Program flow not affected by forward declaration - c++

Question

Here is the example of my forward declaration code which is not working:
void Register();
void Menu();
void Register() {
string username, password, hospitalname;
ofstream file;
file.open("Hospital Creds.txt");
cout << "Hospital Name: ";
cin >> hospitalname;
cout << "Username: ";
cin >> username;
cout << "Password: ";
cin >> password;
cout << endl;
file << hospitalname << endl;
file << username << endl;
file << password << endl;
system("pause");
system("cls");
Menu();
}
void Menu() {
int choice;
cout << " 1. Register Hospital " << endl;
cout << " 2. Register Donor " << endl;
cout << " 3. Exit " << endl;
cout << endl;
cout << "Please enter a number of your choice: ";
cin >> choice;
switch(choice) {
case 1:
system("cls");
Register();
break;
}
}
void main() {
Menu();
}
It's like this, when the program runs, it goes to the Menu(), then when selected, it goes to Register(), but when it wanna call back the function Menu(), the program exited. What's wrong with it?

Answer 1

If you call Menu() from Register(), the calling chain is never ending. The function are nested too deep. You're soon blowing up the call stack.
In practice, when I write a menu system, I never call previous-level functions from any function. Instead I use a loop to run the main menu infinitely, so I can keep my function call relationship like a tree.
void Sub1() {
// Prompt for input
if (option == 0) {
return; // End the call tree
}
else if (option == 1) {
Sub11();
}
else if (option == 2) {
Sub12();
}
...
}
int MainMenu() {
// Prompt for input
if (option == 0) {
return 0; // Quit program
}
else if (option == 1) {
Sub1();
}
else if (option == 2) {
Sub2();
}
}
....
int main() {
while (MainMenu() != 0);
}
Here's a sample project that I wrote. Note how I arranged the main() function.