我可以/应该从 STL 迭代器继承吗?

For LWG convenience, nine STL iterators are depicted as deriving from std::iterator to get their iterator_category/etc. typedefs. Unfortunately (and unintentionally), this also mandates the inheritance, which is observable (not just through is_base_of, but also overload resolution). This is unfortunate because it confuses users, who can be misled into thinking that their own iterators must derive from std::iterator, or that overloading functions to take std::iterator is somehow meaningful. This is also unintentional because the STL's most important iterators, the container iterators, aren't required to derive from std::iterator. (Some are even allowed to be raw pointers.) Finally, this unnecessarily constrains implementers, who may not want to derive from std::iterator. (For example, to simplify debugger views.)

总而言之，我错了，@aschepler 是对的:它可以使用，但肯定不是必需的 - 也不气馁.整个让我们删除 std::iterator"的东西存在是为了标准不限制标准库实现者.

#include <iterator>

template <typename T>
struct MyIterator : std::iterator<std::random_access_iterator_tag, T> {
   value_type data;  // Error: value_type is not found by name lookup 

   // ... implementations details elided ...
};