std::variant<Types...>::operator=
From cppreference.com
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) (constexpr since C++20) |
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()>(*std::get_if<rhs.index()>(std::addressof(rhs))). *this may become
valueless_by_exception
if an exception is thrown on the copy-construction insideemplace
. - Otherwise, equivalent to this->operator=(variant(rhs)).
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::move(*std::get_if<j>(std::addressof(rhs))) to the value contained in *this, with
j
beingindex()
. 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()>(std::move(*std::get_if<rhs.index()>(std::addressof(rhs)))). If an exception is thrown by
T_i
's move constructor, *this becomesvalueless_by_exception
.
This overload participates in overload resolution only 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 everyT_i
fromTypes...
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
;
- An overload F(T_i) is only considered if the declaration T_i x[] = { std::forward<T>(t) }; is valid for some invented variable
- 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)). *this may become
valueless_by_exception
if an exception is thrown on the initialization insideemplace
. - Otherwise, equivalent to this->emplace<j>(T_j(std::forward<T>(t))).
This overload participates in overload resolution only 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<std::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
Contents |
[edit] Parameters
rhs | - | another variant
|
t | - | a value convertible to one of the variant's alternatives |
[edit] Return value
*this
[edit] 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>) && ...))
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>)
std::is_nothrow_constructible_v<T_j, T>)
[edit] Notes
Feature-test macro | Value | Std | Feature |
---|---|---|---|
__cpp_lib_variant |
202106L | (C++20) (DR) |
Fully constexpr std::variant (3)
|
[edit] Example
Run this code
#include <iomanip> #include <iostream> #include <string> #include <type_traits> #include <variant> std::ostream& operator<<(std::ostream& os, std::variant<int, std::string> const& va) { os << ": { "; std::visit([&](auto&& arg) { using T = std::decay_t<decltype(arg)>; if constexpr (std::is_same_v<T, int>) os << arg; else if constexpr (std::is_same_v<T, std::string>) os << std::quoted(arg); }, va); return os << " };\n"; } int main() { std::variant<int, std::string> a{2017}, b{"CppCon"}; std::cout << "a" << a << "b" << b << '\n'; std::cout << "(1) operator=( const variant& rhs )\n"; a = b; std::cout << "a" << a << "b" << b << '\n'; std::cout << "(2) operator=( variant&& rhs )\n"; a = std::move(b); std::cout << "a" << a << "b" << b << '\n'; std::cout << "(3) operator=( T&& t ), where T is int\n"; a = 2019; std::cout << "a" << a << '\n'; std::cout << "(3) operator=( T&& t ), where T is std::string\n"; std::string s{"CppNow"}; std::cout << "s: " << std::quoted(s) << '\n'; a = std::move(s); std::cout << "a" << a << "s: " << std::quoted(s) << '\n'; }
Possible output:
a: { 2017 }; b: { "CppCon" }; (1) operator=( const variant& rhs ) a: { "CppCon" }; b: { "CppCon" }; (2) operator=( variant&& rhs ) a: { "CppCon" }; b: { "" }; (3) operator=( T&& t ), where T is int a: { 2019 }; (3) operator=( T&& t ), where T is std::string s: "CppNow" a: { "CppNow" }; s: ""
[edit] 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 |
LWG 3585 | C++17 | converting assignment was sometimes unexpectedly ill-formed because there was no available move assignment |
made well-formed |
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 |
P2231R1 | C++20 | converting assignment (3) was not constexpr while the required operations can be constexpr in C++20 |
made constexpr |
[edit] See also
constructs a value in the variant , in place (public member function) |