std::variant<Types...>::operator= (3) - Linux Man Pages

std::variant<Types...>::operator=: std::variant<Types...>::operator=

NAME

std::variant<Types...>::operator= - std::variant<Types...>::operator=

Synopsis


constexpr variant& operator=(const variant& rhs); (1) (since C++17)
constexpr variant& operator=(variant&& rhs) noexcept(/* see below */); (2) (since C++17)
template <class T> variant& operator=(T&& t) noexcept(/* see below */); (3) (since C++17)


Assigns a new value to an existing variant object.
1) Copy-assignment:


* If both *this and rhs are valueless by exception, does nothing.
* Otherwise, if rhs is valueless, but *this is not, destroys the value contained in *this and makes it valueless.
* Otherwise, if rhs holds the same alternative as *this, assigns the value contained in rhs to the value contained in *this. If an exception is thrown, *this does not become valueless: the value depends on the exception safety guarantee of the alternative's copy assignment.
* Otherwise, if the alternative held by rhs is either nothrow copy constructible or not nothrow move constructible (as determined by std::is_nothrow_copy_constructible and std::is_nothrow_move_constructible, respectively), equivalent to this->emplace<rhs.index()>(get<rhs.index()>(rhs)).
* Otherwise, equivalent to this->operator=(variant(rhs)). Note that *this may become valueless_by_exception as described in (2).


This overload is defined as deleted unless std::is_copy_constructible_v<T_i> and std::is_copy_assignable_v<T_i> are both true for all T_i in Types.... This overload is trivial if std::is_trivially_copy_constructible_v<T_i>,std::is_trivially_copy_assignable_v<T_i> and std::is_trivially_destructible_v<T_i> are all true for all T_i in Types....
2) Move-assignment:


* If both *this and rhs are valueless by exception, does nothing
* Otherwise, if rhs is valueless, but *this is not, destroys the value contained in *this and makes it valueless
* Otherwise, if rhs holds the same alternative as *this, assigns std::get<j>(std::move(rhs)) to the value contained in *this, with j being index(). If an exception is thrown, *this does not become valueless: the value depends on the exception safety guarantee of the alternative's move assignment.
* Otherwise (if rhs and *this hold different alternatives), equivalent to this->emplace<rhs.index()>(get<rhs.index()>(std::move(rhs))). If an exception is thrown by T_i's move constructor, *this becomes valueless_by_exception.


This overload only participates in overload resolution if std::is_move_constructible_v<T_i> and std::is_move_assignable_v<T_i> are both true for all T_i in Types.... This overload is trivial if std::is_trivially_move_constructible_v<T_i>, std::is_trivially_move_assignable_v<T_i>, and std::is_trivially_destructible_v<T_i> are all true for all T_i in Types....
3) Converting assignment.


* Determines the alternative type T_j that would be selected by overload resolution for the expression F(std::forward<T>(t)) if there was an overload of imaginary function F(T_i) for every T_i from Types... in scope at the same time, except that:


      * An overload F(T_i) is only considered if the declaration T_i x[] = { std::forward<T>(t) }; is valid for some invented variable x;
      * If T_i is (possibly cv-qualified) bool, F(T_i) is only considered if std:remove_cvref_t<T> is also bool.


* If *this already holds a T_j, assigns std::forward<T>(t) to the value contained in *this. If an exception is thrown, *this does not become valueless: the value depends on the exception safety guarantee of the assignment called.
* Otherwise, if std::is_nothrow_constructible_v<T_j, T> || !std::is_nothrow_move_constructible_v<T_j> is true, equivalent to this->emplace<j>(std::forward<T>(t));
* Otherwise, equivalent to this->operator=(variant(std::forward<T>(t))).


This overload only participates in overload resolution if
std::decay_t<T>
(until C++20)
std::remove_cvref_t<T>
(since C++20) is not the same type as variant and std::is_assignable_v<T_j&, T> is true and std::is_constructible_v<T_j, T> is true and the expression F(std::forward<T>(t)) (with F being the above-mentioned set of imaginary functions) is well formed.


  std::variant<string> v1;
  v1 = "abc"; // OK
  std::variant<std::string, std::string> v2;
  v2 = "abc"; // Error
  std::variant <std::string, bool> v3;
  v3 = "abc"; // OK, chooses string; bool is not a candidate
  std::variant<float, long, double> v4; //holds float
  v4 = 0; // OK, holds long; float and double are not candidates

Parameters


rhs - another variant
t - a value convertible to one of the variant's alternatives

Return value


*this

Exceptions


1) May throw any exception thrown by assignment and copy/move initialization of any alternative
2)
noexcept specification:
noexcept(((std::is_nothrow_move_constructible_v<Types> && std::is_nothrow_move_assignable_v<Types>) && ...))
3)
noexcept specification:
noexcept(std::is_nothrow_assignable_v<T_j&, T> && std::is_nothrow_constructible_v<T_j, T>)

Example


 This section is incomplete
 Reason: no example


Defect reports


The following behavior-changing defect reports were applied retroactively to previously published C++ standards.


DR Applied to Behavior as published Correct behavior
LWG_3024 C++17 copy assignment operator doesn't participate in overload resolution if any member type is not copyable defined as deleted instead
P0602R4 C++17 copy/move assignment may not be trivial even if underlying operations are trivial required to propagate triviality
P0608R3 C++17 converting assignment blindly assembles an overload set, leading to unintended conversions narrowing and boolean conversions not considered

See also


        constructs a value in the variant, in place
emplace (public member function)