Python Decorators I encounter a Strange error

Question

My code is very simple. it's for a test, but i get a big error :nonetype is not callable:

def deco(when):
    def wrapper(fun):
        print 'before call myfun'
        fun()
        print 'after call myfun' 
    return wrapper 
@deco('a')
def fun():
    print 'in fun1'
fun()

But when it is slightly modified,the error is removed:

def deco(when):
    def wrapper(fun):
        def dec():
            print 'before call myfun'
            fun()
            print 'after call myfun'
        return dec
    return wrapper

@deco('a')
def fun():
    print 'in fun'
fun()

Can you tell the reason? I was totally confused.

Morever, in the second code block, how can the methon wrapper() visit the variable ‘fun’,the variable ‘fun’ is not in the context(the arg of upper methond is 'when' instead of 'fun'), i am confused too.

Thanks your help

Answer 1

When a decorator is provided with arguments, it is treated as a 'decorator generator', and must return a proper decorator.

See also python decorators with parameters.

Answer 2

A decorator is a function that takes a function and returns a new function. It typically looks like

def deco(fun):
    def wrapped(*args, **kwargs):
        # something... 
        return fun(*args, **kwargs)
    return wrapped

The expression after @ has to return a decorator. If that expression is a function call, then the function it calls has to return a decorator. Therefore, it would have to look like

def deco_with_args(arg):
    def deco(fun): # The decorator function that will be returned
        def wrapped(*args, **kwargs): # The wrapper function that will finally be assigned to the decorated name
            # something... 
            return fun(*args, **kwargs)
        return wrapped
     return deco