I had always assumed that if cin >> x fails, where x is a built-in type such as int, then x is left at its value prior to the input. However the following code produces different output depending on how cin fails:
#include <iostream>
using namespace std;
int main(){
int x = 10, y = 10;
cin >> x;
cin >> y;
cout << x << " " << y;
}
The value of y seems to vary depending on the type of failure (^Z representing EOF)
Input 20 ^Z outputs 20 0
Input 20 c output 20 0
Input 20[ENTER]^Z outputs 20 10
Input 20[ENTER] ^Z outputs 20 0
So it seems if there is a failure then int gets set to the value 0. However I can't make sense of the third case, where I enter 20, press enter and then Cntrl+Z, the value of y doesn't get changed at all in this circumstance.
How are values set when cin fails? Should I make any assumptions about their values? Also, when creating my own input operator for classes, should cin failures generally set the class object back to the pre-input value or to a default value instead (Class())?
