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体。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;
}
最佳答案
以下将适用于变体中任意数量的类型:
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);
}
要注意的是,比较必须始终采用 value == variant
的形式或 value != variant
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 static_assert
,而且我们没有办法让全局运营商比 variant<>
更专业的内置的。
关于c++ - boost::variant 与包含值的比较,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38400344/