Python Functions Can Have Attributes? -

i'm new python , don't understand how functions can seemingly have attributes. in code below, there function called f, , later in code, name of f.count referenced. how can function, namely f, have .count? i'm getting error message of: 'nonetype' object has no attribute 'count' on line, doesn't have attribute yet. how give attribute?

def fcount(n):     print n.__name__   @fcount def f(n):     return n+2  n in range(5):     print n     #print f(n)  print 'f count =',f.count #the line causing error mentioned above  @fcount def g(n):     return n*n  print 'g count =',g.count print g(3) print 'g count =',g.count 

edit: added fcount(), doesn't much, , details error.

let’s start definition of f:

@fcount def f(n):     return n+2 

this defines f return value of call function fcount, used decorator (the leading @) here.

this code equivalent with

def f(n):     return n+2  f = fcount(f) 

since decorator – fcount – not return anything, f none and not function @ call site.

in case fcount should return function , add count attribute returned function. useful (?) might be

def fcount(fn):     def wrapper(n):         wrapper.count += 1         return fn(n)     wrapper.count = 0     return wrapper 

edit

as @jonrsharpe pointed out, generalized decorator can forward positional , keyword arguments capturing them *args , **kwargs in signature , expanding them in same way when calling function. names args , kwargs used convention.

python has helper function (a decorator itself) can transfer information (name, docstring , signature information) 1 function another: functools.wraps. complete example looks this:

from functools import wraps  def decorator(func):     @wraps(func)     def wrapper(*args, **kwargs):         return func(*args, **kwargs)     return wrapper  @decorator def f(a, b, c=none):    "the useful f function"    pass  print f.__name__ # `f` rather `wrapper` print help(f) # `f(*args, **kwargs) useful f function` rather `wrapper(*args, **kwargs)`