On 10/03/2012 01:54 AM, Demian Brecht wrote: > On 12-10-02 07:26 PM, Dave Angel wrote: >> >> if you're stuck with Python2.x, you can use a mutable object for a, and >> mutate it, rather than replace it. For example, >> >> >> def foo(): >> a = [3] >> def bar(): >> b=2 >> a.append(b) #this mutates a, but doesn't assign it >> print (a) >> a[0] += b #likewise, for a number within the list >> print (a) >> bar() >> >> That should work in either 2.x or 3.2 >> > > Alternatively, you can restructure your code by simply adding a > parameter to bar(). Nice thing about this is that if you ever move > bar() out into another module, then you don't have to worry about > documenting the side effects on 'a' so users (including yourself) > aren't confused later: > > >>> def foo(): > ... a = 1 > ... def bar(n): > ... b = 2 > ... return n + b > ... a = bar(a) > ... print a > ... > >>> foo() > 3 > >
One problem with short examples is they mask the reason for the code to be structured that way. I assumed that the OP was really talking about a closure, and that sharing that variable was deliberate. I seldom write nested functions otherwise. -- DaveA -- http://mail.python.org/mailman/listinfo/python-list
