Il Sun, 22 Mar 2009 16:52:02 +0000, R. David Murray ha scritto: > mattia <ger...@gmail.com> wrote: >> Can you explain me this behaviour: >> >> >>> s = [1,2,3,4,5] >> >>> g = (x for x in s) >> >>> next(g) >> 1 >> >>> s >> [1, 2, 3, 4, 5] >> >>> del s[0] >> >>> s >> [2, 3, 4, 5] >> >>> next(g) >> 3 >> >>> >> >>> >> Why next(g) doesn't give me 2? > > Think of it this way: the generator is exactly equivalent to the > following generator function: > > def g(s): > for x in s: > yield x > > Now, if you look at the documentation for the 'for' statement, there is > a big "warning" box that talks about what happens when you mutate an > object that is being looped over: > > There is a subtlety when the sequence is being modified by the loop > (this can only occur for mutable sequences, i.e. lists). An > internal counter is used to keep track of which item is used next, > and this is incremented on each iteration. When this counter has > reached the length of the sequence the loop terminates. This means > that if the suite deletes the current (or a previous) item from the > sequence, the next item will be skipped (since it gets the index of > the current item which has already been treated). Likewise, if the > suite inserts an item in the sequence before the current item, the > current item will be treated again the next time through the loop. > > As you can see, your case is covered explicitly there. > > If you want next(g) to yield 3, you'd have to do something like: > > g = (x for x in s[:]) > > where s[:] makes a copy of s that is then iterated over.
Ok, thanks. Yes, I had the idea that a counter was used in order to explain my problem. Now I know that my intuition was correct. Thanks. -- http://mail.python.org/mailman/listinfo/python-list
