How decorator works?

Question

I'm trying to figure out how decorator works in python. But, there are two things that I can't make clear, so I appreciate it if anyone helps me understand how decorator works in python!

This is the sample code I've just written to see how it works.

In [22]: def deco(f):
   ....:     def wrapper():
   ....:         print("start")
   ....:         f()
   ....:         print("end")
   ....:     return wrapper

In [23]: @deco
   ....: def test():
   ....:     print("hello world")
   ....:

OUTPUT 1

In [24]: test()
start
hello world
end

The first thing I don't understand is why it outputs "start","hello world","end" when I call test(). I learned that when I call test(), it calls "deco(test)" internally. If so, it should return a "wrapper" function object instead of outputting strings. But, it outputs strings as the result. I'm wondering how it's doing the job internally.

OUTPUT 2

In [28]: i = deco(test)

In [29]: i
Out[29]: <function __main__.wrapper>

In [30]: i()
start
start
hello world
end
end

I called "deco(test)" just to see what it outputs as the result. As it's shown above, it returns "wrapper" function object and after I assign it to a variable and call "wrapper" function, it outputs two "start"s and one "hello world" and two "end"s. What is going on internally? Why "start" and "end" are outputted twice respectively?

Could anyone please help me understand how this is working?

Like Preet said, please check the linked thread for explanation of decorators. Regarding the second output, start and end are printed twice since you are wrapping the test function in the decorator once again. It was already decorated on line 23. — vivekagr, Mar 11 '14 at 05:30

score 5 · Accepted Answer · answered Mar 11 '14 at 05:35

I learned that when I call test(), it calls "deco(test)" internally. If so, it should return a "wrapper" function object instead of outputting strings. But, it outputs strings as the result. I'm wondering how it's doing the job internally.

Not quite, applying a decorator is syntatic sugar (meaning a nice way of doing something that is otherwise possible). The equivalent operation is

def test():
    print("Hello, world!")
test = deco(test) # note that test is overwritten by the wrapper that deco returns

To illustrate this, consider the following example

>>> def deco(f):
...     print 'applying deco' # this will print when deco is applied to test
...     def wrapper():
...             print("start")
...             f()
...             print("end")
...     return wrapper
...
>>> @deco
... def test():
...     print("Hello world!")
...
applying deco

Notice that the decorator is applied as soon as the function is defined. This matches up with the above "equivalent operation".

In the second case, you are seeing the double print statements as you are manually applying deco to an already decorated test function.

oh,,, so, test function is overwritten by wrapper function when the decorator is applied to test function. That's why when test() is called, it outputs strings. I see!!! Thanks! — crzyonez777, Mar 11 '14 at 06:19
applying a decorator is Syntactic sugar of this "test = deco(test)". OK!! I got it! it totally makes sense to me, now!! Thank you so much!! — crzyonez777, Mar 11 '14 at 06:23

How decorator works?

1 Answers1