On 5 April 2013 14:49, Candide Dandide <c.cand...@laposte.net> wrote: > Until now, I was quite sure that the is operator acts the same as the id > builtin function, or, to be more formal, that o1 is o2 to be exactly > equivalent to id(o1) == id(o2). This equivalence is reported in many books, > for instance Martelli's Python in a Nutshell. > > But with the following code, I'm not still sure the equivalence above is > correct. Here's the code : > > > #-------------------------------------------------------- > class A(object): > def f(self): > print "A" > > a=A() > print id(A.f) == id(a.f), A.f is a.f > #-------------------------------------------------------- > > > outputing: > > True False > > So, could someone please explain what exactly the is operator returns ? The > official doc says : > > The ‘is‘ operator compares the identity of two objects; the id() function > returns an integer representing its identity (currently implemented as its > address).
And the doc is right! >>> Af = A.f >>> af = a.f >>> print id(Af) == id(af), Af is af False False You've fallen victim to the fact that CPython is very quick to collect garbage. More precisely, when Python interprets `id(A.f) == id(a.f)`, it does the following: 1. Create a new unbound method (A.f) 2. Calculate its id 3. Now the refcount of A.f is down to 0, so it's garbage collected 4 Create a new bound method (a.f) **and very probably use the same memory slot as that of A.f** 5 Calculate its id 6 ... -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
