没有使用重载模式对 std：：visit（） 的匹配函数调用

问：

我正在尝试在具有访问者重载模式的变体中调度变体。但是，我似乎无法将参数拟合到预期值。我得到这个（截断，有关更多输出，请参阅编译器资源管理器链接）：

error: no matching function for call to 'visit ...'

这是我的代码。我忽略了什么？

演示

#include <variant>
#include <cstdio>

template <typename... Ts>
struct overload : Ts...
{
    using Ts::operator() ...;
};

int main() {

    struct StateA{};
    struct StateB{};
    struct StateC{};

    using state_type = std::variant<StateA, StateB, StateC>;
    auto state = state_type{StateA{}};

    struct EventA{};
    struct EventB{};

    struct WifiConnectedNote{};
    struct ErrorNote{};

    using QueueVariantType = std::variant<std::monostate, WifiConnectedNote, ErrorNote>;
    
    using event_type = std::variant<std::monostate, QueueVariantType, EventA, EventB>;
    auto event = event_type{EventA{}};

    for (size_t i=0; i < 10; i++) {
        state = std::visit(overload{
            [&](QueueVariantType&& variant, auto&& state_tmp) -> state_type {
                std::visit(overload{
                    [&](WifiConnectedNote&&, auto&&){
                        printf("WifiConnectedNote A\n");
                    },
                    [&](ErrorNote&&, auto&&){
                        printf("ErrorNote A\n");
                    },
                    [&](auto&&, auto&&){
                        printf("Other in Queue variant\n");
                    },
                }, std::forward<decltype(variant)>(variant), std::forward<decltype(state_tmp)>(state_tmp));
                return StateC{};
            },
            [&](auto&&, const StateA&) -> state_type {
                printf("State A\n");
                return StateC{};
            },
            [&](auto&&, const StateB&) -> state_type {
                printf("State B\n");
                return StateA{};
            },
            [&](auto&&, const StateC&) -> state_type {
                printf("State C");
                return StateB{};
            },
        }, std::move(event), state);
    }
}