Should I use QScopedPointer or std::unique_ptr?

Question

I\'m starting a new project, using Qt5 and QMAKE_CXXFLAGS += -std=c++1y. I\'m not sure whether I should prefer QScopedPointer or std::unique_ptr

Answer 1

Even though the main intention of both classes is similar,one important difference between those two which makes you to take some business decision.

QScopedPointer does not have an assignment operator. Where as std::unique_ptr has assignment operator.

If you want to be strict on your QT object/control and does not want an assignment use QScopedPointer.

Answer 2

QScopedPointer is strictly weaker than unique_ptr as it does not support move semantics.

Its functionality is otherwise extremely similar.

Move semantics are extremely useful, and accidentally using them incorrectly to cause problems is extremely rare. So they vary from harmless to (more typically) helpful.

About the only reason you should use QScopedPointer is interoperability with existing code bases; and even there, given how similar they are, an adapter would be pretty easy.

So if you don't need to adapt, use unique_ptr.

---

I will now discuss adapting.

The tricky part is the 2nd parameter to QScopedPointer. It very roughly corresponds to the 2nd parameter of unique_ptr.

In unique_ptr stateful deleters are permitted. In QScopedPointer they are not. The

static void cleanup(T* pointer)

corresponds to the

void operator()(T* pointer)const

in the unique_ptr in a pretty one-to-one basis. So:

template<class QDelete>
struct std_deleter {
  template<class T>
  void operator()(T* target) const {
    QDelete::cleanup(target);
  }
};

maps a Qt deleter to a std deleter. The other way is limited by the deleter being stateless:

template<class Std_deleter>
struct Qt_deleter {
  template<class T>
  static void cleanup(T* target) {
    static_assert(std::is_empty<Std_deleter>{}, "Only works with stateless deleters");
    Std_deleter{}(target);
  }
};

we can now convert:

template<class T, class D>
QScopedPointer<T, Qt_deleter<D>>
to_qt( std::unique_ptr<T, D>&& src ) {
  return src.release();
}
template<class T, class D>
QScopedPointer<T, Qt_deleter<D>>
to_qt( std::unique_ptr<T[], D>&& src ) {
  return src.release();
}
template<class T>
QScopedPointer<T>
to_qt( std::unique_ptr<T>&& src ) {
  return src.release();
}
template<class T>
QScopedPointer<T, QScopedPointerArrayDeleter>
to_qt( std::unique_ptr<T[]>&& src ) {
  return src.release();
}
template<
  class T, class D, class R=std::unique_ptr<T, std_deleter<D> >
>
to_std( QScopedPointer<T, D>&& src ) {
  return R(src.take()); // must be explicit
}
template<class T, class R=std::unique_ptr<T>>
to_std( QScopedPointer<T>&& src ) {
  return R(src.take()); // must be explicit
}
template<class T, class R=std::unique_ptr<T[]>>
to_std( QScopedPointer<T,QScopedPointerArrayDeleter >&& src ) {
  return R(src.take()); // must be explicit
}

which covers about the only reason why you'd use QScopedPointer. There are a few corner cases -- the default deleter QScopedPointer should be converted to a default std::unique_ptr and vice versa.

The array delete QScopedPointer should be converted to a unique_ptr<T[]> and vice versa.

In other cases, I simply wrap up the deleter. In theory, a really fancy trick would be to notice if the incoming deleter was already wrapped up and reverse the wrapping, but if your code is doing that many round-trips there is probably already something wrong.

Answer 3

Why would you use something that's not from the standard library compared to something from the standard library?

To me there's only one reason any good programmer would do that: If the external library provides something that the standard library doesn't provide. Is that the case?

Consider your program's portability and updates in the future, and then make that decision.