Is there a generic way for a function to reference itself?

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There is no generic way for a function to refer to itself. Consider using a decorator instead. If all you want as you indicated was to print information about the function that can be done easily with a decorator:

from functools import wraps
def showinfo(f):
    @wraps(f)
    def wrapper(*args, **kwds):
         print(f.__name__, f.__hash__)
         return f(*args, **kwds)
    return wrapper

@showinfo
def aa():
    pass

If you really do need to reference the function, then just add it to the function arguments:

def withself(f):
    @wraps(f)
    def wrapper(*args, **kwds):
        return f(f, *args, **kwds)
    return wrapper

@withself
def aa(self):
      print(self.__name__)
      # etc.

Edit to add alternate decorator:

You can also write a simpler (and probably faster) decorator that will make the wrapped function work correctly with Python’s introspection:

def bind(f):
    """Decorate function `f` to pass a reference to the function
    as the first argument"""
    return f.__get__(f, type(f))

@bind
def foo(self, x):
    "This is a bound function!"
    print(self, x)


>>> foo(42)
<function foo at 0x02A46030> 42
>>> help(foo)
Help on method foo in module __main__:

foo(self, x) method of builtins.function instance
    This is a bound function!

This leverages Python’s descriptor protocol: functions have a __get__ method that is used to create bound methods. The decorator simply uses the existing method to make the function a bound method of itself. It will only work for standalone functions, if you wanted a method to be able to reference itself you would have to do something more like the original solution.

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