std::variant 和不完整的类型：它是如何工作的？

Question

我确实知道 std::variant 适用于不完整的类型。但是，我不明白它是如何工作的，因为根据我的理解，std::variant 必须需要它所拥有的类型的最大大小。

那么，为什么这段代码不能用 s1 和 s2 编译。如何让它像 std::variant 一样工作？

#include <variant>
#include <vector>
#include <type_traits>
#include <typeinfo>
#include <iostream>

struct Rect;
struct Circle;

using Shape = std::variant<Rect, Circle>;

template<typename C>
struct S {static constexpr auto s = sizeof(C);};

constexpr auto s1 = S<Rect>::s;
constexpr auto s2 = sizeof(Rect);

struct Circle{};
struct Rect{
    std::vector<Shape> shapes;
};

int main() {}

Answer 1

I do understand that std::variant works with incomplete type.

我不认为你知道。它没有。

However, I don't understand how it can works because

有道理。它不能工作，因为：

in my unstanding, std::variant must need the maximum size of the types it holds.

标准是这样说的：

[res.on.functions]

In certain cases (replacement functions, handler functions, operations on types used to instantiate standard library template components), the C++ standard library depends on components supplied by a C++ program. If these components do not meet their requirements, this document places no requirements on the implementation.

In particular, the effects are undefined in the following cases:

...

if an incomplete type ([basic.types]) is used as a template argument when instantiating a template component or evaluating a concept, unless specifically allowed for that component.

[variant] 部分没有特定规则允许不完整的类型。

std::variant 和不完整的类型：它是如何工作的？

std::variant and incomplete type: how does it work?

c++

variant