Can a class deriving from QSharedData use the default copy constructor?

Asperamanca

In the example in Qt docs, a copy constructor is explicitly defined for a trivial class with only copyable members.

Is this intentional? Or is it fine to declare the copy constructor as "=default"?

VRonin

As long as the QSharedData copy constructor is called (which should be guaranteed by the standard) =default works just as well

? Offline

In the example, Employee and EmployeeData are defined in the same header file. But now imagine you split the two parts and you have:

employee.h

#ifndef EMPLOYEE_H
#define EMPLOYEE_H

#include <QString>
#include <QSharedDataPointer>

class EmployeeData;

class Employee
{
  public:
    Employee();
    Employee(int id, const QString &name);
    Employee(const Employee &other);
    ~Employee();

    int id() const;
    void setId(int id);

    QString name() const;
    void setName(const QString &name);

private:
    QSharedDataPointer<EmployeeData> d;
};

#endif // EMPLOYEE_H

Now the compiler couldn't generate the copy constructor as the member QSharedDataPointer<EmployeeData> d is not fully known. So in this case, the usual case I'd say, you actually have to define / implement the (trivial) copy-constructor.

VRonin

I think we are talking about two different things.
Did you mean to =default the copy constructor for EmployeeData or the one for Employee?

? Offline

Oh. If you were talking about EmployeeData, yes, you can reduce that code to:

struct EmployeeData : public QSharedData
{
    int id = -1;
    QString name;
};

Asperamanca

Yes, I was talking about the private data class. Since all I was doing was faithfully copy every single member, "=default" deemed a viable alternative.

I have tried it and so far have not run into any issues. Together with your input, I feel confident this is safe to do so, and the example looks as it does just because it was written pre-C++11

Thanks!