CareerCup

E. Becoz GetName() returns const char* . So compiler should try to type cast the person object in const char* .

D.

Conversion function will not have return type.. Answer is E

the question is wrong...there has to be 'operator<<' which is overloaded here...
I think this should work...
char* operator<<(ostream &o,Person &p){return p.GetName();}

friend std::ostream& operator<< (std::ostream& o, Person const& p){ return o << p.m_szName;};

Its E.
We dont need to overload << operator. Person object has already been converted by operator const char*() conversion function.
I ran the program and its working fine.

E is correct answer.


Here is complete program.

class Person{
public:
Person(const char* szName)
{
m_szName = new char();

strcpy_s(m_szName,strlen(szName)+1, szName);
}
const char* GetName() const
{
return m_szName;
}

operator const char*() const { return GetName(); };

private:
char *m_szName;
};




int main(int argc, char **argv)
{
Person person("John");
std::cout <<person;
getchar();
}

output

It will print "John"