Why happens to variable when cin breaks? C++

Question

I have the following code snippet:

    int a = 1;
    double b = 3.14;
    string c = "hi";

    cin >> a >> b >> c;
    cout << a << " " << b << " " << c << endl;

If I enter apple 11 tammy, why does it cout: 0 3.14 hi instead of: 1 3.14 hi?

Why does the value of a change when cin is broken?

Answer 1

Why does the value of a change when cin is broken?

This is the expected behavior of std::basic_istream::operator>> since C++11; If extraction fails the value will be set to 0.

If extraction fails, zero is written to value and failbit is set. If extraction results in the value too large or too small to fit in value, std::numeric_limits<T>::max() or std::numeric_limits<T>::min() is written and failbit flag is set.

Note that after failbit being set, the following input won't be performed; that means b and c will remain their original values.

BTW: Before C++11 the value will be left unmodified when extraction fails.

If extraction fails (e.g. if a letter was entered where a digit is expected), value is left unmodified and failbit is set.