Checking input on a char variable - c++

Question

I'm trying to validate input for a quit/return question in my program so the user is forced to enter either 'r' or 'q'.
I have managed to get it almost working. The problem is if the user enters 'r' or 'q' at the beginning followed by random letters then the program accepts that input. Any ideas on how to get the program to allow only a single 'r' or 'q' ?
void exit()
{
char choice;
bool badInput;
do
{
cout << "Press 'r' to return to the menu\nPress 'q' to quit the program\n\n" << endl;
cin >> choice;
badInput = cin.fail();
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
} while (badInput == true && choice == 'r' && choice == 'q' && (cin.peek() == EOF));
if (choice == 'q')
{
system("CLS");
cout << "Bye!\n";
system("PAUSE");
}
else if (choice == 'r')
{
system("CLS");
main();
}
else
{
exit();
}
}

Answer 1

You've got a very strange way of approaching this problem that's got a lot of issues. In particular, creating a function called exit() is problematic since that's a core function, and calling it recursively to try and get input is likewise not a good plan. You already have a loop, you just need to use it more effectively.
Likewise, main() is called automatically and you should never have reason to call it manually.
Here's a first pass rewrite:
void getInput()
{
char choice;
while (true)
{
cout << "Press 'r' to return to the menu\nPress 'q' to quit the program\n\n" << endl;
cin.clear();
cin.ignore(numeric_limits<streamsize>::max(), '\n');
cin >> choice;
switch (choice)
{
case 'q':
system("CLS");
cout << "Bye!\n";
system("PAUSE");
exit(0);
break;
case 'r':
system("CLS");
doMain();
break;
}
}
}
This obviously needs more work, but at least should theoretically function. In your original code you're demanding that the input value be simultaneously two different things which is an impossibility, that code will never work.
I'd also encourage you to stop doing things like system("CLS") and system("PAUSE") and instead do something in C++ natively. That's not portable code and it's terribly clunky as it depends on commands from 1980s DOS.

Answer 2

Handling the input
Most implementation use a buffered cin, so that input will only be returned once the user press enter. If this is not acceptable to you, you'll have to use OS-dependent functions.
If it's ok for you, then if you read a char, only the first one will be handed over: the remaining chars until the enter will wait for subsequent reads. Therefore, instead of reading a single char, I propose you to read a full line into a string:
void want_exit()
{
const string message="Press 'r' to return to the menu\nPress 'q' to quit the program\n";
string line;
cout << message << endl;
while (getline(cin, line) && line.length()!=1
&& tolower(line[0])!='y' && tolower(line[0])!='n')
{
cout << "Invalid input" <<endl << message <<endl;
line.resize(0);
}
Now line contains either a single valid char, or it is empty (in case of premature eof, meaning, that there was an input redirection and that there will anyway be no more input).
Processing the input
You can't call recursively main(): you should instead return from the function, and organize the calling function so to continue the process
if (line.empty() || tolower(line[0])!='y' ) {
system("CLS"); // This is non portable
cout << "Bye!\nPress enter...";
cin.getch(); // as good as system("PAUSE");
std::exit(0); // be aware of the naming issue !
}
return;
}
The calling function (main() ?) would then use it in a loop:
while (...) {
...
if (...)
want_exit();
}