为什么带有指针子对象的文字类类型的constexpr表达式不能是非类型模板参数

查看 the 关于非类型模板参数的帖子，我对该帖子中的示例感到困惑，在此引用示例:

After looking at the post about non-type template argument,I have a confusion for an example in that post,I cite the example here:

struct VariableLengthString {
   const char *data_ = nullptr;
   constexpr VariableLengthString(const char *p) : data_(p) {}
   auto operator<=>(const VariableLengthString&) const = default;
};

template<VariableLengthString S>
int bar() {
   static int i = 0;
   return ++i;
}

int main() {
   int x = bar<"hello">(); // ERROR
}

帖子说相关的措辞是[temp.arg.nontype]/2"，所以我看了一下这个规则，它限制了什么可以是非类型模板参数.

The post says "the relevant wording is [temp.arg.nontype]/2",So I took a look at that rule,It constrains what can be a non-type template argument.

非类型模板参数的模板参数应为模板参数类型的转换常量表达式.

所以，我看了一下转换后的常量表达式，它的定义在这里:

So,I took a look at converted constant expression,Its definitions are here:

类型T的转换后的常量表达式是隐式转换为类型T的表达式，其中转换后的表达式是常量表达式，并且隐式转换序列仅包含...

A converted constant expression of type T is an expression, implicitly converted to type T, where the converted expression is a constant expression and the implicit conversion sequence contains only...

什么是常量表达式?这些规则在这里:

What is a constant expression?These rules are here:

表达式e是核心常量表达式，除非按照抽象机的规则对e求值将对以下表达式之一求值:

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

(2.2)调用除文字类的constexpr构造函数以外的函数，constexpr函数或平凡的析构函数的隐式调用.

(2.2) an invocation of a function other than a constexpr constructor for a literal class, a constexpr function, or an implicit invocation of a trivial destructor.

那么，类类型 VariableLengthString 是文字类吗?是的，可以在这里证明哪些规则:

So,Is the class type VariableLengthString a literal class?Yes,it is.what rules can prove that are here:

如果是，则类型是文字类型:
1.合格的或
2.标量类型；或
3.参考类型；或
4.文字类型的数组；或
可能具有cv资格的类类型，具有以下所有属性:

A type is a literal type if it is:
1. possibly cv-qualified void; or
2. a scalar type; or
3. a reference type; or
4. an array of literal type; or
a possibly cv-qualified class type that has all of the following properties:

1. 它有一个琐碎的析构函数
2. 它是闭包类型，聚合类型，或者至少具有一个constexpr构造函数或构造函数模板(可能从基类继承)，而不是复制或移动构造函数，
3. 如果不是联合，则其所有非静态数据成员和基类均为非易失性文字类型.

关键是，这些类型为 VariableLengthString 的对象的子对象类型是否都是文字类型?类 VariableLengthString 是否至少有一个constexpr构造函数?是的.因为这些规则:

The key point is that,Are these types of sub-objects of ojbects of type VariableLengthString all literal types? Does the class VariableLengthString have at least one constexpr constructor?Yes they are.Because of these rules:

算术类型，枚举类型，指针类型，指向成员类型的指针([basic.compound])，std :: nullptr_t和这些类型的cv限定版本统称为标量类型.

Arithmetic types, enumeration types, pointer types, pointer to member types ([basic.compound]), std​::​nullptr_­t, and cv-qualified versions of these types are collectively called scalar types.

因此，子对象 data _ 是指针类型.因此，它是标量类型，也是文字类型.满足项目符号3.是否 constexpr VariableLengthString(const char *p)一个constexpr构造函数?是的，因为这些规则:

So,the sub-object data_,it's of type pointer.hence,it's scalar type,also a literal type.The bullet 3 is satisfied.Does constexpr VariableLengthString(const char *p) a constexpr constructor?Yes,it is,Because of these rules:

constexpr构造函数的定义应满足以下要求:

The definition of a constexpr constructor shall satisfy the following requirements:

1. 该类不应具有任何虚拟基类；
2. 每种参数类型应为文字类型；
3. 每个不变的非静态数据成员和基类子对象都应初始化

对于 constexpr VariableLengthString(const char * p)，这三个规则都得到满足.总之， VariableLengthString 类是一个文字类型和constexpr表达式类型 VariableLengthString 可以用作非类型模板参数，因为它满足转换后的常量表达式的要求.为什么上面的代码格式错误?错过了一些，请帮助我找出它们.

For constexpr VariableLengthString(const char *p),these three rules are all be satisfied.In summary,the class VariableLengthString is a literal type and a constexpr expression of type VariableLengthString could be used as a non-type template argument,because it satisfies the requirement of being a converted constant expression.why the code above is ill-formed?If I miss something,please help me find out them.