Basic derived class member override. Challenge Activity 11.3 - c++

Question

I have a challenge for the school that states
Define a member function PrintAll() for class PetData that prints output as follows. Hint: Make use of the base class's PrintAll() function.
Name: Fluffy, Age: 5, ID: 4444
Then they give me the code below with a small snippet that I can alter. Between
// FIXME: Add PrintAll() member function
and
/* Your solution goes here */
is the code I added but I'm not getting the correct result.
MyCode
// FIXME: Add PrintAll() member function
void PrintAll() {
AnimalData data;
data.PrintAll();
cout << ", ID: " << idNum;
}
/* Your solution goes here */
Full Code
#include <iostream>
#include <string>
using namespace std;
class AnimalData {
public:
void SetName(string givenName) {
fullName = givenName;
};
void SetAge(int numYears) {
ageYears = numYears;
};
// Other parts omitted
void PrintAll() {
cout << "Name: " << fullName;
cout << ", Age: " << ageYears;
};
private:
int ageYears;
string fullName;
};
class PetData: public AnimalData {
public:
void SetID(int petID) {
idNum = petID;
};
// FIXME: Add PrintAll() member function
void PrintAll(){
AnimalData data;
data.PrintAll();
cout << ", ID: " << idNum;
}
/* Your solution goes here */
private:
int idNum;
};
int main() {
PetData userPet;
userPet.SetName("Fluffy");
userPet.SetAge (5);
userPet.SetID (4444);
userPet.PrintAll();
cout << endl;
return 0;
}
Results I'm getting
Name: , Age: -502747520, ID: 4444
Results I want
Name: Fluffy, Age: 5, ID: 4444

Answer 1

in
void PrintAll(){
AnimalData data;
data.PrintAll();
cout << ", ID: " << idNum;
}
AnimalData data; creates a new, default-initialized AnimalData that is entirely separate from the AnimalData that is part of the current PetData. Since you want to use the AnimalData you already have, discard this line and remove data from the next. This is where things can go very wrong, so I'm going to skip directly to the correct answer: To get the PrintData function for AnimalData, you need to be explicit:
void PrintAll(){
AnimalData::PrintAll();
cout << ", ID: " << idNum;
}
Why You Cannot Simply Remove data
If you remove the object to invoke PrintAll on, data,
void PrintAll(){
PrintAll();
cout << ", ID: " << idNum;
}
this is assumed. this is a PetData, and for PetData the best match for the PrintAll function is the current function. The result is infinite recursion.
The real lesson to be learned here is to be careful when re-using an identifier. In this case, PetData::PrintAll shadows AnimalData::PrintAll, replacing it inside PetData. The function is not virtual, and you do not want virtual in this case, so you do not get an override. You can easily and accidentally do this with functions and variables, leading to confusion about which one is being used by the program.
A (possibly) Better Way to Structure This Program
Take advantage of both inheritance and polymorphism
#include <iostream>
#include <string>
using namespace std;
class AnimalData
{
public:
virtual ~AnimalData() {}; // with polymorphism you must make sure the
// correct destructor is always called.
// Derived classes will override this
// destructor whether you explicitly define
// them or not.
void SetName(string givenName)
{
fullName = givenName;
}
void SetAge(int numYears)
{
ageYears = numYears;
}
// virtual function. Derived classes can, but do not have to, replace
// this function with a version better suited to the derived class
virtual void PrintAll()
{
cout << "Name: " << fullName;
cout << ", Age: " << ageYears;
}
private:
int ageYears;
string fullName;
};
class PetData: public AnimalData
{
public:
void SetID(int petID)
{
idNum = petID;
}
// Replacing virtual function. Note if the base class function is
// virtual, then child class overrides are automatically virtual
void PrintAll() override // override keyword notifies with a compiler
// error if the function does NOT override when
// it should.
{
AnimalData::PrintAll(); // call base class version for basic data
cout << ", ID: " << idNum; // adding derived class-specific stuff
}
private:
int idNum;
};
// can add WildData here to handle wild animals.
int main()
{
PetData userPet;
userPet.SetName("Fluffy");
userPet.SetAge(5);
userPet.SetID(4444);
userPet.PrintAll();
cout << endl;
// add an vanilla animal for demo
AnimalData generic;
generic.SetName("Fang");
generic.SetAge(7);
generic.PrintAll();
cout << endl;
// demonstrate polymorphism
AnimalData * ptr = & generic;
ptr->PrintAll();
cout << endl;
ptr = & userPet;
ptr->PrintAll(); // runtime calls the correct PrintAll function
cout << endl;
return 0;
}

Answer 2

You appear to have a small misunderstanding about how to invoke methods of the superclass AnimalData in PetData:
void PrintAll(){
AnimalData data;
data.PrintAll();
cout << ", ID: " << idNum;
}
This creates a new instance of the class AnimalData, and invokes PrintAll() on that object. Because the fullName and ageYears aren't initialised in this new object, you get the unexpected output. To invoke the same method of the superclass, the syntax is Superclass::Methodname(). So the correct version of AnimalData::PrintAll() is as follows:
void PrintAll(){
AnimalData::PrintAll();
cout << ", ID: " << idNum;
}

Answer 3

Here's your problem.
void PrintAll(){
AnimalData data;
data.PrintAll();
cout << ", ID: " << idNum;
}
This function creates a new object of class AnimalData (which will be constructed according to the default constructor) and then calls the AnimalData::PrintAll method on that object. What you wanted to do was call the AnimalData::PrintAll method on this object.
void PrintAll(){
AnimalData::PrintAll(); // call base class method
cout << ", ID: " << idNum;
}

Answer 4

So this one was alot simpler than I thought. It's really a matter of calling the printAll method from PetData, and appending the ID num.
public void printAll(){
super.printAll();
System.out.print(", ID: " + idNum);
}