std::variant constexpr 中的修改

Question

考虑以下两个程序：

#include<variant>
#include<iostream>

constexpr auto f() {
    using T = std::variant<bool, int>;
    T t(false);
    t = T(true);
    return std::get<bool>(t);
}

template<auto V> 
void print() { std::cout << V << "\n"; }

int main() {
    print<f()>();
}

和

#include<variant>
#include<iostream>

constexpr auto f() {
    using T = std::variant<bool, int>;
    T t(false);
    t = T(42);
    return std::get<int>(t);
}

template<auto V> 
void print() { std::cout << V << "\n"; }

int main() {
    print<f()>();
}

GCC 编译这两个并输出预期的结果。在这两种情况下，Clang 都不会使用以下错误消息编译它们中的任何一个：

<source>:4:16: error: constexpr function never produces a constant expression [-Winvalid-constexpr]
constexpr auto f() {
               ^
<source>:7:7: note: non-constexpr function 'operator=' cannot be used in a constant expression
    t = T(42);
      ^
/opt/compiler-explorer/gcc-8.2.0/lib/gcc/x86_64-linux-gnu/8.2.0/../../../../include/c++/8.2.0/variant:1095:16: note: declared here
      variant& operator=(variant&&) = default;

这两个程序的格式是否正确？如果不是，为什么？

此外，如果它们格式不正确，Clang 给出的错误消息是否合适？根据 [variant.assign] 移动赋值运算符应该是 constexpr.

进一步根据(7.4) the assignment in the second example should behave equivalent to emplace<int>(...) which is not declared constexpr ([variant.mod])。这是否意味着第二个例子是错误的，因为模板参数不能被评估为常量表达式，或者措辞 allow/require 是否有这种行为？

编辑：

根据评论，如果使用 libc++，Clang 似乎编译并输出正确的结果，并且错误仅发生在 libstdc++ 中。这是标准库和编译器不兼容吗？

在 https://godbolt.org/:

在两种情况下都有效：

GCC 8.2.0“-std=c++17”
Clang 7.0.0“-std=c++17 -stdlib=libc++”

在这两种情况下都不起作用：

Clang 7.0.0“-std=c++17”

Answer 1

这看起来像是一个 clang bug we can see from the libstdc++ variant header，移动赋值运算符确实没有标记为 constexpr:

variant& operator=(variant&&) = default;

但是默认和隐式定义的移动赋值运算符仍然可以是 constexpr，我们可以从 [class.copy.assign]p10 中看到这一点（强调我的）：

A copy/move assignment operator for a class X that is defaulted and not defined as deleted is implicitly defined when it is odr-used ([basic.def.odr]) (e.g., when it is selected by overload resolution to assign to an object of its class type), when it is needed for constant evaluation ([expr.const]), or when it is explicitly defaulted after its first declaration. The implicitly-defined copy/move assignment operator is constexpr if

(10.1) X is a literal type, and

(10.2) the assignment operator selected to copy/move each direct base class subobject is a constexpr function, and

(10.3) for each non-static data member of X that is of class type (or array thereof), the assignment operator selected to copy/move that member is a constexpr function.

据我所知，libstdc++ 实现应该适合所有这些情况，它是文字类型，它没有非静态数据成员，并且它所有基的赋值运算符也应该是 constexpr。

std::variant constexpr 中的修改

std::variant modification in constexpr

c++

variant

constexpr

c++17