Passing a parameter to the decorator in python

Question

Why is this decorator with a parameter not working?

def decAny( f0 ):
    def wrapper( s0 ):
        return "<%s> %s </%s>" % ( any, f0(), any )
    return wrapper

@decAny( 'xxx' )
def test2():
    return 'test1XML'

print( test2() )

always gives me an error saying "str is not callable" it is trying to execute the return string inside the wrapper() instead of processing it and return the result string

Answer 1

Decorators are functions that return functions. When "passing a parameter to the decorator" what you are actually doing is calling a function that returns a decorator. So decAny() should be a function that returns a function that returns a function.

It would look something like this:

import functools

def decAny(tag):
    def dec(f0):
        @functools.wraps(f0)
        def wrapper(*args, **kwargs):
            return "<%s> %s </%s>" % (tag, f0(*args, **kwargs), tag)
        return wrapper
    return dec

@decAny( 'xxx' )
def test2():
    return 'test1XML'

Example:

>>> print(test2())
<xxx> test1XML </xxx>

Note that in addition to fixing the specific problem you were hitting I also improved your code a bit by adding *args and **kwargs as arguments to the wrapped function and passing them on to the f0 call inside of the decorator. This makes it so you can decorate a function that accepts any number of positional or named arguments and it will still work correctly.

You can read up about functools.wraps() here:
http://docs.python.org/2/library/functools.html#functools.wraps