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";// OKstd::variant<std::string, std::string> v2; v2 ="abc";// Errorstd::variant<std::string, bool> v3; v3 ="abc";// OK, chooses string; bool is not a candidatestd::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 constexprstd::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)>;ifconstexpr(std::is_same_v<T, int>) os << arg;elseifconstexpr(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) |