如何使用 std::variant 打印 map 键/值？

Question

I'm trying to print the key of a certain value in my dictionary. I defined my map like this:

std::map<std::string, std::variant<float,int,bool,std::string>> kwargs; 
kwargs["interface"] = "probe"; 
kwargs["flag"] = true; 
kwargs["Height"] = 5; 
kwargs["length"] = 6; 

I tried to print the value normally but a (no operator "<<" matches these operands) error occurs.

std::cout << kwargs["flag"] << std::endl; 

Can someone help me with this error?

Answer 1

There's no operator<< for std::variant , which can't be printed out directly. You need to read the value from the variant (by index or type), eg

std::cout << std::get<bool>(kwargs["flag"]) << std::endl; 

Or

std::cout << std::get<2>(kwargs["flag"]) << std::endl; 

Answer 2

You can define operator<< for your type:

std::ostream& operator<<(std::ostream& os, std::variant<float,int,bool,std::string> const& v) {
    std::visit([&os](auto const& x) { os << x; }, v);
    return os;
}

You can generalize the solution to work with any variant specialization:

template <class Var, class = std::variant_alternative_t<0, Var>>
std::ostream& operator<<(std::ostream& os, Var const& v) {
    std::visit([&os](auto const& x) { os << x; }, v);
    return os;
}

Note, the second template parameter exists to disable the definition if the Var is not a std::variant specialization.

Answer 3

I would use std::visit with a generic lambda that prints any type to std::cout , as follows:

std::map<std::string, std::variant<float,int,bool,std::string>> kwargs; 
kwargs["interface"] = "probe"s;
kwargs["flag"] = true; 
kwargs["Height"] = 5; 
kwargs["length"] = 6; 

for (const auto& [k, v] : kwargs){
    std::cout << k << " : ";
    std::visit([](const auto& x){ std::cout << x; }, v);
    std::cout << '\n';
}

This works because all of float , int , bool , and std::string have overloads for the output stream operator << . For other types, or to override this behaviour, you could use a custom functor class instead of a lambda:

struct PrintVisitor {
    template<typename T>
    void operator()(const T& t) const {
        std::cout << t;
    }

    // prints std::string in quotes
    void operator(const std::string& s) const {
        std::cout << '\"' << s << '\"';
    }
};

[...]

std::visit(PrintVisitor{}, kwargs);

NOTE : unfortunately, the line kwargs["interface"] = "probe"; doesn't actually select the std::string constructor for the variant type, because the string literal preferentially converts to a bool rather than a std::string ( further reading ). You can avoid this by explicitly making a std::string using, for example, std::string{"probe"} or the standard std::string user-defined literal, as in "probe"s .

Live demo