Is it possible to "inherit" a derived function from a base class function?

Question

Firstly, let me apologise for the title, I honestly couldn't think how to word it better. The example below should make my question a little clearer:

class Foo    {
    public:
      Foo(int x);
    };

class Bar : public Foo
{
public:
  Bar(int y) : Foo(y);
};

Demonstrates code that would force the Bar constructor to call the Foo constructor with parameter Y. My question is, is there a similar method for inheriting a derived class function to call a base class function?

For example all calls to Bar.Func(); would also automatically call Foo.Func();?

This has already been answered: stackoverflow.com/questions/357307/… — onit
– onit, Commented Nov 10, 2011 at 14:38

John Dibling · Accepted Answer · 2011-11-10 14:37:47Z

1

class Foo    {
    public:
      Foo(int x);
  void DoIt();
    };

class Bar : public Foo
{
public:
  Bar(int y) : Foo(y);
  void DoIt();
};

void Bar::DoIt()
{
  Foo::DoIt();
}

If Foo is intended to be derived from and used polymorphicly, you should declare DoIt as virtual. In addition, you will also want a virtual base class destructor.

answered Nov 10, 2011 at 14:37

John Dibling

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2 Comments

TVOHM Over a year ago

I apologise I wasn't exactly clear on this, I'm currently solving my problem exactly as you suggested with a manual call in Bar::DoIt to Foo::DoIt. As with the constructor example in my question, is there a way to do this in the class interface similar to the constructor Bar(int y) : Foo(y) Say for example void DoIt() : Foo::DoIt() ?

John Dibling Over a year ago

If I understand you correctly, then no. There is no initializer list for functions. You must call the function explicitly in the body of the function.

Cat Plus Plus · Accepted Answer · 2011-11-10 14:38:40Z

0

Automatically, no. You can use Base::member() to call it, though.

struct foo {
    void frob() { std::cout << "foo::frob" << std::endl; }
};

struct bar : foo {
    void frob() { foo::frob(); std::cout << "bar::frob" << std::endl; }
};

answered Nov 10, 2011 at 14:38

Cat Plus Plus

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Comments

NPE · Accepted Answer · 2011-11-10 14:37:55Z

0

Any public function declared in Foo is also accessible via Bar (the same function will get called). If you may need to override the function in Bar to alter its behaviour, make it virtual.

answered Nov 10, 2011 at 14:37

NPE

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Comments

Ian Haggerty · Accepted Answer · 2011-11-10 14:55:45Z

This is the code you'll need if you want to 'inherit' the function statements for a function F() from its base. Declare F() virtual in Foo, implement some statements there. Then declare F() in Bar and make a call to Foo::F() at the beginning of the implementation of Bar::F():

class Foo {
public:
  Foo(int i) { cout << "Foo(" << i << ") called." << endl; }
  virtual void F() { cout << "foo::F() called." << endl; }

  virtual ~Foo() {};
};

class Bar : public Foo {
public:
  Bar(int i) : Foo(i) { cout << "Bar(" << i << ") called" << endl; }
  void F()  { Foo::F(); cout << "bar::F() called" << endl; }
};

int main(int argc, char** argv) {
  Bar b(3);
  b.F();
}

gives

Foo(3) called.
Bar(3) called
foo::F() called.
bar::F() called

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