为什么这段简短的模板代码有效?

问题描述

按照标题，当has_type_struct<no_type>肯定是无效类型时，我不理解下面的代码如何编译.

As per the title, I do not understand how can the following code compile when has_type_struct<no_type> is certainly an invalid type.

template<typename T>
using my_int = int;

struct no_type {};

template<typename T>
struct has_type_struct { using type = typename T::type; };

template<typename T>
using has_type_using = typename T::type;

int main() {
   my_int<has_type_struct<no_type>> a; // why does this compile?
   //my_int<has_type_using<no_type>>(); // this rightfully does not compile
   return 0;
}

推荐答案

该程序有效，因为未实例化has_type_struct<no_type>.

The program is valid because has_type_struct<no_type> is not instantiated.

[temp.inst]/1 :

my_int<has_type_struct<no_type>>的使用不需要完成has_type_struct<no_type>，因此不会实例化后者，并且不会检查其定义中从属名称的有效性.

The use of my_int<has_type_struct<no_type>> does not require has_type_struct<no_type> to be complete, therefore the latter is not instantiated and the validity of the dependent name in its definition is not checked.

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