The most simple way is to switch based on the current std::variant::index(). This approach requires your types (std::monostate, A, B, C) to always stay in the same order.
// I omitted C to keep the example simpler, the principle is the same
using my_variant = std::variant<std::monostate, A, B>;
void foo(my_variant &v) {
switch (v.index()) {
case 0: break; // do nothing because the type is std::monostate
case 1: {
doSomethingWith(std::get<A>(v));
break;
}
case 2: {
doSomethingElseWith(std::get<B>(v));
break;
}
}
}
If your callable works with any type, you can also use std::visit:
void bar(my_variant &v) {
std::visit([](auto &&arg) -> void {
// Here, arg is std::monostate, A or B
// This lambda needs to compile with all three options.
// The lambda returns void because we don't modify the variant, so
// we could also use const& arg.
}, v);
}
If you don't want std::visit to accept std::monostate, then just check if the index is 0. Once again, this relies on std::monostate being the first type of the variant, so it is good practice to always make it the first.
You can also detect the type using if-constexpr inside the callable. With this approach, the arguments don't have to be in the same order anymore:
void bar(my_variant &v) {
std::visit([](auto &&arg) -> my_variant {
using T = std::decay_t<decltype(arg)>;
if constexpr (std::is_same_v<std::monostate, T>) {
return arg; // arg is std::monostate here
}
else if constexpr (std::is_same_v<A, T>) {
return arg + arg; // arg is A here
}
else if constexpr (std::is_same_v<B, T>) {
return arg * arg; // arg is B here
}
}, v);
}
Note that the first lambda returns void because it just processes the current value of the variant. If you want to modify the variant, your lambda needs to return my_variant again.
You could use an overloaded visitor inside std::visit to handle A or B separately. See std::visit for more examples.
