Clarification requested for decorators - python-3.x

Question

Got big doubts on decorators.
Below a simple test code, read in some book for beginner.
# -*-coding:Latin-1 -*
import sys
# The decorator.
def my_decorator(modified_function):
"""My first decorator."""
# Modifying modified_function.
def modifying_function():
print("--> before")
ret = modified_function()
print("<-- after={}".format(ret))
return ret
print("Decorator is called with modified_function '{0}'".format(modified_function))
# Return the modifying function.
return modifying_function
#my_decorator
def hello_world():
"""Decorated function."""
print("!! That's all folks !!")
return (14)
print("Python version = {}".format(sys.version))
# We try to call hello_world(), but the decorator is called.
hello_world()
print("--------------------------------------------------------------")
my_decorator(hello_world)
print("--------------------------------------------------------------")
# Found this other way on the WEB, but does not work for me
my_hello = my_decorator(hello_world)
my_hello()
print("--------------------------------------------------------------")
For this code, the output is rather strange, to me.
Maybe it's stupid, but ...
Decorator is called with modified_function '<function hello_world at 0x0000011D5FDCDEA0>'
Python version = 3.6.2 (v3.6.2:5fd33b5, Jul 8 2017, 04:57:36) [MSC v.1900 64 bit (AMD64)]
--> before
!! That's all folks !!
<-- after=14
--------------------------------------------------------------
Decorator is called with modified_function '<function my_decorator.<locals>.modifying_function at 0x0000011D5FDCDF28>'
--------------------------------------------------------------
Decorator is called with modified_function '<function my_decorator.<locals>.modifying_function at 0x0000011D5FDCDF28>'
--> before
--> before
!! That's all folks !!
<-- after=14
<-- after=14
--------------------------------------------------------------
the python version is printed after the decorator trace.
in 2nd and 3rd call, print("Decorator is called with modified_function ...") gives me some strange value for the function. At least, not what I expected.
the traces ("--> ...") and ("<-- ...") are doubled.
Any clarification for a newbie is welcome.

Answer 1

Your confusion has to do with the fact that you're using the decorator manually, but you also used Python's special syntax to apply decorators to functions as you define them: #my_decorator. You usually don't want to do both of those for the same function.
Try this example and it might help you understand things better:
# after defining my_decorator as before...
def foo(): # define a function for testing with manual decorator application
print("foo")
print("----")
foo() # test unmodified function
print("----")
decorated_foo = my_decorator(foo) # manually apply the decorator (assigning to a new name)
print("----")
decorated_foo() # test decorated function
print("----")
foo() # confirm that the original function is still the same
print("----")
foo = my_decorator(foo) # apply the decorator again, replacing the original name this time
print("----")
foo() # see that the original function has been replaced
print("----")
#my_decorator # this line applies the decorator for you (like the "foo = ..." line above)
def bar():
print("bar")
print("----")
bar() # note that the decorator has already been applied to bar
print("----")
double_decorated_bar = my_decorator(bar) # apply the decorator again, just for fun
print("----")
double_decorated_bar() # get doubled output from the doubled decorators

Answer 2

Oh ... OK.
Light came when I saw :
double_decorated_bar = my_decorator(bar) # apply the decorator again, just for fun
Did not pay attention that the #decorator was not there in example.
Thx again for clarification.