No match for 'operator<<' in std::cout

Question

I am developing gsoap web service where I am retrieving vectors of objects in return of a query. I have two ways to do it: first by simple loop and by iterator. None of them working.

The error is:

error: no match for 'operator<<' in 'std::cout mPer.MultiplePersons::info.std::vector<_Tp, _Alloc>::at<PersonInfo, std::allocator<PersonInfo> >(((std::vector<PersonInfo>::size_type)i))'

MultiplePersons mPer; // Multiple Person is a class, containing vector<PersonInfo> info
std::vector<PersonInfo>info; // PersonInfo is class having attributes Name, sex, etc.
std::vector<PersonInfo>::iterator it;

cout << "First Name: \t";
cin >> firstname;
if (p.idenGetFirstName(firstname, &mPer) == SOAP_OK) {
    // for (int i = 0; i < mPer.info.size(); i++) {
    //    cout << mPer.info.at(i); //Error
    //}
    for (it = info.begin(); it != info.end(); ++it) {
        cout << *it; // Error
    }

} else p.soap_stream_fault(std::cerr);

}

It's obvious that operator overloading operator<< in cout is the problem. I have looked at several problems related to this, but no one helped me out. If someone can provide a concrete example on how to solve it, it would be very appreciated. (Please do not talk in general about it, I am new to C++ and I have spent three days on it searching for solution.)

Solution 1

You need to provide an output stream operator for PersonInfo. Something like this:

struct PersonInfo
{
  int age;
  std::string name;
};

#include <iostream>
std::ostream& operator<<(std::ostream& o, const PersonInfo& p)
{
  return o << p.name << " " << p.age;
}

This operator allows expressions of the type A << B, where A is an std::ostream instance (of which std::cout is one) and B is a PersonInfo instance.

This allows you do do something like this:

#include <iostream>
#include <fstream>
int main()
{
  PersonInfo p = ....;
  std::cout << p << std::endl; // prints name and age to stdout

  // std::ofstream is also an std::ostream, 
  // so we can write PersonInfos to a file
  std::ofstream person_file("persons.txt");
  person_file << p << std::endl;
}

which in turn allows you to print the de-referenced iterator.

Solution 2

The result of *it is an L-value of type PersonInfo. The compiler is complaining that there is no operator<< which takes a right-hand side argument of type PersonInfo.

For the code to work, you need to provide such an operator, for example like this:

std::ostream& operator<< (std::ostream &str, const PersonInfo &p)
{
  str << "Name: " << p.name << "\nAge: " << p.age << '\n';
  return str;
}

The exact implementation of the operator depends on your needs for representing the class in output, of course.