如何将变体作为构造函数参数或函数参数传递

Question

我正在阅读有关使用变体实现状态机的信息。我尝试创建一个采用变体参数来初始化状态的结构。然而，当构造函数被定义时，我收到一条警告，指出这被识别为函数声明。

此外，当我尝试为状态定义 setter 时，尝试调用此

时会生成编译器错误

这是代码

#include "pch.h"
#include <iostream>
#include <variant>
#include <cassert>

struct DoorState
{
    struct DoorOpened {};
    struct DoorClosed {};
    struct DoorLocked {};

    using State = std::variant<DoorOpened, DoorClosed, DoorLocked>;
    DoorState()
    {
    }
    DoorState(State & state)
    {
        m_state = state;
    }
    void open()
    {
        m_state = std::visit(OpenEvent{}, m_state);
    }

    void close()
    {
        m_state = std::visit(CloseEvent{}, m_state);
    }

    void lock()
    {
        m_state = std::visit(LockEvent{}, m_state);
    }

    void unlock()
    {
        m_state = std::visit(UnlockEvent{}, m_state);
    }

    struct OpenEvent
    {
        State operator()(const DoorOpened&) { return DoorOpened(); }
        State operator()(const DoorClosed&) { return DoorOpened(); }
        // cannot open locked doors
        State operator()(const DoorLocked&) { return DoorLocked(); }
    };

    struct CloseEvent
    {
        State operator()(const DoorOpened&) { return DoorClosed(); }
        State operator()(const DoorClosed&) { return DoorClosed(); }
        State operator()(const DoorLocked&) { return DoorLocked(); }
    };

    struct LockEvent
    {
        // cannot lock opened doors
        State operator()(const DoorOpened&) 
        { 
            std::cout << "DoorOpened" << std::endl;
            return DoorOpened(); 
        }
        State operator()(const DoorClosed&) { return DoorLocked(); }
        State operator()(const DoorLocked&) { return DoorLocked(); }
    };

    struct UnlockEvent
    {
        // cannot unlock opened doors
        State operator()(const DoorOpened&) { return DoorOpened(); }
        State operator()(const DoorClosed&) { return DoorClosed(); }
        // unlock
        State operator()(const DoorLocked&) { return DoorClosed(); }
    };
    void set(State state)
    {
    }
    State m_state;
};

int main()
{
    //DoorState s(DoorState::DoorOpened);
    DoorState s; 
    s.set(DoorState::DoorOpened);
    s.lock();
    return 0;
}

Answer 1

在

s.set(DoorState::DoorOpened);

你正在传递一个类型，你应该传递一个类型的实例，试试

s.set(DoorState::DoorOpened{});

此更改后，我能够在 MSVC 2019 (16.1.3) 中编译

编辑：这是针对 Scheff 和 Jarod24 评论的编辑，如果我们要取消对构造函数的注释并编写

，就会出现最令人烦恼的解析情况

DoorState s(DoorState::DoorOpened());

这可以使用统一的初始化语法来解决，例如参见 https://arne-mertz.de/2015/07/new-c-features-uniform-initialization-and-initializer_list/

DoorState s{DoorState::DoorOpened{}};

这本来可以解决最棘手的解析问题，但却产生了一个新问题：DoorState::DoorOpened{}本来是一个临时问题，它可能永远不会绑定到 ctor 的输入参数：

DoorState(State& state)

然后我们需要将其更改为

DoorState(const State& state)

再次感谢 Jarod 和 Scheff 指出问题。

如何将变体作为构造函数参数或函数参数传递

How to pass a variant as constructor parameter or as function argument

c++

variant