Why is this code breaking on string input? - c++14

Question

#include <iostream>
using namespace std;
string flip();
int x;
int main()
{
for(int i=0;i<10;++i){
cout<<"Press 1 and enter"<<endl;
cin>>x;
if(isalpha(x)==true){
cout<<"int pls"<<endl;
}
else if(x==1){
flip();
cout<<"flipped"<<endl;
}
else if(x!=1){
cout<<"try again"<<endl;
}
}
system("pause>0");
return 0;
}
string flip(){
string ans;
int y=rand()%2;
if(y==0){
string ans = "Heads";
cout<<ans<<endl;
}
else{
string ans = "Tails";
cout<<ans<<endl;
}
return ans;
}
whenever I put 2 instead of 1 it works and says try again but when i write some string like "fa"
the code closes instead of writing try again
if I change x to int and then if I try to input some string it just prints press 1 and enter 10 times instead of asking for an input again

Answer 1

The type of variable 'x'(=The argument of isalpha()) has to be char. If the type of 'x' is integer, the function(=isalpha()) recognizes 'x' as an ASCII value. How about trying this code?
char x;
int main()
{
for(int i=0;i<10;++i){
cout<<"Press 1 and enter"<<endl;
cin>>x;
if((bool)isalpha(x)==true){
cout<<"int pls"<<endl;
}
else if(x=='1'){
flip();
cout<<"flipped"<<endl;
}
else {
cout<<"try again"<<endl;
}
}
system("pause>0");
return 0;
}