Why can't Python decorators be chained across definitions?

Question

Why arn't the following two scripts equivalent?

(Taken from another question: Understanding Python Decorators)

def makebold(fn):
    def wrapped():
        return "<b>" + fn() + "</b>"
    return wrapped

def makeitalic(fn):
    def wrapped():
        return "<i>" + fn() + "</i>"
    return wrapped

@makebold
@makeitalic
def hello():
    return "hello world"

print hello() ## returns <b><i>hello world</i></b>

and with a decorated decorator:

def makebold(fn):
    def wrapped():
        return "<b>" + fn() + "</b>"
    return wrapped

@makebold
def makeitalic(fn):
    def wrapped():
        return "<i>" + fn() + "</i>"
    return wrapped

@makeitalic
def hello():
    return "hello world"

print hello() ## TypeError: wrapped() takes no arguments (1 given)

Why do I want to know? I've written a retry decorator to catch MySQLdb exceptions - if the exception is transient (e.g. Timeout) it will re-call the function after sleeping a bit.

I've also got a modifies_db decorator which takes care of some cache-related housekeeping. modifies_db is decorated with retry, so I assumed that all functions decorated with modifies_db would also retry implicitly. Where did I go wrong?

Answer 1

The problem with the second example is that

@makebold
def makeitalic(fn):
    def wrapped():
        return "<i>" + fn() + "</i>"
    return wrapped

is trying to decorate makeitalic, the decorator, and not wrapped, the function it returns.

You can do what I think you intend with something like this:

def makeitalic(fn):
    @makebold
    def wrapped():
        return "<i>" + fn() + "</i>"
    return wrapped

Here makeitalic uses makebold to decorate wrapped.