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std::hash<std::variant>

Defined in header <variant>.

template< class... Types >
struct hash<std::variant<Types...>>; // (since C++17)

The template specialization of std::hash for the std::variant template allows users to obtain hashes of variant objects.

The specialization std::hash<std::variant<Types...>> is enabled (see std::hash) if every specialization in std::hash<std::remove_const_t<Types>>... is enabled, and is disabled otherwise.

The member functions of this specialization are not guaranteed to be noexcept.

Template parameters

Types - the types of the alternatives supported by the variant object

Notes

Unlike std::hash<std::optional>, hash of a variant does not typically equal the hash of the contained value; this makes it possible to distinguish std::variant<int, int> holding the same value as different alternatives.

Example

#include <iostream>
#include <string>
#include <variant>

using Var = std::variant<int, int, int, std::string>;

template<unsigned I>
void print(Var const& var)
{
std::cout << "get<" << var.index() << "> = "
<< std::get<I>(var)
<< "\t" "# = "
<< std::hash<Var>{}(var) << '\n';
}

int main()
{
Var var;
std::get<0>(var) = 2020;
print<0>(var);
var.emplace<1>(2023);
print<1>(var);
var.emplace<2>(2026);
print<2>(var);
var = "C++";
print<3>(var);
}
Possible Result
get<0> = 2020   # = 2020
get<1> = 2023 # = 2024
get<2> = 2026 # = 2028
get<3> = C++ # = 15518724754199266859

std::hash<std::variant>

Defined in header <variant>.

template< class... Types >
struct hash<std::variant<Types...>>; // (since C++17)

The template specialization of std::hash for the std::variant template allows users to obtain hashes of variant objects.

The specialization std::hash<std::variant<Types...>> is enabled (see std::hash) if every specialization in std::hash<std::remove_const_t<Types>>... is enabled, and is disabled otherwise.

The member functions of this specialization are not guaranteed to be noexcept.

Template parameters

Types - the types of the alternatives supported by the variant object

Notes

Unlike std::hash<std::optional>, hash of a variant does not typically equal the hash of the contained value; this makes it possible to distinguish std::variant<int, int> holding the same value as different alternatives.

Example

#include <iostream>
#include <string>
#include <variant>

using Var = std::variant<int, int, int, std::string>;

template<unsigned I>
void print(Var const& var)
{
std::cout << "get<" << var.index() << "> = "
<< std::get<I>(var)
<< "\t" "# = "
<< std::hash<Var>{}(var) << '\n';
}

int main()
{
Var var;
std::get<0>(var) = 2020;
print<0>(var);
var.emplace<1>(2023);
print<1>(var);
var.emplace<2>(2026);
print<2>(var);
var = "C++";
print<3>(var);
}
Possible Result
get<0> = 2020   # = 2020
get<1> = 2023 # = 2024
get<2> = 2026 # = 2028
get<3> = C++ # = 15518724754199266859