Does "cin" reset variable to some default value if input type differs from destination type? [duplicate]

Question

I have an issue with behavior of "cin" (I do not understand). My IDE is Netbeans under Windows OS (with Cygwin).

Here's a code example:

int main()
{
    int temp = -1;
    std::cin >> temp;    // here user enters string of characters (string) or a single character

    if (temp == 0)
        std::cout << "temp = " << temp << ".\n";
    if (temp == -1)
        std::cout << "temp = " << temp << ".\n";

    return 0;
}

This code shows message temp = 0 if I enter some sort of a character/string of characters. It's like there is conversion of char to int and conversion always ending by value 0.

Thank you if you can explain this behavior.

Answer 1

This is expected behavior of std::basic_istream::operator>>; since C++11 if extraction fails, the variable will be set to 0. Before C++11, the variable won't be modified then its original value remains.

If extraction fails (e.g. if a letter was entered where a digit is expected), value is left unmodified and failbit is set. (until C++11)

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::max() or std::numeric_limits::min() is written and failbit flag is set. (since C++11)

Answer 2

If the read fails operator>> will set the value to zero (cppreference):

If extraction fails, zero is written to value and failbit is set.