while loop runs forever when a character is entered by user instead of an integer. How to stop it?

Question

Here's my program

#include<iostream>
#include<stdlib.h>//for system pause
int wish,i;
float x,y,z;
using namespace std;
int main()
{
while(true)
{ cout<<"\n\n Hello User what is your wish today\n\n";
cout<<" For Basic Maths enter 1\n\n";
cout<<" to exit enter 2\n\n";
cin>>wish;

cout<<"\nyou choose "<<wish<<endl;
if (wish==2)
{
cout<<"\n\n\n BYE BYE\n\n\n";
break;
}

switch (wish)
{
case 1:
cout<<"\n\n\nFor addition enter 1\n\n"<<
"for subtraction enter 2\n\n"<<
"for multiplication enter 3\n\n"<<
"for division enter 4\n\n"<<
"to return to previous menu 5\n\n";
cin>>i;
cout<<"\nYour Choice: "<<i;
switch (i){
case 1:
cout<<"\n\nEnter first number: ";cin>>x;
cout<<"\nEnter second number: ";cin>>y;
cout<<"\nresult is: "<<x+y;
break;
case 2:
cout<<"\n\nEnter first number: ";cin>>x;
cout<<"\nEnter second number: ";cin>>y;
cout<<"\nresult is: "<<x-y;
break;
case 3:
cout<<"\n\nEnter first number: ";cin>>x;
cout<<"\nEnter second number: ";cin>>y;
cout<<"\nresult is: "<<x*y;
break;
case 4:
cout<<"\n\nEnter first number: ";cin>>x;
cout<<"\nEnter second number: ";cin>>y;
cout<<"\nresult is: "<<x/y;
break;
case 5:
break;
default :
cout<<" which is Not Available";
}
break;
default :
cout<<" which is Not Available";
}
}
system("pause");
}

Answer 1

You could always do

if (wish != 1)
wish = 2;

After cin >> wish

If wish isn't 1, it would set wish to 2. Even if the user types in 2, which isn't 1, it's only going to set wish to 2 anyway, the exact same value.

And since wish is 2 at this point, it's going to end the program because of the if statement that you have afterwards 😁