On Tue, 09 Oct 2007 19:23:37 +0200, Diez B. Roggisch wrote: > Your believes aside, this is simply wrong. The statement > > a += x > > always leads to a rebinding of a to the result of the operation +.
Not true. >>> L = [] >>> id(L) 3083496716L >>> L += [1] >>> id(L) 3083496716L It's the same L, not rebound at all. > I presume you got confused by the somewhat arbitrary difference between > > __add__ > > and > > __iadd__ > > that somehow suggest there is an in-place-modification going on in case > of mutables. The __iFOO__ methods are supposed to do in-place modification if possible, but it is up to the class writer to make them do so. In the case of your example, you specifically created an __iadd__ method that didn't even attempt in-place modification. What did you expect it to do? > but as the following snippet shows - that's not the case: > > > class Foo(object): > def __add__(self, o): > return "__add__" > > def __iadd__(self, o): > return "__iadd__" > > > a = Foo() > a += 1 > print a > > a = Foo() > b = Foo() > c = a + b > print c > > So you see, the first += overrides a with the returned value of > __iadd__. That's because you told it to do that. If you told it to do something more sensible, it would have done so. Lists know how to do in-place modification: >>> id(L) # from above 3083496716L >>> L *= 5 >>> id(L) 3083496716L >>> L = L*5 >>> id(L) 3083496972L -- Steven. -- http://mail.python.org/mailman/listinfo/python-list
