std :: enable_if或SFINAE用于迭代器或指针 [英] std::enable_if or SFINAE for iterator or pointer

问题描述

我想为 MyClass 写一个构造函数，它接受一个参数，只有当参数是一个指针时，我才需要编译。 code>或迭代器（有 iterator_traits ）。如何实现这个？

I would like to write a constructor for MyClass that take an argument and I want this to compile only if the argument is a pointer or an iterator (something having iterator_traits). How to achieve this ?

推荐答案

遗憾的是，没有标准的方法来检测类模型 Iterator 。最简单的检查是 * it 和 ++ it 都是语法有效的;您可以使用标准的SFINAE技术：

Regrettably, there is no standard way to detect whether a class models Iterator. The simplest check would be that *it and ++it are both syntactically valid; you can do this using standard SFINAE techniques:

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

考虑 Iterator 要求从24.2.2 ：2：

Considering the Iterator requirements from 24.2.2:2:

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（记住SFINAE仅适用于直接替换失败）。 libstdc ++具有符合的扩展，借此实例化 iterator_traits 在非迭代器类型将产生一个空类型;您可以通过检查类型上是否存在 iterator_category 来执行类似的操作：

The problem with trying to use iterator_traits is that it is a template defined for all types, and its instantiation will fail in a non-SFINAE context (recall that SFINAE only applies for direct substitution failure). libstdc++ has a conforming extension whereby instantiating iterator_traits on non-iterator types will produce an empty type; you can do a similar trick by checking for the existence of iterator_category on the type:

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);

然而，对于本身并不暴露 iterator_category 但是已经通过单独的 iterator_traits 专门化进行了修改;在这种情况下，简单的SFINAE方法更有意义（您可以在构造函数中实例化 iterator_traits ，以确认类型是类迭代器）。

This will however not work for types that do not themselves expose iterator_category but have been adapted by a separate iterator_traits specialisation; in that case the simple SFINAE method makes more sense (and you can instantiate iterator_traits within the constructor to confirm that the type is iterator-like).

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