std::hash<std::variant>
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                    |   Defined in header  <variant>
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|   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
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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
Run this code
#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 output:
get<0> = 2020 # = 2020 get<1> = 2023 # = 2024 get<2> = 2026 # = 2028 get<3> = C++ # = 15518724754199266859
See also
|    (C++11)  | 
  hash function object  (class template)  |