How would I make two ways to output something with <<?

Question

For example, if I wanted to have a default version of outputting with << and a detailed version.

myClass myObject(//Constructor parameters);

cout << myObject << "\n";
cout << myObject.detailed << "\n";

I tried making a modifier within my class but that didn't seem to work.

class myClass {
public:
   friend std::ostream& operator<<(std::ostream& output, const myClass& myObject);
   std::ostream& detailed(std::ostream& output);
}

That gave the error "Reference to non-static member function must be called".

Answer 1

I think my approach is the same as what @RemyLebeau but more generic.

#include <iostream>
#include <string>

//---------------------------------------------------------------------------------------------
// class templated to be able to apply
// detailed mechanism to different types

template<typename type_t>
class detailed
{
public:
    detailed(const type_t& object) : 
        m_object{ object } 
    {
    }

    friend std::ostream& operator<<(std::ostream& os, const detailed<type_t>& detailed_object)
    {
        detailed_object.m_object.detailed_output(os);
        return os;
    }

private:
    const type_t& m_object;
};

//---------------------------------------------------------------------------------------------
// like your class

struct test_class_t
{
public:
    friend std::ostream& operator<<(std::ostream& os, const test_class_t& object)
    {
        os << object.basic;
        return os;
    }

    void detailed_output(std::ostream& os) const // <- const is important
    {
        os << basic << ", " << details;
    }

private:
    std::string basic{ "basic" };
    std::string details{ "details" };
};

//---------------------------------------------------------------------------------------------

int main()
{
    test_class_t object;
    std::cout << detailed{object} << "\n";
}