[Tutor] scoping rules

[Karthik Gurumurthy]

Just wanted to confirm the scoping rules.

>>> a = 1298
>>> def func(x):
	a = 908
	def func(x):
		print "isnide func"
		print "\tlocals",locals()
		print "\tglobals",globals()
		i =123
		print "\tlocals again",locals()
		a+=1 		###statement
		print a
	print "ok",locals()
	func(87)
	print a

i observed the following.

When i commented the marked statement inside the inner func,
a (908 from the enclosing function's local namespace) was present in the
inner function's local namespace.
the global a was present in the global namespace.fine.

The moment i uncomment it (ie a = a + 1), the reference to "enclosing
function's a" from locals() of
inner func was removed.

BUT, reference to a in the global namespace remained. Then why did'nt it
increment the global a?
since it is supposed to look up the locals() first and when not found search
globals()?
instead i got an unbound error.

Does it mean that moment we try to add to / manipulate a variable, python
expects the variable to
be initialized in the function's own local namespace? I mean as long as we
are "reading", it is all fine.

it worked when i put the flag global a right in the begining of the
function.

Does this mean that i will not be able to increment a (908) (defined in the
enclosing function)
inside the inner function?

hope i have framed my question properly.

thanks
karthik.