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Home/ Questions/Q 8137627
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T11:11:52+00:00 2026-06-06T11:11:52+00:00

I don’t know if Nested python decorator? is the right way to state this

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I don’t know if “Nested python decorator?” is the right way to state this question, so let me know if it’s not.

Anyway, I’m taking a class at udacity and have just encountered some code that involves python decorator and looks like voodoo magic, so now I want to ask a generalized question to see if I can figure that code out.

Suppose that I have the following code:

def A(f):
    print 'blah'
    return f

@A
def B(f):
    return f

@B
def C():
    pass

Now, I understand that from the above code, the decorator causes B to turn into:

B = A(B)

and that’s what a decorator does. However, what is C like?
From some small sample codes I have seen, somehow C is affected by A because A changes B and B changes C. But I have two problems understanding this:

  1. The exact nature of C. Is it C = A(B)(C) or C = A(B(C))?
  2. If C is indeed affected by A, why is ‘blah’ only printed once when I run the above code?

Personal guess

Actually, now that I have typed it out, I hypothesize that what happens is we first get:
B = A(B) and then C = B(C). It means that overall, we get C = A(B)(C), and it would explain why ‘blah’ is only printed once.

But it’s best that I make sure.

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  1. Editorial Team
    Editorial Team
    2026-06-06T11:11:54+00:00Added an answer on June 6, 2026 at 11:11 am

    Your personal guess is correct.

    A decorator is “called” at definition-time, when the @-line and its following function definition are evaluated.

    @foo
    def bar():
        pass
    

    is just syntactic sugar for…

    def bar():
        pass
    bar = foo(bar)
    
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