I hope that someone familiar with Python's compilation / run-time procedures could shed some light on my question relating to how Python compiles decorator functions.
Within my sample code, I've included a testing print statement in the "writeit" decorator just before the logtofile closure is defined. If you run the entire code that I've provided, the "testing" print statement in writeit is called for each @writeit decorator defined in the Customer class-- before writeit is ever used.
Why is logtofile being called at compile time? Could someone please explain this behavior?
def logtofile(customer, *arg, **kwargs):
result = func(customer, *arg, **kwargs)
with open('dictlog.txt','w') as myfile:
self._custinfo = dict(firstname=firstname,lastname=lastname,address=address,city=city,state=state,zipcode=zipcode)
cust1 = Customer('Joe','Shmoe','123 Washington','Washington DC','DC','12345')
if(__name__ == '__main__'): main()
Your code runs when the module is imported. Python executes all top-level statements, including class definitions at that time.
A class definition body is executed as a function, with the local namespace becoming the class attributes. This means that class bodies are executed on import, provided the class is defined at the top-level of the module.
When Python encounters a decorated function when executing, it'll define the class, then execute the decorator function, passing in the function object and binding the return value of the decorator to the name of the function. Since the class body is executed during import, this means your decorator is executed at that time.