C++ Academy

🔓Что выведет следующий код?

#include <iostream>
using namespace std;

class Base {
public:
    Base() {
        cout << "Base constructor: ";
        call();
    }
    virtual void call() { cout << "Base::call\n"; }
};

class Derived : public Base {
    int x = init();

    int init() {
        cout << "Derived::init\n";
        return 42;
    }

public:
    Derived() {
        cout << "Derived constructor\n";
    }

    void call() override {
        cout << "Derived::call, x = " << x << "\n";
    }
};

int main() {
    Derived d;
    return 0;
}

🔢Варианты ответа:

A)

Derived::init
Derived constructor

B)

Derived::init
Derived constructor

C)

Base constructor: Derived::call, x = 42
Derived constructor

D)

Derived::call, x = <undefined>
Derived::init
Derived constructor

✅Правильный ответ: B

💡Почему?
В момент вызова конструктора Base, объект ещё не стал Derived. Виртуальная функция вызывается в контексте Base.

www.tg-me.com/es/C Academy/com.cpluspluc/1034

4.2K viewsApr 16 at 11:01

🔓Что выведет следующий код?

#include <iostream>
using namespace std;

class Base {
public:
    Base() {
        cout << "Base constructor: ";
        call();
    }
    virtual void call() { cout << "Base::call\n"; }
};

class Derived : public Base {
    int x = init();

    int init() {
        cout << "Derived::init\n";
        return 42;
    }

public:
    Derived() {
        cout << "Derived constructor\n";
    }

    void call() override {
        cout << "Derived::call, x = " << x << "\n";
    }
};

int main() {
    Derived d;
    return 0;
}

Derived::init
Derived constructor

Derived::init
Derived constructor

Base constructor: Derived::call, x = 42
Derived constructor

Derived::call, x = <undefined>
Derived::init
Derived constructor

BY C++ Academy