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🔓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


Derived::init
Derived constructor


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


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.

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Create: 2025-04-16
Last Update: 2025-06-18 10:44:57

🔓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


Derived::init
Derived constructor


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


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

✅

BY C++ Academy

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Date: 2025-06-18| C Academy

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