Slicing and Object

Questions 1 to 5 of 7

1. 1. What does this program print? ``cpp #include <iostream> #include <string> struct Animal { std::string name = "unnamed"; virtual ~Animal() = default; virtual std::string sound() const { return "..."; } }; struct Dog : Animal { std::string breed = "mixed"; std::string sound() const override { return "Woof " + breed; } }; int main() { Dog d; d.name = "Ace"; d.breed = "Pug"; Animal a = d; Animal& r = d; std::cout << a.name << ": " << a.sound() << "\n"; std::cout << r.name << ": " << r.sound() << "\n"; } ``

   Ace: Woof Pug then Ace: Woof PugAce: ... then Ace: ...Ace: ... then Ace: Woof Pugunnamed: ... then Ace: Woof Pug
2. 2. Why does a by-value Animal parameter slice a Dog argument?

   The function must be declared virtual before it can take a copyA new Animal is built from the Dog, so only the Animal part is copiedDog has no copy constructor, so the call is refusedThe compiler refuses to call sound on a copied object
3. 3. Which standard smart pointer has exactly one owner and deletes its object when that owner goes away? The lesson uses it to keep each whole Dog in a vector of Animal pointers.

   Type your answer.

4. 4. What does this program print? ``cpp #include <iostream> #include <memory> #include <string> #include <vector> struct Animal { std::string name = "unnamed"; virtual ~Animal() = default; virtual std::string sound() const { return "..."; } }; struct Dog : Animal { std::string breed = "mixed"; std::string sound() const override { return "Woof " + breed; } }; void show_value(Animal a) { std::cout << a.sound() << "\n"; } void show_ref(const Animal& a) { std::cout << a.sound() << "\n"; } int main() { Dog d; d.breed = "Beagle"; show_value(d); show_ref(d); std::vector<Animal> values; values.push_back(d); std::cout << values[0].sound() << "\n"; std::vector<std::unique_ptr<Animal>> owned; owned.push_back(std::make_unique<Dog>(d)); std::cout << owned[0]->sound() << "\n"; } ``

   Woof Beagle, Woof Beagle, Woof Beagle, Woof Beagle..., ..., ..., Woof Beagle..., Woof Beagle, ..., ......, Woof Beagle, ..., Woof Beagle
5. 5. Statement: C++ gives every class a default == that compares its fields, so a == b compiles with no code written in the class.

   True, because every C++ class inherits operator== from a hidden base classFalse, because C++ has no universal base class, so == exists only when the class author writes operator==