I do not really understand the behaviour of the following code. The goal of this code was to allow only one processing of the function decorated by the once decorator. Since I return onceResult in the try block I do not get why on the second call I don't get a second "Hello william". Therefor i tried to print onceResult and get None as output. I expected a string containing "Hello william".
Could someone explain me this behaviour?
Here is the output:
Before first call
Exception
Hello William !
Before second call
None
Finished
Here is the code:
def once(func):
"""once is a decorator which allows only one execution of the function"""
def nested(*args,**kargs):
try:
print(nested.onceResult)
return nested.onceResult
except AttributeError:
print("### EXCEPTION###")
nested.onceResult = func(*args,**kargs)
return nested.onceResult
return nested
@once
def hello(name):
print(f"Hello {str(name)} !")
print("Before first call")
hello("William")
print("Before second call")
hello("Roger")
print("Finished")
I would recommend to use a class to achieve your goal.
Here are my approaches:
class ExecuteOnceClass:
called = []
@staticmethod
def execute(function):
def wrapper(*args, **kwargs):
if function.__name__ in ExecuteOnceClass.called:
print("Function already called once")
else:
function(*args, **kwargs)
ExecuteOnceClass.called.append(function.__name__)
print("called function")
return wrapper
@ExecuteOnceClass.execute
def hello(name):
print(f"Hello {name}.")
hello("Neuneu")
hello("Sofian")
Expected output:
Hello Neuneu.
called function
Function already called once
class ExecuteOnceClass:
called = {}
@staticmethod
def execute(function):
def wrapper(*args, **kwargs):
if function.__name__ in ExecuteOnceClass.called and {"args": args, "kwargs": kwargs} in ExecuteOnceClass.called[function.__name__]:
print("Function already called once with the same arguments")
else:
function(*args, **kwargs)
if ExecuteOnceClass.called.get(function.__name__, None):
ExecuteOnceClass.called[function.__name__].append({"args": args, "kwargs": kwargs})
else:
ExecuteOnceClass.called[function.__name__] = [{"args": args, "kwargs": kwargs}]
print("called function")
return wrapper
@ExecuteOnceClass.execute
def hello(name):
print(f"Hello {name}.")
hello("Neuneu")
hello("Sofien")
hello("Neuneu")
hello("Sofien")
Expected output:
Hello Neuneu.
called function
Hello Sofien.
called function
Function already called once with the same arguments
Function already called once with the same arguments
EDIT: if your function returns something you would need to edit the code a bit:
class ExecuteOnceClass:
called = {}
@staticmethod
def execute(function):
def wrapper(*args, **kwargs):
if function.__name__ in ExecuteOnceClass.called and {"args": args, "kwargs": kwargs} in ExecuteOnceClass.called[function.__name__]:
print("Function already called once with the same arguments")
else:
functionResult = function(*args, **kwargs)
if ExecuteOnceClass.called.get(function.__name__, None):
ExecuteOnceClass.called[function.__name__].append({"args": args, "kwargs": kwargs})
else:
ExecuteOnceClass.called[function.__name__] = [{"args": args, "kwargs": kwargs}]
print("called function")
return functionResult
return wrapper
@ExecuteOnceClass.execute
def hello(name):
return f"Hello {name}."
print(hello("Neuneu"))
Expected Output:
Hello Neuneu.
good luck!
SECOND EDIT: for a better understanding of python at all (and of course decorators too) I would highly recommend you this video:
What Does It Take To Be An Expert At Python?
I have learned a lot watching this video.
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