Re: Confusion about inheritance and the allocating memory via new
<tharringtonan@netscape.net> wrote in message
news:1185735763.960124.213490@19g2000hsx.googlegroups.com...
On Jul 29, 2:41 pm, Robert Bauck Hamar <roberth+n...@ifi.uio.no>
wrote:
tharringto...@netscape.net wrote:
I am compiling the following code, main.cpp, as follows:
gcc main.cpp
I get the following compile error:
main.cpp: In function `int main()':
main.cpp:28: no matching function for call to `CPolygon::area()'
The code is listed below:
#include <iostream>
#include <ostream>
using namespace std;
class CPolygon {
public:
virtual void set_values (int a, int b) { width=a; height=b; };
protected:
int width, height;
};
class CRectangle: public CPolygon {
public:
int area () { return (width * height); };
};
int main ()
{
CPolygon * ppoly1 = new CRectangle;
This creates a ppoly1 to be a pointer. It has two types:
Its _static_ type is pointer to CPolygon.
Its _dynamic_ type is pointer to CRectangle.
The difference between static and dynamic types, is that static types is
what the compiler knows at compile time. The dynamic type is a run time
property.
ppoly1->set_values (4,5);
cout << ppoly1->area() << endl;
Here you call a member function of *ppoly. To accomplish this, the
compiler
searches only the static type of *ppoly, that is CPolygon, which has no
member 'area'.
return 0;
}
Why does this not compile?
If ppoly1 had been changed to point to a CPolygon object, what should
have
happened then?
--
rbh- Hide quoted text -
- Show quoted text -
This is the explanation that I was looking for.
Here is my problem in general.
I am writing a molecular dyanmics code. The user of the program will
set in the input file the type of boundary conditions. The boundary
conditions are represented by objects in the code. The important fact
are that there are several types of boundary conditions and they all
share many similar methods. In addition each boundary condition has a
few distinct methods.
So I figured I would write a base class so I could reuse the methods
that are shared by the numerous boundary conditions. However, I seem
to have to declare the methods that aren't shared by the boundaruy
conditions as virtual ...=0 in the base class.
Is there any way around of having to declare the nonshared methods as
virtual ... = 0 in the base class?
As BobR has shown, the way to call a method in a derived class that is not
in the base class is by using dynamic_cast to the derived class and ensure
that the returned pointer is not 0.
Such as the code he has shown, or:
CRectangle Foo* = dynamic_cast<CRectangle*>( ppoly1 );
if ( Foo )
cout << Foo->area() << endl;
However... this puts the burden on checking on the derived class type to the
calling code. It is forseen that you may have rectangle, square, octogon,
hexagon, triangle, polygon, etc.. and it is not a good thing to have to
check for each differnet type.
But, consider, all polygons have an area. Even though each may be
calculated differently, each differnet polygon will probably have an area()
method. This being the case, it makes sense to put a pure virtual method
("virtual ...=0") in the base class. Then the calling code doesn't have to
care about what actual shape it is, only that it is a polygon. Simply
adding a method
int area() = 0;
to your base class will allow your code to compile and is the appropriate
OOP thing to do.