'std::ios_base::ios_base(const std::ios_base&)’ is private' error while overloading operator<< for std::ostram

17,105

Solution 1

std::ostream is not copy constructible, and when you return by value you are copy constructing. Although return value optimization means the copy might not actually be made, the compiler still requires that copying be possible.

The canonical return value for this operator is by non-const reference:

std::ostream& operator<<(std::ostream& o, const SomeType& t);

Solution 2

Return by reference:

std::ostream& operator<<(...)
          //^

Otherwise an attempt is being made to copy s, and ostreams are non-copyable (the error message is stating an attempt to access a private copy constructor).

Solution 3

You forgot reference in return type:

std::ostream &operator<<(std::ostream & s, person & os)
{
    return s << os.surname;
}
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Kacper Kołodziej
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Kacper Kołodziej

I'm student of Computer Science at Łódź University of Technology (Poland). I'm interested in C++, Python, Linux and embedded systems. I write about programming on my devblog in Polish and English

Updated on June 17, 2022

Comments

  • Kacper Kołodziej
    Kacper Kołodziej almost 2 years

    I've got a struct which looks like this:

    sturct person
    {
        string surname;
        person(string n) : surname(n) {};
    }
    

    I need to overload operator<< for std::ostream and person. I wrote this function:

    std::ostream operator<<(std::ostream & s, person & os)
    {
        s << os.surname;
        return s;
    }
    

    but I receive this errors:

    /usr/include/c++/4.6/bits/ios_base.h|788|error: ‘std::ios_base::ios_base(const std::ios_base&)’ is private|

    /usr/include/c++/4.6/bits/basic_ios.h|64|error: within this context

    /usr/include/c++/4.6/ostream|57|note: synthesized method ‘std::basic_ios::basic_ios(const std::basic_ios&)’ first required here |

  • Jarrod Cabalzar
    Jarrod Cabalzar over 10 years
    Thanks for that, helped me out :) +1