This snippet of code is taken from https://en.cppreference.com/w/cpp/utility/variant/visit
using var_t = std::variant<int, long, double, std::string>;
template<class... Ts> struct overloaded : Ts... { using Ts::operator()...; };
std::vector<var_t> vec = {10, 15l, 1.5, "hello"};
for (auto& v: vec) {
// 4. another type-matching visitor: a class with 3 overloaded operator()'s
// Note: The `(auto arg)` template operator() will bind to `int` and `long`
// in this case, but in its absence the `(double arg)` operator()
// *will also* bind to `int` and `long` because both are implicitly
// convertible to double. When using this form, care has to be taken
// that implicit conversions are handled correctly.
std::visit(overloaded {
[](auto arg) { std::cout << arg << ' '; },
[](double arg) { std::cout << std::fixed << arg << ' '; },
[](const std::string& arg) { std::cout << std::quoted(arg) << ' '; }
}, v);
}
Can someone explain what using Ts::operator()...; means here?
And in the following, what constructor is this calling? with the 3 lambda functions?
overloaded {
[](auto arg) { std::cout << arg << ' '; },
[](double arg) { std::cout << std::fixed << arg << ' '; },
[](const std::string& arg) { std::cout << std::quoted(arg) << ' '; }
}
I think the concrete overloaded instance is deriving from all 3 of these function types, and then the visitor is picking the right one to use depending on the type of the variant. Is that right?
I just don't fully understand this example.