Re: factoring book question and cast question.
On Jan 31, 4:31 pm, ricec...@gehennom.invalid (Marcus Kwok) wrote:
Well, the idea is that since both kid1 and kid2 both inherit from base,
then they both should have a common interface, thus eliminating the need
to know the specific type:
#include <iostream>
#include <vector>
class Base {
public:
virtual void do_something() = 0;
virtual ~Base();
};
Base::~Base() { }
class Kid1 : public Base {
public:
virtual void do_something() { std::cout << "I am a Kid1\n"; }
};
class Kid2 : public Base {
public:
virtual void do_something() { std::cout << "I am a Kid2\n"; }
};
int main()
{
using std::vector;
typedef vector<Base*> Container;
Container container;
container.push_back(new Kid1);
container.push_back(new Kid2);
typedef Container::const_iterator ConstIter;
for (ConstIter it = container.begin(); it != container.end(); ++it) {
(*it)->do_something(); // no cast needed here!
}
typedef Container::iterator Iter;
for (Iter it = container.begin(); it != container.end(); ++it) {
delete *it;
}
return 0;
}
Output:
I am a Kid1
I am a Kid2
--
Marcus Kwok
That's neat, but strange.
If Kid1 and Kid2 are nodes in a list that do different things, I can
disguise what they do to make them have the same interface.
But the interface I invent, will branch to the different
functionality. Well, maybe this is good; (if done right) thanks!
todd.
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