自定义QVariant麻烦

Question

I am discovering Custom Types with QVariant and tried to implement it in my project. The problem is, it seems like it creates an object with the default constructor when I thought it would use the copy constructor...

Maybe there is something I don't understand...

Here is my custom class :

Parameter.h

#ifndef PARAMETER_H
#define PARAMETER_H
#include <QDebug>
#include <QMetaType>

#include "variable.h"
#include <QAbstractTableModel>

class Parameter : public Variable, public QAbstractTableModel
{
public:
    Parameter();
    Parameter(QString name, int min, int max, int val, QObject *parent);
    Parameter(const Parameter &source);

private:
    int m_value;
};
Q_DECLARE_METATYPE(Parameter)

#endif // PARAMETER_H

Parameter.cpp

include "parameter.h"

Parameter::Parameter() : Variable()
{
    qDebug()<<"Default constructor";
    m_value = 0;
    setName("Test");
}

Parameter::Parameter(QString name, int min, int max, int val, QObject     *parent)
: Variable(name, min, max), QAbstractTableModel(parent)
{
    qDebug()<<"constructor";
    m_value = val;
}

 Parameter::Parameter(const Parameter &source)
     : Variable(source.getName(), source.getMin(), source.getMax()),
      QAbstractTableModel(),
      m_value(source.m_value)
{
    qDebug()<<"copy constructor";  
}

I create an instance of Parameter in a class MainWindow. The following code is an extract of the constructor of this class :

Parameter *param = new Parameter("Param",0,100,10, this);
QVariant v = QVariant::fromValue(param);
Parameter op = v.value<Parameter>();
qDebug()<< op.getName();

The output of this code is :

constructor
Default constructor 
Default constructor
"Test"

I would like to understand why the default constructor is called twice. And what I should do so that it calls the copy constructor instead (in order to get the object I created, whose name is "Param)

Thank you a lot for your answers :)

Answer 1

You store pointer in your QVariant instead of value. But want to retrieve value.

If the value cannot be converted, a default-constructed value will be returned.

so v.value<Parameter>(); returns a default-constructed object as Parameter* cannot be converted to Parameter .