At 14:04 21.10.2003 -0700, Guido van Rossum wrote:
... BTW, while Alex has shown that a generator function with no free variables runs quite fast, a generator expression that uses variables from the surrounding scope will have to use the nested scopes machinery to access those, unlike a list comprehension; not only does this run slower, but it also slows down all other uses of that variable in the surrounding scope (because it becomes a "cell" throughout the scope).
The implementation could synthesize a generator function abusing default arguments to give the generator's frame locals with the same names.
Yes, I think that could work -- I see no way that something invoked by the generator expression could possibly modify a variable binding in the surrounding scope.
so this, if I understand:
def h(): y = 0 l = [1,2] it = (x+y for x in l) y = 1 for v in it: print v
will print 1,2 and not 2,3
def h(): y = 0 l = [1,2] def gen(S): for x in S: yield x+y it = gen(l) y = 1 for v in it: print v