| Issue |
63829
|
| Summary |
clang wrongly believes that std::variant is not constexpr constructible if any Ti is not constexpr constructible
|
| Labels |
new issue
|
| Assignees |
|
| Reporter |
wanghan02
|
Example could be found below or on [Compiler Explorer](https://godbolt.org/z/bx86vEa9e). clang trunk (and previous versions) does not compile while gcc and VC++ does.
```
#include <variant>
struct S1 {
constexpr S1(int) {};
};
struct S2 {
S2() {}
};
constexpr std::variant<S1, S2> var(S1(2)); // clang error: constexpr variable cannot have non-literal type 'const std::variant<S1, S2>'
constexpr std::variant<S1, S2> var2(std::in_place_type<S1>, 2); // clang error: constexpr variable cannot have non-literal type 'const std::variant<S1, S2>'
```
According to [variant.ctor#19](https://timsong-cpp.github.io/cppwp/n4868/variant.ctor#19) and [variant.ctor#34](https://timsong-cpp.github.io/cppwp/n4868/variant.ctor#34):
> Remarks: The _expression_ inside noexcept is equivalent to is_nothrow_constructible_v<Tj, T>[.](https://timsong-cpp.github.io/cppwp/n4868/variant.ctor#19.sentence-1) If Tj's selected constructor is a constexpr constructor, this constructor is a constexpr constructor[.](https://timsong-cpp.github.io/cppwp/n4868/variant.ctor#19.sentence-2)
> Remarks: If TI's selected constructor is a constexpr constructor, this constructor is a constexpr constructor[.](https://timsong-cpp.github.io/cppwp/n4868/variant.ctor#34.sentence-1)
If the selected constructor is a constexpr constructor, std::variant's converting constructor is also a constexpr constructor. IMO clang is wrong here. I haven't checked other constructors, probably have the same behavior as these 2 above.
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