c++ - 自 C++20 以来，是否允许对分配的存储进行指针运算？

Question

在 C++20 标准中，说数组类型是隐式生命周期类型。
这是否意味着可以隐式创建非隐式生命周期类型的数组？这种数组的隐式创建不会导致数组元素的创建？
考虑这个案例:

//implicit creation of an array of std::string 
//but not the std::string elements:
void * ptr = operator new(sizeof (std::string) * 10);
//use launder to get a "pointer to object" (which object?)
std::string * sptr = std::launder(static_cast<std::string*>(ptr));
//pointer arithmetic on not created array elements well defined?
new (sptr+1) std::string("second element");

从 C++20 开始，这段代码不再是 UB 了吗？

也许这种方式更好？

//implicit creation of an array of std::string 
//but not the std::string elements:
void * ptr = operator new(sizeof (std::string) * 10);
//use launder to get a "pointer to object" (actually not necessary)
std::string (* sptr)[10] = std::launder(static_cast<std::string(*)[10]>(ptr));
//pointer arithmetic on an array is well defined
new (*sptr+1) std::string("second element");

TC 回答 + 评论结论:

未创建数组元素但已创建数组

launder的使用在第一个例子中导致 UB，并且是
在第二个例子中没有必要。

正确的代码是:

    //implicit creation of an array of std::string 
    //but not the std::string elements:
    void * ptr = operator new(sizeof (std::string) * 10);
    //the pointer already points to the implicitly created object
    //so casting is enough 
    std::string (* sptr)[10] = static_cast<std::string(*)[10]>(ptr);
    //pointer arithmetic on an array is well defined
    new (*sptr+1) std::string("second element");

Answer 1

Does it means that an array to a non implicit lifetime type can be implicitly created?

是的。

The implicit creation of such an array would not cause creation of the array's elements?

是的。

这就是 std::vector可以在普通 C++ 中实现。