Determine if a type is an STL container at compile time

问题

I would like to write a template that will determine if a type is an stl container at compile time.  

I\'ve got the following bit of code:

struct is_cont{};
struct not_cont{};

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

but I\'m not sure how to create the necessary specializations for std::vector<T,Alloc>, deque<T,Alloc>, set<T,Alloc,Comp> etc...

回答1:

First, you define your primary template, which will have a member which is false in the default case:

template <typename T>
struct is_cont {
  static const bool value = false;
};

Then you will define partial specializations for your container types which have a value of true instead:

template <typename T,typename Alloc>
struct is_cont<std::vector<T,Alloc> > {
  static const bool value = true;
};

Then for a type X that you want to check, use it like

if (is_cont<X>::value) { ... } 

回答2:

Note: the following code is taken from an excellent utility called pretty-print written by @Kerrek SB (a topic on it at stackoverflow).

Disclaimer : I don't know if I'm allowed to copy and paste this code here without taking permission from the original author. @Kerrek, let me know if you've any issue. :-)

---

You can use this classs template:

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

Usage:

 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

Note that is_container needs following helper class templates:

template<typename T>
struct has_const_iterator
{
private:
    typedef char                      yes;
    typedef struct { char array[2]; } no;

    template<typename C> static yes test(typename C::const_iterator*);
    template<typename C> static no  test(...);
public:
    static const bool value = sizeof(test<T>(0)) == sizeof(yes);
    typedef T type;
};

template <typename T>
struct has_begin_end
{
    template<typename C> static char (&f(typename std::enable_if<
      std::is_same<decltype(static_cast<typename C::const_iterator (C::*)() const>(&C::begin)),
      typename C::const_iterator(C::*)() const>::value, void>::type*))[1];

    template<typename C> static char (&f(...))[2];

    template<typename C> static char (&g(typename std::enable_if<
      std::is_same<decltype(static_cast<typename C::const_iterator (C::*)() const>(&C::end)),
      typename C::const_iterator(C::*)() const>::value, void>::type*))[1];

    template<typename C> static char (&g(...))[2];

    static bool const beg_value = sizeof(f<T>(0)) == 1;
    static bool const end_value = sizeof(g<T>(0)) == 1;
};

回答3:

Many of the already proposed solutions are verbose for detecting STL containers.
They focus on the characteristics that all containers possess, instead of explicitly stating what the containers are.

If you wanted to create your own containers and have them evaluated with a true type, I'd recommend the other solutions. If you only want to validate legitimate STL containers, and not STL-like containers, consider using the following implementation, as it provides precise STL container detection:

#include <deque>
#include <forward_list>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <type_traits>

//specialize a type for all of the STL containers.
namespace is_stl_container_impl{
  template <typename T>       struct is_stl_container:std::false_type{};
  template <typename T, std::size_t N> struct is_stl_container<std::array    <T,N>>    :std::true_type{};
  template <typename... Args> struct is_stl_container<std::vector            <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::deque             <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::list              <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::forward_list      <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::set               <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::multiset          <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::map               <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::multimap          <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::unordered_set     <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::unordered_multiset<Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::unordered_map     <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::unordered_multimap<Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::stack             <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::queue             <Args...>>:std::true_type{};
  template <typename... Args> struct is_stl_container<std::priority_queue    <Args...>>:std::true_type{};
}

//type trait to utilize the implementation type traits as well as decay the type
template <typename T> struct is_stl_container {
  static constexpr bool const value = is_stl_container_impl::is_stl_container<std::decay_t<T>>::value;
};

Note the use of std::decay to avoid incorrect type deduction based on type qualifiers. Also, we've utilized inheriting std::true_type and std::false_type to avoid specifying the ::type types ourselves. C++11 variadic templates are used to deduce the n amount of template type parameters needed to construct the containers.

Using the implementation is as you would expect:

  std::cout << std::boolalpha;
  std::cout << is_stl_container<std::vector<int>>::value << '\n';
  std::cout << is_stl_container<std::vector<int>const&>::value << '\n';
  std::cout << is_stl_container<int>::value << '\n';

prints:

true
true
false

回答4:

Pursuing the suggestion that a generic compiletime test for has-an-stl-container-like-interface would be an appropriate solution, this one defines an stl-like container T by the interface:

T::iterator T::begin();
T::iterator T::end();
T::const_iterator T::begin() const;
T::const_iterator T::end() const;

*T::iterator is T::value_type &
*T::const_iterator is T::value_type const &

Additional requirements, e.g. a size() method, could be added in an obvious fashion, or other canonical type interfaces probed at compiletime in an obvious similar way.

#ifndef IS_STL_CONTAINER_LIKE_H
#define IS_STL_CONTAINER_LIKE_H

#include <type_traits>

template<typename T>
struct is_stl_container_like
{
    typedef typename std::remove_const<T>::type test_type;

    template<typename A>
    static constexpr bool test(
        A * pt,
        A const * cpt = nullptr,
        decltype(pt->begin()) * = nullptr,
        decltype(pt->end()) * = nullptr,
        decltype(cpt->begin()) * = nullptr,
        decltype(cpt->end()) * = nullptr,
        typename A::iterator * pi = nullptr,
        typename A::const_iterator * pci = nullptr,
        typename A::value_type * pv = nullptr) {

        typedef typename A::iterator iterator;
        typedef typename A::const_iterator const_iterator;
        typedef typename A::value_type value_type;
        return  std::is_same<decltype(pt->begin()),iterator>::value &&
                std::is_same<decltype(pt->end()),iterator>::value &&
                std::is_same<decltype(cpt->begin()),const_iterator>::value &&
                std::is_same<decltype(cpt->end()),const_iterator>::value &&
                std::is_same<decltype(**pi),value_type &>::value &&
                std::is_same<decltype(**pci),value_type const &>::value;

    }

    template<typename A>
    static constexpr bool test(...) {
        return false;
    }

    static const bool value = test<test_type>(nullptr);

};

#endif

Here is a test program, built with GCC 4.7.2, clang 3.2, Intel C++ 13.1.1:

#include "is_stl_container_like.h"

// Testing ...

#include <iostream>
#include <vector>
#include <array>
#include <functional>

using namespace std;

template<class C>
struct polymorphic : private C
{
    typedef typename C::value_type value_type;
    typedef typename C::iterator iterator;
    typedef typename C::const_iterator const_iterator;

    virtual ~polymorphic(){}

    virtual const_iterator begin() const {
        return C::begin();
    }

    virtual iterator begin()  {
        return C::begin();
    }

    virtual const_iterator end() const {
        return C::end();
    }

    virtual iterator end()  {
        return C::end();
    }   
};

template<class C>
struct reject : private C
{
    typedef typename C::value_type value_type;
    typedef typename C::iterator iterator;
    typedef typename C::const_iterator const_iterator;


    const_iterator begin() {
        return C::begin();
    }

    iterator begin() const {
        return C::begin();
    }

    const_iterator end() {
        return C::end();
    }

    iterator end() const {
        return C::end();
    }
};

int main()
{
    cout << is_stl_container_like<vector<int> const >::value << endl; // Yes
    cout << is_stl_container_like<array<int,42>>::value << endl; // Yes
    cout << is_stl_container_like<polymorphic<vector<int>>>::value << endl; // Yes
    cout << is_stl_container_like<function<int(int)>>::value << endl; // No
    cout << is_stl_container_like<int>::value << endl; // No
    cout << is_stl_container_like<reject<vector<int>>>::value << endl; //No
}

回答5:

There is is_container in boost http://www.boost.org/doc/libs/1_51_0/libs/spirit/doc/html/spirit/advanced/customize/is_container.html

is_container<C>::type --- Result of the metafunction that evaluates to mpl::true_ if a given type, C, is to be treated as a container, mpl::false_ otherwise Generally, any implementation of is_container needs to behave as if if was a MPL Boolean Constant..

回答6:

This code defines traits for container. It's originally from prettyprint library:

//put this in type_utils.hpp 
#ifndef commn_utils_type_utils_hpp
#define commn_utils_type_utils_hpp

#include <type_traits>
#include <valarray>

namespace common_utils { namespace type_utils {
    //from: https://raw.githubusercontent.com/louisdx/cxx-prettyprint/master/prettyprint.hpp
    //also see https://gist.github.com/louisdx/1076849
    namespace detail
    {
        // SFINAE type trait to detect whether T::const_iterator exists.

        struct sfinae_base
        {
            using yes = char;
            using no  = yes[2];
        };

        template <typename T>
        struct has_const_iterator : private sfinae_base
        {
        private:
            template <typename C> static yes & test(typename C::const_iterator*);
            template <typename C> static no  & test(...);
        public:
            static const bool value = sizeof(test<T>(nullptr)) == sizeof(yes);
            using type =  T;

            void dummy(); //for GCC to supress -Wctor-dtor-privacy
        };

        template <typename T>
        struct has_begin_end : private sfinae_base
        {
        private:
            template <typename C>
            static yes & f(typename std::enable_if<
                std::is_same<decltype(static_cast<typename C::const_iterator(C::*)() const>(&C::begin)),
                             typename C::const_iterator(C::*)() const>::value>::type *);

            template <typename C> static no & f(...);

            template <typename C>
            static yes & g(typename std::enable_if<
                std::is_same<decltype(static_cast<typename C::const_iterator(C::*)() const>(&C::end)),
                             typename C::const_iterator(C::*)() const>::value, void>::type*);

            template <typename C> static no & g(...);

        public:
            static bool const beg_value = sizeof(f<T>(nullptr)) == sizeof(yes);
            static bool const end_value = sizeof(g<T>(nullptr)) == sizeof(yes);

            void dummy(); //for GCC to supress -Wctor-dtor-privacy
        };

    }  // namespace detail

    // Basic is_container template; specialize to derive from std::true_type for all desired container types

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

    template <typename T, std::size_t N>
    struct is_container<T[N]> : std::true_type { };

    template <std::size_t N>
    struct is_container<char[N]> : std::false_type { };

    template <typename T>
    struct is_container<std::valarray<T>> : std::true_type { };

    template <typename T1, typename T2>
    struct is_container<std::pair<T1, T2>> : std::true_type { };

    template <typename ...Args>
    struct is_container<std::tuple<Args...>> : std::true_type { };

}}  //namespace
#endif

For more explanation see my blog post.

Related question: c++ template class; function with arbitrary container type, how to define it?

来源：https://stackoverflow.com/questions/9407367/determine-if-a-type-is-an-stl-container-at-compile-time