On Mon, Sep 7, 2009 at 7:21 PM, Henry 'Pi' James<henrypija...@gmail.com> wrote: > I've just found out that a subclass shares the class variables of its > superclass until it's instantiated for the first time, but not any > more afterwards: > > Python 3.1 (r31:73574, Jun 26 2009, 20:21:35) [MSC v.1500 32 bit > (Intel)] on win32 > Type "help", "copyright", "credits" or "license" for more information. >>>> class A: > ... n = 0 > ... def __init__(self): > ... type(self).n += 1 >>>> class B(A): > ... pass <snip> > This makes no sense to me at all. Could it possibly be a bug?
Makes sense to me. To step through what's happening: >>> A.n, B.n (0, 0) Here, the lookup on B.n fails (that is, B itself has no variable n), and thus falls back to A.n Thus, at this point, the expressions `A.n` and `B.n` are equivalent. >>> (A().n, A.n, B.n) (1, 1, 1) A.__init__() gets called, incrementing A.n; again, B.n falls back to A.n >>> (A().n, A.n, B.n) (2, 2, 2), <same thing> >>> (B().n, A.n, B.n) (3, 2, 3) A.__init__() gets called since B did not define one of its own and this inherited A's. Therein, type(self) evaluates to B (not A as before). B.n += 1 is in this case equivalent to: B.n = B.n +1 Evaluating the right side, B.n falls back A.n once again, and we add 1. Now the assignment takes place, creating a new variable B.n, *separate* from A.n From hereon in, lookups of B.n don't fall back to A.n since B now has its own variable 'n'. Thus, the values of A.n and B.n differ and the expressions now refer to 2 distinct variables. The rest falls out from this and is left as an exercise for the reader. Cheers, Chris -- http://blog.rebertia.com -- http://mail.python.org/mailman/listinfo/python-list
