Is possible in C++ to make this?
class Base {
int a(Derived d) { d.b(); }
};
class Derived : public Base {
int b();
};
Should i include Derived.hpp even in Base.hpp?
3 Answers
Is possible in C++ to make this?
Yes, it's very easy and a basic pattern (called polymorphism or Template Method Pattern) used in the c++ language:
class Base {
int a() { b(); } // Note there's no parameter needed!
// Just provide a pure virtual function declaration in the base class
protected:
virtual int b() = 0;
};
class Derived : public Base {
int b();
};
4 Comments
paweldac
You have to remember not to call such functions in constructor or desctructor, otherwise you'll end up with pure function call.
πάντα ῥεῖ
@paweldac Fortunately that's not the case in the OP's sample. If there's need to do so your answer fixes that problem.
paweldac
just wanted to make it clear to other readers that might use proposed code. True that in OPs question, this error won't come up :)
πάντα ῥεῖ
@paweldac I'm out of votes for today, sorry. Otherwise I would have upvoted your answer.
The following compiles:
class Derived;
class Base {
public:
int a(Derived d);
};
class Derived : public Base {
public:
int b() { return 42; }
};
int Base::a(Derived d) { d.b(); }
1 Comment
πάντα ῥεῖ
It compiles, yes. Though I'm afraid that's not what's being asked here (as you mentioned in the comment).
It's possible to call functions from Derived class from Base class with C++ idiom called: "Curiously recurring template pattern" CRTP. Please find call below:
template <class Child>
struct Base
{
void foo()
{
static_cast<Child*>(this)->bar();
}
};
struct Derived : public Base<Derived>
{
void bar()
{
std::cout << "Bar" << std::endl;
}
};
1 Comment
πάντα ῥεῖ
That's another way to do it, yes.
b()a pure virtual function inBase.