Wrong Inputs will cause the program to exit

Question

Look at this code:

#include <iostream>
using namespace std;
int main()
{
    string s;
    int n;
    float x;
again:
    cout << "Please Type this: ABC456 7.8 9 XYZ\n";
    cin >> s >> n >> x >> s;
    cout << "\nDo you want to try Again(y/n)? ";
    char t;
    cin >> t;
    if (t=='y' || t=='Y')
        goto again;
    return 0;
}

Just try to type "ABC456 7.8 9 XYZ" and press enter, it will cause the program exit before prompting the user to try again. I know the inputs are wrong and they are not in their types, but why it causes the exit? and how to avoid such exit?

Answer 1

Change

cin >> s >> n >> x >> s;

to

cin >> s >> x >> n >> s;

As you are entering 7.8 as 2nd input but you are collecting it in a integer variable instead of a floating point variable. As a result of this when you enter:

ABC456 7.8 9 XYZ

s gets ABC456, n gets 7 (as it is of int type and the input buffer still has .8 9 XYZ\n in it). Next n gets .8 and finally s gets "9". Now the input buffer has XYZ\n in it. The next time when you read the input into t to get the user's choice, X gets read into t and since it is not y or Y, the loop exits.