Calling derived class' function from base class pointer

Question

If a function is overloaded in a derived class but the function signature is unchanged, is there any way to call the derived class' function from a base class pointer? For instance:

Given the following base and derived classes

class Base {
public:
    Base(int n): n_(n) {}
    ~Base() {}
    int get_n() {
    return n_;
    }
private:
    const int n_;
};

class Derived : public Base {
public:
    Derived(int n): Base(n), n_(n) {}
    ~Derived() {}
    double get_n() {
    return 1.5*n_;
    }
private:
    const int n_;
};

Is there any way to get the following to output 1.5 given that two get_n() functions differ only in their return types?

#include <iostream>
int main() {
Base *b = new Derived(1);
std::cout << b->get_n() << std::endl;
}

You are changing the return type of `get_n()` so even if you made it a virtual function it will still not work. You should look at templates maybe or change the classes design — , Jun 06 '14 at 01:38
http://stackoverflow.com/questions/21991133/forcing-late-method-resolution-in-case-of-class-inheritance-in-c — chris, Jun 06 '14 at 01:39
@Deduplicator, The return type is not part of the signature. — chris, Jun 06 '14 at 01:40
@glinka, Do note that as I say in my answer there, changing the design is a vastly superior solution. — chris, Jun 06 '14 at 01:42
Yes, I suspected this was the case. Thanks for finding the link. — glinka, Jun 06 '14 at 01:43

score 0 · Answer 1 · answered Jun 06 '14 at 01:37

0

Make your get_n virtual in base class. Like,

virtual int get_n() 
{
    return n_;
}

answered Jun 06 '14 at 01:37

Pranit Kothari

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Calling derived class' function from base class pointer

1 Answers1