std::monostate
From cppreference.com
Defined in header <variant>
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struct monostate { }; |
(since C++17) | |
Unit type intended for use as a well-behaved empty alternative in std::variant. In particular, a variant of non-default-constructible types may list std::monostate
as its first alternative: this makes the variant itself default-constructible.
Member functions
(constructor) (implicitly declared) |
trivial implicit default/copy/move constructor (public member function) |
(destructor) (implicitly declared) |
trivial implicit destructor (public member function) |
operator= (implicitly declared) |
trivial implicit copy/move assignment (public member function) |
Non-member functions
std::operator==, !=, <, <=, >, >=, <=>(std::monostate)
constexpr bool operator==(monostate, monostate) noexcept { return true; } |
(since C++17) | |
constexpr bool operator!=(monostate, monostate) noexcept { return false; } constexpr bool operator<(monostate, monostate) noexcept { return false; } |
(since C++17) (until C++20) |
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constexpr std::strong_ordering operator<=>(monostate, monostate) noexcept { return std::strong_ordering::equal; } |
(since C++20) | |
All instances of std::monostate
compare equal.
The |
(since C++20) |
Helper classes
std::hash<std::monostate>
template <> struct std::hash<monostate>; |
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Specializes the std::hash algorithm for std::monostate
.
Example
Run this code
#include <variant> #include <iostream> #include <cassert> struct S { S(int i) : i(i) {} int i; }; int main() { // Without the monostate type this declaration will fail. // This is because S is not default-constructible. std::variant<std::monostate, S> var; assert(var.index() == 0); try { std::get<S>(var); // throws! We need to assign a value } catch(const std::bad_variant_access& e) { std::cout << e.what() << '\n'; } var = 12; std::cout << std::get<S>(var).i << '\n'; }
Possible output:
std::get: wrong index for variant 12
See also
constructs the variant object (public member function) |