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

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< cpp‎ | utility‎ | variant

constexpr variant& operator=( const variant& rhs ); | (1) | (since C++17)
constexpr variant& operator=( variant&& rhs ) noexcept(/* see below */); | (2) | (since C++17)
| (3) |
template< class T > variant& operator=( T&& t ) noexcept(/* see below */); |  | (since C++17) (until C++20)
template< class T > constexpr variant& operator=( T&& t ) noexcept(/* see below */); |  | (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()>(get<rhs.index()>(rhs)). *this may become `valueless_by_exception` if an exception is thrown on the copy-construction inside `emplace`.
- 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::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 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 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 *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 inside `emplace`.
- 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

### 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

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: ""

### 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
P2231R1 | C++20 | converting assignment was not constexpr while the required operations can be in C++20 | made constexpr
LWG 3585 | C++17 | converting assignment was sometimes unexpectedly ill-formed because there was no available move assignment | made well-formed

### See also

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

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