Object created within conditional If generates c++ build error

Question

I’m having trouble compiling a C++ program in IDE Code::Blocks. What I’m trying to do is access an object that was instantiated within an If parole. I will put down the codes of the main program and also the classes I created.

User class (parent class of Student and Teacher):

#ifndef UTILIZADOR_H
#define UTILIZADOR_H
#include <string>

using namespace std;

class Utilizador{

private:
    string name;
    int access_level;

public:

    //Setters
    void setName(string name);
    void setAccessLevel(int level);

    //Getters
    string getName();
    int getAccessLevel();
};

#endif // UTILIZADOR_H

Teacher class inherits from User:

#ifndef PROFESSOR_H
#define PROFESSOR_H
#include <string>
#include "Utilizador.h"

using namespace std;

class Professor : public Utilizador{

private:

public:
    //Constructor Method
    Professor(string name);

};

#endif // PROFESSOR_H

Student class inherits from User:

#ifndef ALUNO_H
#define ALUNO_H
#include <string>
#include "Utilizador.h"

using namespace std;

class Aluno : public Utilizador{

private:
    double *nota = new double; //Store an array of data with the scholar grades of the student during the year
public:
    //Constructor method
    Aluno(string name);

};

#endif // ALUNO_H

Below the code of the main program, where I try to implement the object:

#include <iostream> //Input and Output library
#include <windows.h>
#include <vector>
#include <string>
#include "include/Aluno.h"
#include "include/Professor.h"

using namespace std;

int main()
{
    //Below, this variables have the function of receive the input of data
    string nome;
    int nivel;

    //First contact with the user, request data entry
    cout << "Bem vindo ao programa de operacoes matematicas, vamos comecar!\n\n";
    cout << "Informe seu nome: ";
    cin >> nome; cout << endl;
    cout << "Agora informe qual o seu tipo de utilizador para poder oferecer-mos a melhor experiencia. (1) Professor / (2) Aluno: ";
    cin >> nivel; cout << endl;

    //Creating the object depending of the data that the user inserted in previous step
    if(nivel == 1){
        Professor *a = new Professor(nome);
        a->setAccessLevel(nivel);
    }else{
        Aluno *a = new Aluno(nome);
        a->setAccessLevel(nivel);
    }

    cout << a->getName(); //Linha que consta o erro de Compilação!

    return 0;
}

In the code of the main program you can notice that I create the object depending on the level of the user so I do it within the conditional if, but when compiling and executing the program the Code::Blocks informs the following:

"error: 'a' was not declared in this Scope".

I didn’t understand why he didn’t recognize that the object will be in any way instantiated in the variable a.

If I create the out-of-parole object works smoothly when I try to access the methods.

Answer 1

You are trying to use the variable a outside its scope. The scope of a local variable is only between opening brackets { and closure } of a block, starting in its statement from.

if(nivel == 1){
    Professor *a = new Professor(nome); // escopo inicia
    a->setAccessLevel(nivel);
} // escopo termina
// 'a' não está disponível aqui

likewise

else{
    Aluno *a = new Aluno(nome); // escopo inicia
    a->setAccessLevel(nivel);
} // escopo termina
// 'a' não está disponível aqui

To solve your problem, two points:

1. You must move your variable declaration to the same scope level as the line that has the code cout << a->getName();. This way, this code will have access to the requested variable.
2. Instead of two pointers for each derived class, you should have only one pointer that points to the base class. Within the conditionals, you then dynamically allocate memory to one of the derived classes.

With the changes:

int main()
{
    Utilizador *a;
    // ...
    // Creating the object depending of the data that the user inserted in previous step
    if (nivel == 1) {
        a = new Professor(nome);
    } else {
        a = new Aluno(nome);
    }

    a->setAccessLevel(nivel);
    cout << a->getName();
}