Question

The purpose of it is the operations within the overloaded method, but what would be it’s true purpose?

class CModeType
{
public:
    CModeType(){m_nModeType=1;}
    ~CModeType(){}
    int m_nModeType;
    CModeType&              operator&&( const CModeType& rModeType );
};
CModeType& CModeType::operator &&( const CModeType& rModeType )
{
    this->m_nModeType += rModeType.m_nModeType;
    return *this;
}
int _tmain(int argc, _TCHAR* argv[])
{
    CModeType Mode;
    CModeType Mod2;
    Mode && Mode;
    cout << Mode.m_nModeType << endl; //output:2
    return 0;
}

I would also ask about || but I believe the answer will help me to understand it so.

Answer 1

For built-in types, the && operator is “logical-and”, so that x && y is true if and only if both x and y are true when evaluated in a boolean context.

For built-in types, the && operator is short-circuited, so that the second argument is never evaluated if the first one is already false, so you can say things like if (p && *p == 10) without risking a null-pointer dereferencing. For the overloaded operator&& of a user-defined type, this is not true, and it is just an ordinary function (so arguments are evaluated in an indeterminate order). Beware!