【枚举】：复合态的处理（ Flag = Flag1

【枚举】：复合态的处理（ Flag = Flag1 | Flag2 )

news2024/9/27 19:25:08

一、引言

枚举 enum是我们用的比较多的类型，而枚举通常适用的场景，也是为了我们在业务逻辑处理时，作为不同的逻辑分支的判断条件；比起直接定义1~N这样魔幻到让人一头雾水的分支 Flag , 定义完美的枚举类型可以见名知意，简直不要太优雅。
但是，我们偶尔也会面临这样的分支选择，我既要执行状态A，还要执行状态B，（好看的姑娘当然都要，小孩子才做选择），那么，在使用枚举作为分支条件时，该如何处理呢？
以下，抛砖引玉，给出简单的示例。

二、代码示例

1、简单版本

#include <iostream>

// 定义位掩码枚举
enum class MultiState : unsigned int {
    NONE      = 0,
    STATE1    = 1 << 0, // 0001
    STATE2    = 1 << 1, // 0010
    STATE3    = 1 << 2, // 0100
    STATE4    = 1 << 3  // 1000
};

// 定义位运算符重载
inline MultiState operator|(MultiState a, MultiState b) {
    return static_cast<MultiState>(static_cast<unsigned int>(a) | static_cast<unsigned int>(b));
}

inline MultiState& operator|=(MultiState& a, MultiState b) {
    a = a | b;
    return a;
}

inline bool operator&(MultiState a, MultiState b) {
    return static_cast<bool>(static_cast<unsigned int>(a) & static_cast<unsigned int>(b));
}

void printMultiState(MultiState state) {
    if (state & MultiState::STATE1) std::cout << "STATE1 ";
    if (state & MultiState::STATE2) std::cout << "STATE2 ";
    if (state & MultiState::STATE3) std::cout << "STATE3 ";
    if (state & MultiState::STATE4) std::cout << "STATE4 ";
    if (state == MultiState::NONE) std::cout << "NONE";
    std::cout << std::endl;
}

int main() {
    MultiState state = MultiState::STATE1 | MultiState::STATE3;
    printMultiState(state);

    state |= MultiState::STATE2;
    printMultiState(state);
    
    return 0;
}

在这里插入图片描述
如果有试图问 enum class MultiState 语法的朋友，请前往《为何推荐首选enum class 而非 enum》

这里有巧妙使用位运算，保证复合态值的唯一，并在判断每个态是否为真时，起到关键的作用；不过缺点也很明显，如果枚举类型的值很多，


void printMultiState(MultiState state) {
    if (state & MultiState::STATE1) std::cout << "STATE1 ";
    if (state & MultiState::STATE2) std::cout << "STATE2 ";
    if (state & MultiState::STATE3) std::cout << "STATE3 ";
    if (state & MultiState::STATE4) std::cout << "STATE4 ";
    if (state == MultiState::NONE) std::cout << "NONE";
    std::cout << std::endl;
}

这里 if ... else ...岂不是要写到死？这肯定是不能接受的。

2、进阶版本

#include <iostream>
#include <vector>

// 定义位掩码枚举
enum class MultiState : unsigned int {
    NONE      = 0,
    STATE1    = 1 << 0, // 0001
    STATE2    = 1 << 1, // 0010
    STATE3    = 1 << 2, // 0100
    STATE4    = 1 << 3, // 1000
    // 添加更多状态
    STATE5    = 1 << 4,
    STATE6    = 1 << 5,
    // ...
};

// 定义位运算符重载
inline MultiState operator|(MultiState a, MultiState b) {
    return static_cast<MultiState>(static_cast<unsigned int>(a) | static_cast<unsigned int>(b));
}

inline MultiState& operator|=(MultiState& a, MultiState b) {
    a = a | b;
    return a;
}

inline bool operator&(MultiState a, MultiState b) {
    return static_cast<bool>(static_cast<unsigned int>(a) & static_cast<unsigned int>(b));
}

// 定义状态和字符串表示的结构体
struct StateInfo {
    MultiState state;
    const char* name;
    // std::function<void()> f;
};

// 定义状态和字符串表示的数组
const StateInfo stateInfos[] = {
    { MultiState::STATE1, "STATE1" },
    { MultiState::STATE2, "STATE2" },
    { MultiState::STATE3, "STATE3" },
    { MultiState::STATE4, "STATE4" },
    { MultiState::STATE5, "STATE5" },
    { MultiState::STATE6, "STATE6" },
    // 添加更多状态
};

void printMultiState(MultiState state) {
    bool hasState = false;
    for (const auto& info : stateInfos) {
        if (state & info.state) {
            std::cout << info.name << " ";
            hasState = true;
        }
    }
    if (!hasState) {
        std::cout << "NONE";
    }
    std::cout << std::endl;
}

int main() {
    MultiState state = MultiState::STATE1 | MultiState::STATE3;
    printMultiState(state);

    state |= MultiState::STATE2;
    printMultiState(state);

    return 0;
}