Explicity pass base type

Question

I have an explicit function that takes a reference to the base type of a class. What is the proper way to pass that in?

I am currently doing a static cast:

#include <iostream>

using namespace std;

struct Base
{
    Base() { cout << "Base Constructor" << endl; }
    Base(Base const& c) { cout << "Base-Base Constructor" << endl; }
};

struct Derived : public Base
{
    Derived() { cout << "Derived Constructor" << endl; }
    explicit Derived(Base const& c) { cout << "Derived-Base Constructor" << endl; }
    Derived(Derived const& c) { cout << "Derived-Derived Constructor" << endl; }
};

int main()
{
    Base B;
    cout << "\n";
    Derived D;
    cout << "\n";
    Base* test1 = new Derived(D);
    cout << "\n";
    Base* test3 = new Derived(static_cast<Base>(D));
    cout << "\n";
    Base* test2 = new Derived(B);
    cout << "\n";


    return 0;
}

but that calls the copy constructor of the base class.

I could pass *static_cast<Base*>(&D), but that seems a bit hackish. I feel like I am just overlooking a simple way to do this. Thanks.

@MadScienceDreams lol. I knew it was simple. Thanks that works perfectly. — Cory-G, Aug 15 '14 at 17:07

score 5 · Accepted Answer · edited Aug 15 '14 at 19:00

5

Use this:

static_cast<Base&>(D)

Or this:

static_cast<const Base&>(D)

edited Aug 15 '14 at 19:00

Remy Lebeau

555,201
31
458
770

answered Aug 15 '14 at 17:13

IdeaHat

7,641
1
22
53

Explicity pass base type

1 Answers1