On 7 dic, 22:53, Schüle Daniel <[EMAIL PROTECTED]> wrote: > In [38]: f = [lambda:i for i in range(10)] > In [39]: ff = map(lambda i: lambda : i, range(10)) > In [40]: f[0]() > Out[40]: 9 > In [41]: f[1]() > Out[41]: 9 > In [42]: ff[0]() > Out[42]: 0 > In [43]: ff[1]() > Out[43]: 1 > > I don't understand why in the first case f[for all i in 0..9]==9
In the first case, i is a free variable. That means that Python will get it from other place (the global namespace, likely [surely?]) >>> f=[lambda:i for i in range(10)] >>> f[0]() 9 >>> i=123 >>> f[0]() 123 >>> print f[0].func_closure None In the second case, the inner i is a free variable, but local to its enclosing scope (outer lambda). It's a closure: >>> ff = map(lambda i: lambda : i, range(10)) >>> ff[4]() 4 >>> ff[4].func_closure (<cell at 0x00ACEA90: int object at 0x00995344>,) >>> i=321 >>> ff[4]() 4 > what is different from (more usefull) > > In [44]: f = ["%i" % i for i in range(10)] > In [45]: f > Out[45]: ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'] This is a simple expression evaluated at each iteration > doing it like this works again > > In [54]: def d(x): > ....: return lambda:x > ....: x inside lambda is a free variable, but since it's local to d, this becomes a closure: >>> d(1) <function <lambda> at 0x00A35EF0> >>> d(1).func_closure (<cell at 0x00A2A4F0: int object at 0x00995338>,) > In [55]: f = [d(i) for i in range(10)] > In [56]: f[0]() > Out[56]: 0 > In [57]: f[1]() > Out[57]: 1 This is similar to the ff example above > in a C programmer sence I would say there seems to be no "sequence > point" which would separate "now" from "next" No, the problem is that C has no way to express a closure; this is a functional concept absolutely extraneous to the C language. -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
