How Can I Determine if a Type is an STL Container at Compile Time?

Determine if a Type is an STL Container at Compile Time

In many programming scenarios, it can be beneficial to know whether or not a specific type is an STL container at compile time. This allows for optimizing algorithms or data structures based on the type of container being used.

One approach is to utilize a template struct to determine the container type:

struct is_cont{};
struct not_cont{};

template <typename T>
struct is_cont { typedef not_cont result_t; };

However, this approach requires creating specializations for each STL container type, such as std::vector and std::deque.

A more comprehensive solution involves using helper class templates:

template<typename T> 
struct is_container : std::integral_constant<bool, has_const_iterator<T>::value &amp;&amp; has_begin_end<T>::beg_value &amp;&amp; has_begin_end<T>::end_value> 
{ };

This class template checks for the following properties:

• has_const_iterator::value ensures the existence of a const_iterator type.
• has_begin_end::beg_value and has_begin_end::end_value check if the container has begin and end methods, respectively.

Usage示例：

std::cout << is_container<std::vector<int>>::value << std::endl; // true
std::cout << is_container<std::list<int>>::value << std::endl; // true 
std::cout << is_container<std::map<int>>::value << std::endl; // true
std::cout << is_container<std::set<int>>::value << std::endl; // true
std::cout << is_container<int>::value << std::endl; // false

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