C++ cin gets skipped, even after I use cin.ignore() - c++

Question

Just to make clear, I am very new to C++.
But I wrote I very small program to test my skill with arrays and ran into a problem with cin.
If the user enters number, like the program expects them to, all is well. But if a string gets entered, all input is skipped and the program ends.
I set up all of my inputs like this: cin >> x;cin.clear();cin.ignore();
So what is awry??
Here is the full code:
#include <iostream>
#include <cstdlib>
using namespace std;
int main()
{
system("cls");
int create = 1;
int entry;
int x;
string chc;
cout << "How long should the array be?" << endl;
cout << ":";
cin >> x;cin.clear();cin.ignore();
if(x<1){x=1;}
int myArray[x];
string askcontinue;
for(int x=0;x<sizeof(myArray)/sizeof(myArray[0]);x++){
system("cls");
cout << "Enter value #" << x+1 << endl;
cout << ":";
cin >> entry;cin.clear();cin.ignore();
myArray[x]=entry;
}
system("cls");
cout << "Index - Value" << endl;
for(int x=0;x<sizeof(myArray)/sizeof(myArray[0]);x++){
cout << x << " ------ " << myArray[x] <<endl;
}
system("cls");
cout << "Restart? [Y/N]" << endl;
cout << ":";
cin >> chc;cin.clear();cin.ignore();
if(chc=="y" || chc=="Y"){main();}
}

Answer 1

cin >> x;cin.clear();cin.ignore();
This thing that you're doing throughout your program is part of the problem. If the user enters something that doesn't meet the formatting requirements for an integer, the stream goes into a failure state. Directly after that happens you clear the stream and discard the next character. If the user entered in more than one character as part of the invalid input, the ignore() call is simply discarding the next character, but not all of the invalid input.
You need to check if the input did not succeed, and then discard the input using the overload of ignore() that takes the number of characters you wish to discard. Do the following if you wish to consistently ask the user for input if he does not provide valid characters:
while (!(std::cin >> x)) {
std::cout << "How long should the array be?" << std::endl;
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
}
But judging from your code, it doesn't look like you want to repeatedly ask the user for input. In that case, you should check for valid input instead and do nothing in the invalid case:
if (std::cin >> x) {
...
}
Also, VLAs (or variable-length arrays) are a non-standard feature of C++, provided as extentions in some compilers. Don't use them. Instead, allocate dynamically by using std::vector:
std::vector<int> myArray(x);

Answer 2

NOTE: you should also change the fact that you defining the variable 'x' three times
the problem you are having, is that c input does not type checking, so it does not care what was entered, so this is up to you. You should input everything as a string, and then make sure that the string contains nothing but numbers, THEN you can use std::stoi, or whatever the appropriate conversion method is. if they DO NOT enter a valid number, then you can just say INVALID, and tell the user to enter a valid number, and go back to the input, you could use something such as:
system("cls");
cout << "Enter value #" << x + 1 << endl;
cout << ":";
cin >> entry; cin.clear(); cin.ignore();
while(!is_valid_integer(entry))
{
system("cls");
cout << "INVALID NUMBER \n Enter value #" << x + 1 << endl;
cout << ":";
cin >> entry; cin.clear(); cin.ignore();
}
myArray[x] = std::stoi(entry);
And then entry is a string.
is_valid_integerwould be defined as:
bool is_valid_integer(std::string str)
{
for(auto it : str)
{
if(!(ch == '0' || ch == '1' || ch == '2' || ch == '3' || ch == '4' || ch == '5' || ch == '6' || ch == '7' || ch == '8' || ch == '9'))
return false;
//OR: this is more efficient, but is reliant on using ascii codes (which in this case we are)
//if(!(ch >=48 && ch <= 57)) return false;
}
return true;//all numbers
}