Asking user for const - c++

Question

There is my friend's code. It works, but we would like to ask our user how many times he wants to type informations. Simplier, We don't know how to ask my user for N ("const int N = 3" line). We've tried changing "const int" into "int", but then an error shows up "expression must have a constant value".
#include <iostream>
using namespace std;
struct T_dane_ksiazki
{
char imie[15];
char nazwisko[30];
char tytul[45];
int rokwydania;
int nrwydania;
};
void WCZYTAJ_dane(T_dane_ksiazki& dane) /*wczytanie informacji o książce*/
{
cout << "\nimie autora: ";
cin >> dane.imie;
cout << "nazwisko autora: ";
cin >> dane.nazwisko;
cout << "tytul ksiazki: ";
cin >> dane.tytul;
cout << "rok wydania: ";
cin >> dane.rokwydania;
cout << "numer wydania: ";
cin >> dane.nrwydania;
}
void WYSWIETL_dane(T_dane_ksiazki dane) /*wczytanie informacji o książce*/
{
cout << "\nimie autora: " << dane.imie;
cout << "\nnazwisko autora: " << dane.nazwisko;
cout << "\ntytul ksiazki: " << dane.tytul;
cout << "\nrok wydania: " << dane.rokwydania;
cout << "\nnumer wydania: " << dane.nrwydania << "\n";
}
const int N = 3;
int Zapytajka()
{
cout << N<< "\n";
return 1;
}
int main()
{
T_dane_ksiazki daneq[N];
Zapytajka();
cout << "Podaj informacje o ksiazkach: \n";
for (int i = 0; i<N; i++)
{
WCZYTAJ_dane(daneq[i]);
}
cout << "\n\nInformacje o ksiazkch: \n";
for (int i = 0; i<N; i++)
{
WYSWIETL_dane(daneq[i]);
}
cout << "\nKoniec programu. Nacisnij ENTER";
cin.ignore(); cin.get();
return 1;
}

Answer 1

C++ only supports arrays whose size is constant. However, if you want a non-constant size, you can use vector, which was designed specifically for this purpose.
Example:
#include <vector>
...
int main()
{
Zapytajka();
cout << "Podaj informacje o ksiazkach: \n";
int n;
cin >> n;
std::vector<T_dane_ksiazki> daneq(N); // moved here and modified
for (int i = 0; i<N; i++)
{
WCZYTAJ_dane(daneq[i]);
}
...
}
I changed an array to a vector, and moved its definition to after the value of n is determined. I also changed N to n because it's no longer a constant, and it's a common convention to allocate lower-case names to variables.

Answer 2

Edit: I missed that it is a C++ question. The answer for C++ is that you should use a C++ container, probably a vector. These containers are the reason that the variable length arrays introduced in C in 1999 discussed below do not exist in C++: There is no real need for them. (There is a discussion whether to introduce something like it but it has non-trivial implications for the type system.)
You can have variable length arrays in C99 programs (for gcc: compile with "-std=c99"). That is, you can make N a non-const and the program should still compile and run properly (I didn't check every detail, but it looks pretty straightforward).
In pre-99 C you have to allocate dynamically with malloc or simply define an array which is big enough for the biggest conceivable number and use only part of it.