What does "Decorated function definitions do not result in temporary name binding" mean?
I am now studying decorators in Python, and this sentence really confused me.
1
-
We're going to need to see the context.user2357112– user23571122017-12-03 04:09:07 +00:00Commented Dec 3, 2017 at 4:09
Add a comment
|
1 Answer
I think it's saying that the decorated function is never actually assigned to anything in the namespace before the decorator is applied. When you write a function with a decorator, the function you write is passed to the decorator as an argument, and the object the decorator returns is then treated as the function. Here's a quick example:
def dec(func):
print('times_two' in globals())
def _inner(*args, **kwargs):
print("Decorated")
return func(*args, **kwargs)
return _inner
@dec
def times_two(x):
return x*2
print('times_two' in globals())
You can see this running here. The print in dec says False, because the name "times_two" isn't bound to anything before the decorator is done "decorating" the function.
7 Comments
Hongli Bu
"A function definition may be wrapped by one or more decorator expressions. Decorator expressions are evaluated when the function is defined, in the scope that contains the function definition. The result must be a callable, which is invoked with the function object as the only argument. The returned value is bound to the function name instead of the function object. Multiple decorators are applied in nested fashion. For example, the following code" Are these two statements same??
Patrick Haugh
@HongliBu Which two statements?
Hongli Bu
The statements in my question and comment
Patrick Haugh
They're both true, but they're not the same. Basically, it's easy to imagine a scenario in which there would be some way of calling
times_two in the above example such that it doesn't also print "Decorated", accessing the function before the decorator is applied. The current implementation of generators guarantees that this will never happen. That's what the statement in your question is saying. The statement in your comment is an explanation of what a decorator is without touching on the above argument.
Paul Panzer
"The returned value is bound to the function name instead of the function object." might be argued as saying the same as the statement in the question, only much clearer. Anyway, excellent answer.
|