On 26/04/12 19:01, André Somers wrote: > Op 23-4-2012 20:44, Nikos Chantziaras schreef: >> Then you're not doing what you think you're doing: >> >> QList< QList<int> > listOfLists; >> QList<int> listOfInts; >> listOfInts.append(10); >> >> listOfLists.append(listOfInts); >> listOfLists[0][0] = 9; >> >> qDebug()<< listOfLists[0][0]<< listOfInts[0]; >> >> You are modifying a copy, so it prints"9 10" instead of"10 10". This: >> >> listOfLists[0][0] = 9; >> >> modifies a copy of listOfInts. Also the reverse is true. If you modify >> listOfInts, then the copy of it inside listOfLists is not updated. >> >> "Implicit sharing" means that data is copied when it's modified. It's >> not a replacement for pointers. >> > Not true, in this case.
I posted code that proves my point. It prints "9 10". You can't argue with that one ;-) > Let's look at the signature of the operator that > Scott is using: > T & QList<T>::operator[](int i); > > Note the little & after the first T? That's right, it returns a > *reference*. And a reference is something you can directly modify. The > documentation for that method also explicitly states that: > > > Returns the item at index position/i/as a *modifiable* reference. > > Your statement that "implicit sharing" means that a copy is made when > data is modified, is way too general. That is only true if you actually > *have* a (shared) copy (which you don't in this case). Yes, it returns a reference to the copy. It doesn't return a reference to the original. References in cases like this are used for optimization purposes, so that less copying is involved. Bottom line is, it's still a copy. Without a pointer, there's no way to modify the original. And if you think your code is correct by having this wrong assumption, you're in for buggy surprises. _______________________________________________ Interest mailing list Interest@qt-project.org http://lists.qt-project.org/mailman/listinfo/interest
