polymorphism with reference and pointer object

[josh]

Hi I noticed that when I make a function with a base class parameter
that is a reference
than at run-time the compiler calls a base class function also if I
have passed a derived object.

But if I make that function with a pointer to a base class and then I
pass the derived class
at run-time the compiler calls a derived class function thanks to
dynamic binding.

Now why this difference if a reference is like a pointer and it
contains also an address of????

Thanks!

How is hard to learn C++ ........................

 

[Rolf Magnus]

Hi I noticed that when I make a function with a base class parameter
that is a reference than at run-time the compiler calls a base class
function also if I have passed a derived object.

Yes, if it's not virtual.

But if I make that function with a pointer to a base class and then I
pass the derived class at run-time the compiler calls a derived class
function thanks to dynamic binding.

Yes, if it's virtual.

Now why this difference if a reference is like a pointer and it
contains also an address of????

There is no difference. If the function is virtual in the base class,
dynamic binding happens, no matter whether you call it through a pointer or
a reference to the object. If the function is not virtual, no dynamic
binding happens.

 

[Jim Langston]

Hi I noticed that when I make a function with a base class parameter
that is a reference
than at run-time the compiler calls a base class function also if I
have passed a derived object.

But if I make that function with a pointer to a base class and then I
pass the derived class
at run-time the compiler calls a derived class function thanks to
dynamic binding.

Now why this difference if a reference is like a pointer and it
contains also an address of????

Thanks!

How is hard to learn C++ ........................

Plese explain. The output of the following program for me is:
Derived
Derived
Derived

What is the output for you?

#include <iostream>
#include <string>

class Base
{
public:
virtual ~Base() {}
virtual void Foo() { std::cout << "Base\n"; }
};

class Derived : public Base
{
public:
virtual void Foo() { std::cout << "Derived\n"; }

};

void Foo( Base* b )
{
b->Foo();
}

void Bar( Base& b )
{
b.Foo();
}

int main()
{

Derived* dp = new Derived();
Derived d;

Foo( dp );
Foo( &d );
// Bar( dp ); // Won't Compile
// error C2664: 'Bar' : cannot convert parameter 1 from 'Derived *' to
'Base &'
Bar( d );

std::string wait;
std::getline( std::cin, wait );
}

 

[josh]

Plese explain. The output of the following program for me is:
Derived
Derived
Derived

What is the output for you?

#include <iostream>
#include <string>

class Base
{
public:
virtual ~Base() {}
virtual void Foo() { std::cout << "Base\n"; }

};

class Derived : public Base
{
public:
virtual void Foo() { std::cout << "Derived\n"; }

};

void Foo( Base* b )
{
b->Foo();

}

void Bar( Base& b )
{
b.Foo();

}

int main()
{

Derived* dp = new Derived();
Derived d;

Foo( dp );
Foo( &d );
// Bar( dp ); // Won't Compile
// error C2664: 'Bar' : cannot convert parameter 1 from 'Derived *' to
'Base &'
Bar( d );

std::string wait;
std::getline( std::cin, wait );

}

Damn! in my code I have two friend functions with overloaded operators
<<
in which I have a derived class parameter. Than I make
out << static_cast<Base Class>(parameter) and than is called the <<
base class
operator. I have forgotten that friend functions cannot be virtual!

Excuse me!