[英]How to define is_iterator type trait?
I'm trying to code a is_iterator<T>
type trait. 我正在尝试编写一个is_iterator<T>
类型的特征。 Where when T
is an iterator type is_iterator<T>::value == true
otherwise is is_iterator<T>::value == false
. 当T
是迭代器类型时, is_iterator<T>::value == true
否则is_iterator<T>::value == false
。
What I tried so far: 到目前为止我尝试了什么:
template <class, class Enable = void>
struct is_iterator : std::false_type {};
template <typename T>
struct is_iterator<T, typename std::enable_if<std::is_pointer<typename
std::iterator_traits<T>::pointer>::value>::type> : std::true_type {};
Q: Is there a more proper way to define a is_iterator
type trait than the one displayed above? 问:是否有更合适的方法来定义is_iterator
类型特征而不是上面显示的特征?
As I said in comments, the solutions presented here rely on non-portable properties of iterators_traits
in some implementations. 正如我在评论中所说,这里介绍的解决方案在某些实现中依赖于iterators_traits
非可移植属性。 According to the C++03 and C++11 standards, iterator_traits
is only defined for iterators (and the special case of pointers) so any other use is undefined behaviour. 根据C ++ 03和C ++ 11标准, iterator_traits
仅为迭代器(以及指针的特殊情况)定义,因此任何其他用途都是未定义的行为。 Specifically, using iterator_traits<T>::pointer
in a SFINAE context isn't guaranteed to work, because instantiating iterator_traits<T>
will refer to T::value_type
, T::pointer
, T::iterator_category
etc. and that happens outside the "immediate context" where SFINAE doesn't apply. 具体来说,在SFINAE上下文中使用iterator_traits<T>::pointer
并不能保证工作,因为实例化iterator_traits<T>
将引用T::value_type
, T::pointer
, T::iterator_category
等,这些都发生在外面SFINAE不适用的“直接背景”。
C++14
will fix that
was supposed to fix that (it happened post-C++14 with DR 2408 ), but for C++11 the safe way to define is_iterator
is to write a trait that checks for all the required operations an iterator must define. C ++ 14
将修复
本应该解决的问题(它发生在使用DR 2408的C ++ 14之后),但是对于C ++ 11,定义is_iterator
的安全方法是编写一个检查所有必需操作的特征迭代器必须定义。 The only operations that all iterators are required to support are operator*
and pre- and post-increment. 所有迭代器都需要支持的唯一操作是operator*
以及前后增量。 Unfortunately, there can be types that define those operations which are not valid iterators, so writing a correct trait is quite hard. 不幸的是,可能存在定义那些不是有效迭代器的操作的类型,因此编写正确的特征非常困难。
Your check fails if std::iterator_traits<T>::pointer
is a type that is not a pointer, for instance if T = std::ostream_iterator<U>
. 如果std::iterator_traits<T>::pointer
是不是指针的类型,则检查失败,例如,如果T = std::ostream_iterator<U>
。
I think a better test might be whether std::iterator_traits<T>::iterator_category
is either std::input_iterator_tag
or a derived type, or std::output_iterator_tag
. 我认为更好的测试可能是std::iterator_traits<T>::iterator_category
是std::input_iterator_tag
还是派生类型,或std::output_iterator_tag
。
template <class, class Enable = void> struct is_iterator : std::false_type {};
template <typename T>
struct is_iterator
<T,
typename std::enable_if<
std::is_base_of<std::input_iterator_tag, typename std::iterator_traits<T>::iterator_category>::value ||
std::is_same<std::output_iterator_tag, typename std::iterator_traits<T>::iterator_category>::value
>::type>
: std::true_type {};
I think there's no need to check for any particular property of iterator_traits
's nested typedefs. 我认为没有必要检查iterator_traits
的嵌套typedef的任何特定属性。 It should be enough to check for mere presence of iterator_traits<T>::value_type
(or any other nested typedef, for that matter), because every iterator has one. 它应该足以检查是否存在iterator_traits<T>::value_type
(或任何其他嵌套的typedef),因为每个迭代器都有一个。
#include <type_traits>
#include <iostream>
#include <iterator>
template<typename>
struct void_ {
typedef void type;
};
// remove typename spam below:
template<typename Discard>
using void_t=typename void_<Discard>::type;
template<typename T>
using decay_t=typename std::decay<T>::type;
// stick helper types into details, so the interface
// for is_iterator is cleaner:
namespace details {
template<typename T, typename Enable=void>
sturct is_iterator : is_iterator2<T, Enable> {};
// special case: void* is not an iterator
// but T* specialization would pick it up
// if there weren't the following:
template<typename V>
struct is_iterator<V*, decay_t<V>> : std::false_type {};
// phase 2: SFINAE pass to std::iterator_traits test
// valid in C++14, and in many C++11 compilers, except
// for above void issue:
template<typename, typename Enable = void>
struct is_iterator2 : std::false_type {};
template<typename T>
struct is_iterator2<T,
void_t< typename std::iterator_traits<T>::value_type>
> : std::true_type {};
}
template<typename T>
struct is_iterator : details::is_iterator<T> {};
int main()
{
std::cout
<< is_iterator<int*>::value
<< is_iterator<double>::value;
}
Unfortunately this isn't guaranteed to work in C++11, but C++14 will fix it (thanks to Jonathan Wakely for pointing it out). 不幸的是,这不能保证在C ++ 11中有效,但是C ++ 14会修复它(感谢Jonathan Wakely指出它)。
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