c – std :: enable_if或SFINAE用于迭代器或指针

template<typename T,typename = decltype(*std::declval<T&>(),void(),++std::declval<T&>(),void())>
    MyClass(T);

考虑24.2.2：2中的迭代器要求：

template<typename T> typename std::enable_if<
    !std::is_void<decltype(*std::declval<T &>())>::value
    && std::is_same<decltype(++std::declval<T &>()),typename std::add_lvalue_reference<T>::type>::value,std::true_type>::type has_iterator_requirements_helper(int);
template<typename T> std::false_type has_iterator_requirements_helper(...);
template<typename T> struct has_iterator_requirements:
    decltype(has_iterator_requirements_helper<T>(0)) {};

template<typename,bool> struct is_iterator_check: std::false_type {};
template<typename T> struct is_iterator_check<T,true>: std::true_type {
    typedef typename std::iterator_traits<T>::difference_type difference_type;
    typedef typename std::iterator_traits<T>::value_type value_type;
    typedef typename std::iterator_traits<T>::iterator_category iterator_category;
    typedef typename std::iterator_traits<T>::reference reference;
    typedef typename std::iterator_traits<T>::pointer pointer;
    static_assert(std::is_same<decltype(*std::declval<T &>()),reference>::value
        || std::is_void<reference>::value,"*r must be of type reference");
};
template<typename T> struct is_iterator: is_iterator_check<T,(std::is_pointer<T>::value
     && !std::is_void<typename std::remove_pointer<T>::type>::value
     && !std::is_function<typename std::remove_pointer<T>::type>::value
     ) || (std::is_copy_constructible<T>::value
     && std::is_copy_assignable<T>::value
     && std::is_nothrow_destructible<T>::value
     // TODO: check lvalues are swappable
     && has_iterator_requirements<T>::value
     )> {};

尝试使用iterator_traits的问题在于它是为所有类型定义的模板,并且其实例化将在非SFINAE上下文中失败(回想一下SFINAE仅适用于直接替换失败). libstdc有一个conforming extension,在非迭代器类型上实例化iterator_traits会产生一个空类型;你可以通过检查类型上是否存在iterator_category来做类似的技巧：

template<typename T> std::true_type has_iterator_category_helper(
    T::iterator_category *);
template<typename T> std::false_type has_iterator_category_helper(...);
template<typename T> struct has_iterator_category<T>:
    decltype(has_iterator_category_helper<T>(0)) { };
template<typename T> struct is_iterator: std::integral_constant<bool,std::is_pointer<T>::value || has_iterator_category<T>::value> {};

template<typename T,typename = std::enable_if<is_iterator<T>::value>>
    MyClass(T);

但是,这对于那些本身不公开iterator_category但已经通过单独的iterator_traits特化进行调整的类型不起作用;在这种情况下,简单的SFINAE方法更有意义(并且您可以在构造函数中实例化iterator_traits以确认类型是类迭代器).