我如何在`std::expected<T,E>上使用`.value()`,其中`T`是只移动的类型?

以下代码:

#include <chrono>
#include <expected>
#include <stdexcept>
#include <string>
#include <thread>
#include <utility>

using namespace std::literals;

class Resource {

  Resource() { /* some setup, might raise */ }

  public:
  static std::expected<Resource, std::string> tryCreate() noexcept {
    try {
      return Resource();
    } catch (const std::runtime_error& exc) {
      return std::unexpected(exc.what());
    }
  }
  static const Resource create() noexcept {
    while (true) {
        const auto resource{Resource::tryCreate()};
        if (resource.has_value()) return resource.value();  // <-- this is no good.
        else std::this_thread::sleep_for(1s);
    }
  }
  Resource(const Resource&) = delete;
  Resource(Resource&& other) noexcept {}
  Resource& operator=(const Resource&) = delete;
  Resource& operator=(Resource&& other) noexcept { return *this; }
  ~Resource() { /* some cleanup */ }

  void use() const noexcept {}
};

int main() noexcept {
    const auto resource{Resource::create()};
    resource.use();
}

…拒绝编译,出现错误:

<source>:25:42: error: call to deleted constructor of 'const Resource'
   25 |         if (resource.has_value()) return resource.value();
      |                                          ^~~~~~~~~~~~~~~~
<source>:29:3: note: 'Resource' has been explicitly marked deleted here
   29 |   Resource(const Resource&) = delete;
      |   ^
1 error generated.
Compiler returned: 1

我没有看到任何解释 cppreference 只允许移动的类型是不允许的,而且我在搜索互联网时无法立即找到其他有类似问题的人(尽管我对此还不熟悉,所以我的C++google fu可能不够)。我认为这可能与“复制省略”有关,但我真的很困惑。

是否不能使用仅限移动的类型 std::expected ?如果是,我如何修改我的示例代码使其正常工作(不定义复制构造函数)?

godbolt -在Clang上使用-std=c++23失败,但我在GCC 14.2.1上遇到了类似的情况,同样是在-std=c++23上。