CODEWITHSUNDEEP
c++boostboost-variant
MWB
MWB

Reputation: 12567

Why is this boost::variant missing operator<<?

I've read that a boost::variant is streamable if all of its variants are streamable. However,

#include <iostream>
#include <vector>
#include <string>
#include <boost/variant.hpp>

std::ostream& operator<<(std::ostream& out, const std::vector<int>& v) {
    for(int i = 0; i < v.size(); ++i)
        out << " " << v[i];
    return out;
}

int main() {
    boost::variant<int, std::string > a(3);
    std::cout << a << '\n'; // OK

    std::vector<int> b(3, 1);
    std::cout << b << '\n'; // OK

    boost::variant<int, std::vector<int> > c(3);
    std::cout << c << '\n'; // ERROR
}

fails to compile. Why?

Versions:

Upvotes: 0

Views: 204

Answers (1)

jrok
jrok

Reputation: 55395

I haven't checked the documentation of serialization, but I'm pretty sure that operator<< for the types of boost::variant needs to be either found by Argument Dependent Lookup or else be present in boost namespace.

This works:

#include <iostream>
#include <vector>
#include <string>
#include <boost/serialization/variant.hpp>

namespace boost {

    std::ostream& operator<<(std::ostream& out, const std::vector<int>& v) {
        for(int i = 0; i < v.size(); ++i)
            out << " " << v[i];
        return out;
    }

}

int main() {
    boost::variant<int, std::string > a(3);
    std::cout << a << '\n';

    {
    using namespace boost;
    std::vector<int> b(3, 1);
    std::cout << b << '\n';
    }

    boost::variant<int, std::vector<int> > c(3);
    std::cout << c << '\n';
}

Output:

3
 1 1 1
3

Upvotes: 6

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