Doubt about inheritance and call to functions in base class

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Doubt about inheritance and call to functions in base class

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#1 by (2 votes)

-1

The doubt has come for Qt, but it is C ++.

If I have a class structure like this: (I only define eventFilter )

class DelegadoBase : public QStyledItemDelegate
{
    Q_OBJECT
public:

    explicit DelegadoBase(QObject* parent=nullptr);
    bool eventFilter(QObject *obj, QEvent* event);
};

And this derived class:

class DelegadoEditorNumeros : public DelegadoBase
{
public:
    explicit DelegadoEditorNumeros(QObject* parent=nullptr);
    QWidget * createEditor(QWidget * parent, const QStyleOptionViewItem&option, const QModelIndex&index) const;
    void setEditorData(QWidget * editor, const QModelIndex&index)const;
    void setModelData(QWidget * editor, QAbstractItemModel * model, const QModelIndex&index) const;
    void paint( QPainter *painter,const QStyleOptionViewItem &option, const QModelIndex &index ) const;
    QSize sizeHint( const QStyleOptionViewItem &option, const QModelIndex &index ) const;
    QString displayText(const QVariant & value, const QLocale & locale) const;

private:
    QRegExp* rx;
};

In the paint() method of the derived class, when I call the base class method, it is assumed that I should call the method paint() of DelegadoBase , and that it calls the method paint() of QStyledItemDelegate , but not being defined in DelegadoBase I get the doubt of how it would be right.

It would be something like this:

void DelegadoEditorNumeros::paint( QPainter *painter,const QStyleOptionViewItem &option, const QModelIndex &index ) const
{ 
if (condicion)
    {              
        haz las cosas asi
    else
    {
        DelegadoBase::paint(painter, option, index);//así
        //QStyledItemDelegate::paint(painter, option, index);o así
    }

c++ qt

asked by user3733164 13.11.2017 в 18:44

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1 answer

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score 2 · Accepted Answer

At the moment when, in a class, a method is declared as virtual , a table of function pointers associated with the class is created. In this table, redeclared calls are stored in each derived class.

If a function is not redeclared in the parent class but in the grandfather class, absolutely nothing happens, the compiler will consult the aforementioned table and call the first Function available:

If the function is redeclared in the parent , it will be called.
If it is not declared in the father but in the grandfather it will be called the grandfather's
If it is not declared in the grandfather but in the great-grandfather ...

Likewise, you can skip the implementation of the parent by calling Abuelo::funcion() directly.

A practical example:

struct A
{
  virtual void func1()
  { std::cout << "A::func1()\n"; }

  virtual void func2()
  { std::cout << "A::func2()\n"; }

  virtual void func3()
  { std::cout << "A::func3()\n"; }
};

struct B : A
{
  void func1() override
  { std::cout << "B::func1()\n"; }

  void func3() override
  { std::cout << "B::func3()\n"; }
};

struct C : B
{
  void func1() override
  {
    std::cout << "C::func1()\n";
    B::func1(); // Llama a la funcion del padre -> existe
  }

  void func2() override
  {
    std::cout << "C::func2()\n";
    B::func2(); // Llama a la funcion del padre -> no existe
  }

  void func3() override
  {
    std::cout << "C::func3()\n";
    A::func3(); // Llama directamente a la funcion del abuelo
  }
};

You can see it working here