QDebug, inheritance QObject issue ambiguous call

Dariusz

Hey

So this one is a bit unexpected. I decided to do another refactor of my class debug system. The base class works fine, but when it gets to printing inherited ones... it kinda flips...
Here is some code :


class base {
public:
     base();
     ~base();
     friend QDebug operator<<(QDebug stream, const base*object);
}

class inherA : public QObject, public base{
public
     inherA();
     ~inherA();
}

say I have

auto *class = inherA();
qDebug()<<class;

I'll get issue as c++ don't know which debug it should go for.... the native Qt one, or the one I implemented...

Woahz... any ideas on how to bite it?

TIA

jeremy_k

C++ friendship isn't inherited.

https://en.cppreference.com/w/cpp/language/friend note #2

Dariusz

@Dariusz said in QDebug, inheritance QObject issue ambiguous call:

friend QDebug operator<<(QDebug stream, const base*object);

Hmmmm thats good to know, but not sure if thats the issue?

This is the problem

base class:
friend QDebug operator<<(QDebug stream, const base*object);
Qt:
QDebug operator<<(QDebug stream, const QObject*object);
They both are viable options for the inheritedA class, and thust cause issue, for now I did qDebug()<<(base*)class; instead and that seems to work, but I do wonder if there is a more "proper" way of doing it.

jeremy_k

I was under the impression that friendship played a role in overload resolution, but am not finding a reference in https://en.cppreference.com/w/cpp/language/overload_resolution

Casting works. static_cast is generally preferred over the C-style (Type *).

Another option is to create an intermediate class that is closer to the final inherited class.

class Derived: public QObject, public base {};

QDebug operator<<(QDebug stream, const * Derived d) { return operator<<(stream, const_cast<const base *>(d)); }

class inherA: public Derived;

void test() {
  inherA instance;
  qDebug() << instance;
}

Dariusz

@jeremy_k Ohh I see, so re-implement debug call inside inherited class and do a cast there to correct type/etc/etc... nice! Will give it a go, thanks!

jeremy_k

Anything to make one definition more specific should work. My ability to recall the full set of rules is clearly less than perfect.