Menu list with choices - c++

Question

I'm trying to make a choice list in a program and now I'm stuck with what happens when user writes incorrect input.
I've tried doing it with do loop (I've looped switch statement). Also, tried to set variable key to integer and char, but both have problems. Char only takes one character and when you write more than one it starts checking every by itself. Using int I get an infinite loop of menu function when inputting incorrectly.
Also, tried using cin.fail(), which I've found here, in stackoverflow, but it didn't work out.
void menu()
{
if(!isEmpty())
{
showStack();
}
else
{
cout << "stekas tuscias!" << endl;
}
cout << "---------------------------" << endl;
cout << "1. Iterpti i steka (push)" << endl;
cout << "2. Pasalinti is steko (pop)" << endl;
cout << "3. Issaugoti steka" << endl;
cout << "4. Issaugoti steka ir uzdaryti programa" << endl;
cout << "---------------------------" << endl;
cout << "Jusu pasirinkimas: ";
char key;
cin >> key;
switch(key)
{
case '1':
{
cout << "Irasykite reiksme i konsole: ";
int data;
cin >> data;
cout << endl;
push(data);
break;
}
case '2':
{
cout << endl;
pop();
break;
}
case '3':
{
write();
cout << "--------" << endl;
cout << "Stekas issaugotas!" << endl;
cout << "--------" << endl;
menu();
break;
}
case '4':
{
write();
break;
}
default:
{
cout << "Tokio pasirinkimo nera!" << endl;
cout << endl;
key = '\0';
menu();
break;
}
}
}
My whole code if it's needed: https://pastebin.com/Xv1HE0Mh

Answer 1

Use a try-catch-throw block. Use decltype or typeof feature to see if the input is an integer.

Answer 2

I'm sure why cin.fail() or (!cin) isn't working for you. But, you can do this. And, I'm also sure this kind of question exists.
template <class T>
int check (T ch)
{
if (ch >= 'a' && ch <= 'Z') return 0;
else return 1;
}
And, then.
main ()
{
.
.
char c;
cin.get (c); //Note here!!
if (check (c)) {...}
else {...} //You can use goto.
.
}