Reaching end of void function yield unexpected results - c++

Question

I wrote a simple program to manage a list of names (a segment of the program is below). I expected the function "choice()" to end and return to main()--and thus end the program--when the user's input for the variable "option" was 4. However, choice() tends to repeat a few times even after assigning 4 to "option", e.g:
Would you like to:
(1) - Add name(s) to the list
(2) - Remove name(s) from the list
(3) - View the list
(4) - Exit the program
(type 1, 2, 3, or 4): 4
Exiting Program...
Would you like to:
(1) - Add name(s) to the list
(2) - Remove name(s) from the list
(3) - View the list
(4) - Exit the program
(type 1, 2, 3, or 4): 4
Exiting Program...
ryan$ _
In the absence of the entire program, what do you think may be causing the problem? Sorry if there is not enough information to make a recommendation.
You can see I commented "exit(0)" from the end of the fourth if statement of choice()--in which choice() should reach its end and return to main(). I would prefer to avoid this statement but it is the only solution I can make work at the moment. Thanks for your help.
-Ryan
int main()
{
cout << endl
<< "WELCOME TO THE NAME LIST PROGRAM" << endl
<< "--------------------------------" << endl;
choice();
return 0;
}
void choice()
{
string option = "0";
cout << "\nWould you like to:" << endl
<< "\t(1) - Add name(s) to the list" << endl
<< "\t(2) - Remove name(s) from the list" << endl
<< "\t(3) - View the list" << endl
<< "\t(4) - Exit the program" << endl
<< "(type 1, 2, 3, or 4): ";
getline(cin,option);
while( option != "1" && option != "2" &&
option != "3" && option != "4" )
{
cout << "\nInvalid input, reenter: ";
getline(cin,option);
}
if(option == "1")
{
appendNames();
choice();
}
else if(option == "2")
{
string flag = "menu";
removeNames(flag);
choice();
}
else if(option == "3")
{
viewNames();
choice();
}
else if(option == "4")
{
cout << "\n\nExiting Program...\n\n";
// exit(0);
}
}

Answer 1

I think the problem comes from a misconception about a how a function works. Say that you have a function that shows your name:
void showName()
{
cout << "Ryan" << endl;
}
Now say that you want to repeat that 10 times. In your particular way of programming, you probably would do something like:
void showName(int times)
{
cout << "Ryan" << endl;
if ( times > 0 ) {
showName( --times );
}
return;
}
int main()
{
showName( 10 );
}
This way of programming is known as recursion, but I am quite sure that you have reached that by chance. Recursion is handy in some situations, but it is not the common way of doing simple things, like repeating something. Let's see the actual program calling the function 10 times:
void showName()
{
cout << "Ryan" << endl;
}
int main()
{
for(int i = 0; i < 10; ++i) {
showName();
}
}
Each time you call the function showName(), the callee, it has to return to the caller. If you do what you have shown in your example:
if( option == "1" )
{
appendNames();
choice();
}
Then you are actually calling choice() as many times as the user has chosen any other option apart from 4 (exit). This explains why you have to call exit( 0 ) in order to get an immediate effect.
Summarizing, your program should be:
int main()
{
cout << endl
<< "WELCOME TO THE NAME LIST PROGRAM" << endl
<< "--------------------------------" << endl;
while( !choice() );
return 0;
}
bool choice()
{
bool toret = false;
string option = "0";
cout << "\nWould you like to:" << endl
<< "\t(1) - Add name(s) to the list" << endl
<< "\t(2) - Remove name(s) from the list" << endl
<< "\t(3) - View the list" << endl
<< "\t(4) - Exit the program" << endl
<< "(type 1, 2, 3, or 4): ";
getline(cin,option);
while( option != "1" && option != "2" &&
option != "3" && option != "4" )
{
cout << "\nInvalid input, reenter: ";
getline(cin,option);
}
if(option == "1")
{
appendNames();
}
else if(option == "2")
{
removeNames();
}
else if(option == "3")
{
viewNames();
}
else if(option == "4")
{
cout << "\n\nExiting Program...\n\n";
toret = true;
}
return toret;
}
Hope this helps.

Answer 2

The problem is in your while loop. The user will have to keep entering inputs until an invalid input it given.
while( option != "1" && option != "2" &&
option != "3" && option != "4" )
{
cout << "\nInvalid input, reenter: ";
getline(cin,option);
}
Changing the loop condition or moving everything within the loop should do the trick.
Hope that helps. :)
P.S: If you're moving everything within the While loop, make sure you add a break; for the exit option.