JuniorTheoryCommon

What happens when you build and run this program?

C++

#include <iostream>

struct Base {
    virtual void print(int x) const { std::cout << "Base " << x; }
};
struct Derived : public Base {
    void print(long x) const override { std::cout << "Derived " << x; }
};

int main() { Derived d; const Base& b = d; b.print(1); }

It does not compile: print(long) overrides nothing in Base

Expecting Base 1: the compiler checks override; a function marked with it must override a virtual function of a base class, or the program does not compile.
Expecting Derived 1: an override needs the same parameter types, so print(long) is a different function from print(int), even though 1 converts to long.

What would the program print if the word override were removed?
Why is final sometimes written next to override?

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JuniorTheoryCommon

What happens when you build and run this program?

C++

#include <iostream>

struct Base {
    virtual void print(int x) const { std::cout << "Base " << x; }
};
struct Derived : public Base {
    void print(long x) const override { std::cout << "Derived " << x; }
};

int main() { Derived d; const Base& b = d; b.print(1); }

It does not compile: print(long) overrides nothing in Base

Expecting Base 1: the compiler checks override; a function marked with it must override a virtual function of a base class, or the program does not compile.
Expecting Derived 1: an override needs the same parameter types, so print(long) is a different function from print(int), even though 1 converts to long.

What would the program print if the word override were removed?
Why is final sometimes written next to override?

Back to questions Open topic in roadmap →Answer it yourself — Quiz