Subject: [boost] [BOOST CORE C++ Feature] The super or base keyword for C++
From: Max Hölzer (max.hoelzer_at_[hidden])
Date: 2011-04-29 17:48:51
Hello,
i was wondering if it makes sens to add the well known super or base keyword to
the C++ language in a generic way.
Here I have a simple example on how it would make inherited base or super class
overloaded function calls easier.
Example 1:
class A
{
public:
virtual void print() {
std::cout << "A" << std::endl;
}
};
class B : public A
{
public:
virtual void print()
{
std::cout << "B" << std::endl;
std::cout << "super call "; A::print();
}
};
class C : public B
{
public:
virtual void print()
{
std::cout << "C" << std::endl;
std::cout << "super call "; B::print();
}
};
int main(int argc, char* argv[])
{
A a;
B b;
C c;
a.print(); // Prints A
b.print(); // Prints B and super call A
c.print(); // Prints C and super call B super call A
return 0;
}
Adding a template class superable would avoid one to write the explicit base or
super class name.
Example 2:
namespace boost
{
template<typename T>
class superable : public T
{
public:
typedef T super; // C++ and java style
typedef T base; // C# style
};
}
class A
{
public:
virtual void print() {
std::cout << "A" << std::endl;
}
};
class B : boost::superable<A>
{
public:
virtual void print()
{
std::cout << "B" << std::endl;
std::cout << "super call "; super::print(); // C++ and java style
call
}
};
class C : boost::superable<B>
{
public:
virtual void print()
{
std::cout << "C" << std::endl;
std::cout << "super call "; base::print(); // C# style call
}
};
int main(int argc, char* argv[])
{
A a;
B b;
C c;
a.print(); // Prints A
b.print(); // Prints B and super call A
c.print(); // Prints C and super call B super call A
return 0;
}
If somebody could add this to the core boost library, that would be great.
Sincerely yours,
Max