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量参数来初始化状态的构造。但是，当定义了构造函数时，我会得到一个警告，它被认为是函数声明。

此外，当我尝试为状态定义setter时，尝试调用此方法会生成编译器错误

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);

您要传递类型，应该传递类型的实例，尝试

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

此更改后，我能够在MSVC 2019（16.1.3）中进行编译

编辑：这是一个处理Scheff和Jarod24注释的编辑，如果取消注释构造函数并编写

，则可能是最令人头疼的解析情况。

DoorState s(DoorState::DoorOpened());

这可以使用统一的初始化语法解决，请参见例如 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，但会带来一个新问题：DoorState::DoorOpened{}将是暂时的，并且永远不会绑定到ctor的输入参数上：

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指出了这个问题。

如何将变量作为构造函数参数或函数参数传递

1 个答案: