std::variant cout in C++ - c++

Question

I am relatively new to CPP and have recently stumbled upon std::variant for C++17.
However, I am unable to use the << operator on such type of data.
Considering
#include <iostream>
#include <variant>
#include <string>
using namespace std;
int main() {
variant<int, string> a = "Hello";
cout<<a;
}
I am unable to print the output. Is there any short way of doing this? Thank you so much in advance.

Answer 1

You can use std::visit if you don't want to use std::get.
#include <iostream>
#include <variant>
struct make_string_functor {
std::string operator()(const std::string &x) const { return x; }
std::string operator()(int x) const { return std::to_string(x); }
};
int main() {
const std::variant<int, std::string> v = "hello";
// option 1
std::cout << std::visit(make_string_functor(), v) << "\n";
// option 2
std::visit([](const auto &x) { std::cout << x; }, v);
std::cout << "\n";
}

Answer 2

use std::get
#include <iostream>
#include <variant>
#include <string>
using namespace std;
int main() {
variant<int, string> a = "Hello";
cout << std::get<string>(a);
}
If you want to get automatically, it can't be done without knowing its type. Maybe you can try this.
string s = "Hello";
variant<int, string> a = s;
cout << std::get<decltype(s)>(a);

Answer 3

#include <iostream>
#include <variant>
#include <string>
int main( )
{
std::variant<int, std::string> variant = "Hello";
std::string string_1 = std::get<std::string>( variant ); // get value by type
std::string string_2 = std::get<1>( variant ); // get value by index
std::cout << string_1 << std::endl;
std::cout << string_2 << std::endl;
//may throw exception if index is specified wrong or type
//Throws std::bad_variant_access on errors
//there is also one way to take value std::visit
}
Here is the description link: https://en.cppreference.com/w/cpp/utility/variant