如何检查迭代器是否是c ++中的output_iterator? [英] How to check if an iterator is an output_iterator in c++?
问题描述
template<typename Iterator>
void put_value(Iterator pos, int n)
{
static_assert(IsOutputIterator<Iterator>);
//
// How to implement IsOutputIterator?
//
*pos = n;
}
std :: iterator_traits< Iterator> :: iterator_category
没有帮助。例如: vector< int> :: iterator
显然是 output_iterator
,但是 std: :iterator_traits<矢量< INT> ::迭代> ::的iterator_category
将返回 random_access_iterator
,这可能不是一个 output_iterator
,比如 const_iterator
。
std::iterator_traits<Iterator>::iterator_category
doesn't help. For example: vector<int>::iterator
is obvious an output_iterator
, but std::iterator_traits<vector<int>::iterator>::iterator_category
will returns random_access_iterator
, which might not be an output_iterator
, say a const_iterator
.
有没有可行的方法来检查如果迭代器是c ++中的output_iterator?
推荐答案
我的第一个回答是问输出迭代器是为了什么? C ++输出迭代器不指定值类型,因为相同的迭代器可能能够输出多个值类型。确定是否可以通过给定的迭代器 o
编写给定表达式 E
的唯一方法是查看是否 * o = std :: declval< decltype((E))>()
是一个有效的表达式。
My first response is to ask "Output iterator for what?" C++ output iterators don't specify a value type, because the same iterator may be able to output multiple value types. The only way to determine if you can write a given expression E
through a given iterator o
is to see if *o = std::declval<decltype((E))>()
is a valid expression.
In C ++ 14,我定义了这样做的特性:
In C++14, I'd define a trait to do so:
template <class...> using void_t = void;
template <class, class, class = void>
constexpr bool is_output_iterator = false;
template <class I, class E>
constexpr bool is_output_iterator<I, E, void_t<
typename std::iterator_traits<I>::iterator_category,
decltype(*std::declval<I>() = std::declval<E>())>> = true;
在带有概念的C ++中 - 我假设您感兴趣,因为您用 c ++ - concepts - 我会从github获取示例Ranges TS实现并使用其 OutputIterator< I,E> ;()
概念。
In C++ with concepts - which I assume you are interested in since you tagged this question with c++-concepts - I would grab the sample Ranges TS implementation from github and use its OutputIterator<I, E>()
concept instead.
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