Internal method in Python -- unable to access to a variable -

i wonder, why have error of unboundlocalerror: local variable 'i' referenced before assignment or nameerror: global name 'i' not defined in following code:

def method1():   = 0   def _method1():     # global -- same error     += 1     print 'i=',    # = 0 -- same error   _method1()  method1() 

how rid of it? i should not visible outside of method1()

one way in py2x pass variable internal method, in py3x has been fixed , can use nonlocal statement there.

the error raised because functions objects , evaluated during definition, , during definition python sees i += 1 (which equivalent i = + 1) thinks i local variable inside function. but, when function called fails find value i (on rhs ) locally , error raised.

def method1():   = 0   def _method1(i):     += 1     print 'i=',     return     #return updated value    i=_method1(i)  #update   print  method1() 

or use function attribute:

def method1():   method1.i = 0       #create function attribute   def _method1():     method1.i += 1     print 'i=', method1.i    _method1()  method1() 

for py3x:

def method1():   =0    def _method1():     nonlocal       += 1     print ('i=', i)   _method1()  method1()