python - A function with itself as a default argument -

defaults parsed @ definition. not work

def f(x, func = f):     if x<1:         return x     else:         return x + func(x-1) 

i did find way, though. start dummy function

def a(x):     return x  def f(x, func=a):     if x < 1:         return x     else:         return x + func(x-1) 

and issue

f.__defaults__ = (f,) 

obviously awkward. there way or bad python? if it's bad, can explain why? break things?

in case, works:

in [99]: f(10) out[99]: 55  in [100]: f(10, f) out[100]: 55  in [101]: f(10, lambda x : 2*x) out[101]: 28  

to elaborate on andrea corbellini's suggestion, can this:

def f(x, func=none):     if func none:         func = f      if x < 1:         return x     else:         return x + func(x-1) 

it pretty standard idiom in python implement actual defaults inside of function , defining default parameter none (or private sentinel object in case none valid input), due problems mutable default values such list objects.