Pass the class itself to the class method decorator

Question

Python

Here is a simple decorator:

def deco(clazz):
  def dec(fn):
    print clazz,fn
    return fn
  return dec

We are trying to use it:

class Foo(object):
  @deco(Foo)
  def Bar():
    print "bar"

On startup it says:

Traceback (most recent call last):
  File "<pyshell#12>", line 1, in <module>
    class Foo(object):
  File "<pyshell#12>", line 2, in Foo
    @deco(Foo)
NameError: name 'Foo' is not defined

Actually, why does he not see the class and how can this be bypassed?

Answer 1

Because at the time of the execution of the class body, it has not yet been formed.
You can for example like this:

import traceback

def deco():
    clazz = traceback.extract_stack()[1][2]
    def dec(fn):
        print clazz,fn
        return fn
    return dec

class Foo(object):
    @deco()
    def Bar():
        print "bar"

Maybe someone will offer something better, something the head does not cook.

Answer 2

Method level decorator - solves the issues of the method itself. And it is not correct to try to extend it to the class level. Understand that a method can be declared anywhere and assigned to any class in any other method. What should the decorator return in this case? Such a python. It is necessary to try to reformulate the problem, perhaps the solution is not in the decorators at all?

Answer 3

It looks like what implements the answer to the question posed is this example:
wiki.python.org/moin/PythonDecoratorLibrary#CA-1f74040af898602b483f2d7691d29b4a5644cce2_1

Answer 4

> I need to get the class object in the decorator and add a couple of properties to it. Metaclasses are more suitable for this. In any case, you won't get the class object at the moment the decorator of its method is called - the class hasn't been created yet.