ should store a reference to the wrapped function as an instance attribute, so that __call__ can call it.

Upvotes: 1

Wikiii122
Wikiii122

Reputation: 486

The output of your code is

inside myDecorator.__init__()
inside aFunction()
Finished decorating aFunction()
inside myDecorator.__call__()

First, do you know, what this @ decorator syntax mean?

@decorator
def function(a):
    pass

is just another way of saying:

def function(a):
    pass
function = decorator(function)

So, in Your case

@myDecorator
def aFunction():
   print ("inside aFunction()")

means just

def aFunction():
    print ("inside aFunction()")
aFunction = myDecorator(aFunction)

At first, You basically create a new instance of myDecorator class, invoking it's constructor (__init__) and passing to it a aFunction function object. Then, it executes print and given function. Also, note that this is happening while the function is loaded by interpreter, not when it's executed, so in case you import something from this file, it will execute then, not on use or call.

Then, executing the aFunction(), when aFunction is still refering to the myDecorator instance, makes it to call the __call__ method of myDecorator, which executes. Note, that f() means the same as f.__call__(f) in this case, as __call__ method is used to enable and override default object's calling behaviour (in simplification, any object is callable when it has __call__ method defined).

If you want to execute aFunction when it's called, then you should assign it to instance variable in __init__ and invoke it in __call__ of myDecorator.

Upvotes: 11

Daniel Roseman
Daniel Roseman

Reputation: 599796

That's the entire purpose of a decorator: to replace (or, more usually, wrap) a function with the function returned by the decorator. In your case, aFunction is being replaced with an instance of myDecorator, so when you call aFunction() you are really calling that instance: and in Python, calling a class instance invokes its __call__ method.

Defining a decorated function is exactly equivalent to this:

def aFunction():
    print("inside aFunction()")
aFunction = myDecorator(aFunction)

Usually, of course, the wrapping function would call the original function after doing whatever it does.

Upvotes: 2

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