boost::variant 与包含值的比较

Question

我正在尝试找到一种方法来比较 boost::variant 与基础值，而无需从该基础值构造变量。该问题在“main()”函数的注释中定义 辅助问题是关于代码中定义的比较运算符。如何减少比较运算符的数量？如果 boost::variant 包含 6 种不同的类型，我是否必须定义 6！运算符能够比较两个变体？

谢谢！

#include <boost/variant.hpp>
namespace test {

    namespace Tag {
        struct Level1{ int t{ 1 }; };
        struct Level2{ int t{ 2 }; };
    }

    template <typename Kind> struct Node;

    using LevelOne = Node<Tag::Level1>;
    using LevelTwo = Node<Tag::Level2>;

    using VariantNode = boost::variant
    <
        boost::recursive_wrapper<LevelOne>,
        boost::recursive_wrapper<LevelTwo>
    >;

    typedef VariantNode* pTree;
    typedef std::vector<pTree> lstTree;

    template <typename Kind> struct Node
    {
        Node(pTree p, std::string n) : parent(p), name(n) {}
        Node(const Node& another) : name(another.name), parent(another.parent) {}
        virtual ~Node() {}
        std::string name;
        pTree parent;
    };

    bool operator == (const LevelOne& one, const LevelTwo& two) {
        return false;
    }
    bool operator == (const LevelTwo& two, const LevelOne& one) {
        return false;
    }
    bool operator == (const LevelOne& one, const LevelOne& two) {
        return true;
    }
    bool operator == (const LevelTwo& one, const LevelTwo& two) {
        return true;
    }
}

int main(int argc, char *argv[])
{
    using namespace test;
    LevelOne l1(nullptr, "level one");
    VariantNode tl2 = VariantNode(LevelTwo(nullptr, "level two"));
    VariantNode tl1 = VariantNode(LevelOne(nullptr, "level one"));
    bool rv = (tl1 == tl2); // this line compiles OK (comparing two variants)
    // comparison below does not compile, because "l1" is not a variant. 
    // Question: How can I compare "variant" value "tl1" 
    // with one of the possible content values "l1"
    bool rv1 = (tl1 == l1); 
    return 1;
}

Answer 1

以下内容适用于变体中的任意数量的类型：

template<typename T>
struct equality_visitor : boost::static_visitor<bool> {
    explicit constexpr equality_visitor(T const& t) noexcept : t_{ &t } { }

    template<typename U, std::enable_if_t<std::is_same<T, U>::value>* = nullptr>
    constexpr bool operator ()(U const& u) const {
        return *t_ == u;
    }

    template<typename U, std::enable_if_t<!std::is_same<T, U>::value>* = nullptr>
    constexpr bool operator ()(U const&) const {
        return false;
    }

private:
    T const* t_;
};

template<
    typename T,
    typename... Ts,
    typename = std::enable_if_t<
        boost::mpl::contains<typename boost::variant<Ts...>::types, T>::value
    >
>
bool operator ==(T const& t, boost::variant<Ts...> const& v) {
    equality_visitor<T> ev{ t };
    return v.apply_visitor(ev);
}

template<
    typename T,
    typename... Ts,
    typename = std::enable_if_t<
        boost::mpl::contains<typename boost::variant<Ts...>::types, T>::value
    >
>
bool operator !=(T const& t, boost::variant<Ts...> const& v) {
    return !(t == v);
}

问题是比较必须始终采用value == variant 或value != variant 的形式，而不是variant == value 或variant != value。这是因为boost::variant<> 本身将这些运算符定义为始终为static_assert，我们无法让全局运算符比variant<> 的内置运算符更专业。

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