Hi, i want to propose new check. Recently i worked with QGraphicsItem and wanted to cast a pointer. Because i use Qt i prefer to use it's cast functions, for example qobject_cast. But somehow qgraphicsitem_cast did not work as i expected it should be for "_cast" function. My first guess was to reimplement the type() function as mentioned in documentation. But this did not help me too. Then after some time i figured out that qgraphicsitem_cast can only cast from QGraphicsItem pointer to real object type, you can't get nothing on "middle" of hierarchy with it. And i was trying to get item's parent. My mistake. To sum up thoughts i propose check if the type() reimplemented for item used with qgraphicsitem_cast. Without it it will not work. And also check if developer doesn't try to cast on "middle". This will also doesn't work. Yes, this is stupid mistake i did, but for someone such a check can potentially save several hours. Hope you will like my idea.
QGraphicsItem *item = ...; if you do qgraphicsitem_cast<Foo>(item); I can warn if Foo::type() is missing, indeed. But what's this about "middle" ? If Foo derives from Bar, and Bar::type() is also implemented, isn't this OK ? Sorry, I've never used qgraphicsitem_cast
See, this cast function is very tricky. Let's imagine such an hierarchy: class Bar : public QGraphicsItem { virtual int type() const Q_DECL_OVERRIDE {return Type;} enum { Type = UserType + 1}; }; class Foo : public Bar { virtual int type() const Q_DECL_OVERRIDE {return Type;} enum { Type = UserType + 2}; }; Foo *f = new Foo(); Bar *b = f; b->type() == f->type(); Because the type() function is virtual in both cases you will get the same result from Foo::type(). This looks obvious. But what if i want use qgraphicsitem_cast like this: QGraphicsItem *p = new Foo(); Bar *parent = qgraphicsitem_cast<Bar *>(p); It will not work! qgraphicsitem_cast can only cast to Foo in this case. If i want to cast to Bar i must not to reimplement the type() function in Foo.