这种元组递归引用安全吗？
Is this recursive reference with tuples safe?
我正在试验并写了这个怪物：
class my_tuple : public std::tuple < std::vector<my_tuple> > {};
它似乎可以编译并实际工作。我发现它很狡猾，因为以下内容无法编译：
using my_other_tuple = std::tuple < std::vector<my_other_tuple> > ;
最终，我想弄明白为什么 my_tuple 有效，以及是否有任何潜在的可怕后果。我正在尝试了解元组的全部内容以及我 can/am 应该如何处理它们。因此，如果有人愿意对此发表评论，提供一些精辟的见解，我将不胜感激。
Windows 7 和 VS 2013。
class my_tuple : public std::tuple < std::vector<my_tuple> > {};
这是当前未定义的行为，因为它实例化了一个标准库容器，std::vector，类型不完整，my_tuple，直到其结束 }定义。但是，有一个proposal to permit instantiating certain standard containers, including std::vector, with incomplete types. Boost.Containers supports incomplete types as well.
using my_other_tuple = std::tuple < std::vector<my_other_tuple> > ;
这是错误的格式。 [dcl.typedef]/p2，强调我的：
A typedef-name can also be introduced by an alias-declaration. The
identifier following the using keyword becomes a typedef-name and the optional attribute-specifier-seq following the identifier
appertains to that typedef-name. It has the same semantics as if it
were introduced by the typedef specifier. In particular, it does not
define a new type and it shall not appear in the type-id.
我正在试验并写了这个怪物：
class my_tuple : public std::tuple < std::vector<my_tuple> > {};
它似乎可以编译并实际工作。我发现它很狡猾，因为以下内容无法编译：
using my_other_tuple = std::tuple < std::vector<my_other_tuple> > ;
最终，我想弄明白为什么 my_tuple 有效，以及是否有任何潜在的可怕后果。我正在尝试了解元组的全部内容以及我 can/am 应该如何处理它们。因此，如果有人愿意对此发表评论，提供一些精辟的见解，我将不胜感激。
Windows 7 和 VS 2013。
class my_tuple : public std::tuple < std::vector<my_tuple> > {};
这是当前未定义的行为，因为它实例化了一个标准库容器，std::vector，类型不完整，my_tuple，直到其结束 }定义。但是，有一个proposal to permit instantiating certain standard containers, including std::vector, with incomplete types. Boost.Containers supports incomplete types as well.
using my_other_tuple = std::tuple < std::vector<my_other_tuple> > ;
这是错误的格式。 [dcl.typedef]/p2，强调我的：
A typedef-name can also be introduced by an alias-declaration. The identifier following the
usingkeyword becomes a typedef-name and the optional attribute-specifier-seq following the identifier appertains to that typedef-name. It has the same semantics as if it were introduced by thetypedefspecifier. In particular, it does not define a new type and it shall not appear in the type-id.