Problem creating an array of "class"

Roberto64_Ge

Hi I'm getting problem running the sw I attach here below. I am trying to make an array of a certain class. In the header I have the class then a .cpp with public methods. In the main I'm doing this:

#include "Anagrafica.h"
#include
#include
#include
using namespace std;

void cod (anagrafica arr,string n, string c,int chiave)
{ arr.codifica(n,c,chiave); }

void decod (anagrafica arr)
{ arr.decodifica(); }

int main()
{
ifstream file;
file.open("input.txt");
int N;
string n,c;

int chiave;
int chiave_input;
file>>N;
cout<>chiave;
file>>n;
file>>c;
//Arr_[i].setChiave(chiave);
//cod(Arr_[i],n,c,chiave);
}
for (int i=0; i>chiave_input;

}

here is header

#include
#include
using namespace std;

class anagrafica
{
private:

string nome_cifrato;
string cognome_cifrato;
int chiave_cifratura;

public :

anagrafica(string ="", string ="", int=0); //costruttore

void setNome(string );

void setCognome(string );

string getNome();

string getCognome();

void codifica(string , string , int );

void decodifica();

void setChiave(int);

int getChiave();

void stampa();

};

I do not attach the public method here because problem is before using methods. Problem occurs when I create instance of array of class anagrafica. At the end that is not an array. Program does not raise any exception in the for with i from 1 to N. As if array was of dimension N. But it isn't. I tried also with the double pointer like this

anagrafica** Arr_ = new anagrafica*[N];

for (int i = 0; i < N; i++)
{
Arr_[i] = new anagrafica("ciao", "miao", i*25);
}

but Arr_ is of only 1 element, the first no other :confused:

Mircea Neacsu

Roberto64_Ge wrote:

but Arr_ is of only 1 element, the first no other

How did you verify that? If you just looked in VisualStudio debugger, it will indeed show only one element because it doesn't know the size of the array. You can enter a length specifier in the inspector window (like Arr_,10) to make it show 10 elements. Alternatively you can just say:

anagrafica *Arr_[10];
for (int i = 0; i < N; i++)
{
Arr_[i] = new anagrafica("ciao", "miao", i*25);
}

In this case the debugger will know the size of the array and show all elements. Otherwise your code looks good as far as array creation goes.

Mircea

Roberto64_Ge

Problem is that I tried also with a costant as in your example but I could only see the first element. In the snippet I posted some parts of the code is commented. But when I remove comments app is working but the behaviour is that as if array was of 1 only element. I will read further replies tomorrow, thank you. ;) :zzz:

Lost User

The array variable needs to point to an array of pointers, so the syntax needs to be:

anagrafica** Arr_ = new anagrafica*[N]; // a pointer to an array of pointers

for (int i = 0; i < N; i++)
{
Arr_[i] = new anagrafica("ciao", "miao", i*25);
}

Remember that Arr_ is a pointer to the actual array so it will appear to contain only one element. But you can check the contents with another simple loop. You could make life much simpler for yourself by using one of the STL containers, such as vector.

Roberto64_Ge

ok, i better checked, so thank you. Now, how is the syntax to access the method of any instance (of the class) pointed by the i element af Arr_. I don't find the way to do that. I mean that I need to do something like Example : Arr_[i].setChiave(3) By the way, may I call a method , example setNome(stringaNome) from within another method?

Greg Utas

Arr_[i]**->**setChiave(3) EDIT: And sure, one method can call another, subject to access controls (public, protected, private).

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Lost User

Roberto64_Ge wrote:

Example : Arr_[i].setChiave(3)

Yes, as a simple test would show you.

Roberto64_Ge wrote:

may I call a method , example setNome(stringaNome) from within another method?

Yes again. All methods* within a class may be called from other methods, whether they are private or public. *But you cannot call non-static from static methods, as the non-statics need an object reference.

Roberto64_Ge

OK thanks. By the way, may I call a method , example setNome(stringaNome) from within another method? I attach here three methods of the class, the greater calling the two small.

void anagrafica::setNome(string n_)
{
nome_cifrato=n_;
}

void anagrafica::setCognome(string c_)
{
cognome_cifrato=c_;
}

void anagrafica::codifica(string nome_chiaro, string cognome_chiaro, int chiave)
{
char car;
nome_cifrato = "";
cognome_cifrato = "";
for (int i=0;i setNome(nome_cifrato);
for (int i=0; isetCognome(cognome_cifrato);
}

The two lines of code hghlighted have no effects.

Roberto64_Ge

if I use double pointer compiler don't let me use that syntax.

Lost User

What do you mean by, "double pointer"? If something does not work then please show the actual code and the error message(s).

Lost User

I have commented out all the parts of your code that are not needed to create a simple test. The result is as follows:

// header file
#include
#include
using namespace std;

class anagrafica
{
private:

string nome_cifrato;
string cognome_cifrato;
int chiave_cifratura;

public :

anagrafica(string a, string b, int c); //costruttore

// void setNome(string );

// void setCognome(string );

// string getNome();

// string getCognome();

// void codifica(string , string , int );

// void decodifica();

void setChiave(int);
void print();

// int getChiave();

// void stampa();

};

// implementation
#include "Anagrafica.h"
#include
#include
#include
using namespace std;

// void cod (anagrafica arr,string n, string c,int chiave)
// { arr.codifica(n,c,chiave); }

// void decod (anagrafica arr)
// { arr.decodifica(); }

anagrafica::anagrafica(string a, string b, int c) //costruttore
{
nome_cifrato = a;
cognome_cifrato = b;
chiave_cifratura = c;
}

void anagrafica::setChiave(int a)
{
chiave_cifratura = a;
}

// added to show the results
void anagrafica::print()
{
cout << nome_cifrato << " " << cognome_cifrato << " " << chiave_cifratura << endl;
}

int main()
{
// ifstream file;
// file.open("input.txt");
int N = 4; // create just four entries
//string n,c;

// int chiave;
// int chiave_input;
// file>>N;
// cout<print();
}

// for (int i=0; i>chiave;
// file>>n;
// file>>c;
// //Arr_[i].setChiave(chiave);
// //cod(Arr_[i],n,c,chiave);
// }
// for (int i=0; i>chiave_input;

}

Roberto64_Ge

i do not know how to quote a previous answer. Anyway you suggested me to use a pointer to an array of pointers so we did anagrafica **Arr_ = new anagrafica*[N]; But in this case compiler didn't let me access the method in this way : Arr_[i].setChiave(3); For that reason I asked for the correct syntax. Meanwhile I was waiting your reply I decided to try and so I went back to the first version temporarily and this is working, with respect to my first question " why I cannot see all the array elements?". With this I mean that, also with my first version, if I access Arr_[i] during the debug (within the relevant window "Control expression". I guess this is the correct translation from Italian) I see all the different instances of the class in each element of the array Arr_. I attach here the code of the first version. With this version running then I get a new malfunction that is that when I call the method for cifrating the true name and surname, from within this method I call the two methods to set the name and the surname, setNome and setCognome but this has no effects. If preferred I will attach the complete solution.

anagrafica* Arr_ = new anagrafica[N];

for (int i = 0; i < N; i++)
{
Arr_[i] = anagrafica("ciao", "miao", i*25);
}

for (int i=0; i>chiave;
file>>n;
file>>c;
Arr_[i].setChiave(chiave);
cod(Arr_[i],n,c,chiave);
}
for (int i=0; i

Lost User

Arr_[i].setChiave(3);

Sorry, I misread that. Arr_[i] is a pointer so you need to use the correct pointer dereferencing syntax:

Arr_[i]**->**setChiave(3);

Roberto64_Ge

Ok , so the syntax is not Arr_[i].setChiave(3) but

Arr_[i]->setChiave(3)

as per Greg Utas and your new indication. As I wrote I would say that also the version

anagrafica* Arr_ = new anagrafica[N];

for (int i = 0; i < N; i++)
{
Arr_[i] = anagrafica("ciao", "miao", i*25);
}

is working the same. Then I have the further problem of the two methods for setting cifrated name and surname tha have no effects. I see that there is some problem of synchronizations with the replies. But ok syntax has been clarified.

Roberto64_Ge

Ok i realized that. thank you very much. What about the other problem? Methods ....

Roberto64_Ge

Clarified the syntax problem I resume here. The code works also in this way ::

anagrafica* Arr_ = new anagrafica[N];

for (int i = 0; i < N; i++)
{
Arr_[i] = anagrafica("ciao", "miao", i*25);
}

for (int i=0; i>chiave;
file>>n;
file>>c;
Arr_[i].setChiave(chiave);
cod(Arr_[i],n,c,chiave);
}
for (int i=0; i

Because I can see all the Arr_ elements containing instances of the class. But two methods recalled from one other method do not heve effects.

Lost User

You have declared a number of methods in your header file but you have not added their implementation to the .cpp source. In my sample code above you can see the implementation for the setChiave method. You need to do the same for all the other methods declared in the class.

Roberto64_Ge

I attach the method file .cpp

#include "Anagrafica.h"
#include
#include

using namespace std;

anagrafica::anagrafica(string name,string surname, int key)
{
nome_cifrato=name;
cognome_cifrato=surname;
chiave_cifratura=key;
}

void anagrafica::setNome(string n_)
{
nome_cifrato=n_;
}
void anagrafica::setCognome(string c_)
{
cognome_cifrato=c_;
}

string anagrafica::getNome()
{
return nome_cifrato;
}
string anagrafica::getCognome()
{
return cognome_cifrato;
}

int anagrafica::getChiave()
{
return chiave_cifratura;
}

void anagrafica::setChiave(int c)
{
chiave_cifratura=c;
}

void anagrafica::stampa()
{
string nom = nome_cifrato;
string cog = cognome_cifrato;
cout<<" nome cifrato "<

I can't understand where is the

Lost User

I think this thread is getting confused (I certainly am). I suggest you start a new thread and post the actual code that you are now using, and explain what methods do not work.

Greg Utas

Those lines have no effect because the default in C++ is to pass an argument by value. That is, to make a copy of it. If you want to modify the data that is passed to the function, it must be passed by reference; another option is to pass a pointer to the argument. To pass by reference, you append a & to the argument's type. So, you need to change your two small functions like this:

void anagrafica::setNome(string**&** n_)
{
nome_cifrato=n_;
}

void anagrafica::setCognome(string**&** c_)
{
cognome_cifrato=c_;
}

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