問題描述

是否可以編寫一個值對所有常見 STL 結(jié)構(gòu)(例如，vector、set、map、map、...)?

首先，我想編寫一個類型特征，它對 vector 為真，否則為假.我試過這個，但它不能編譯:

template結(jié)構(gòu) is_vector {static bool const value = false;};模板<T類，U類>struct is_vector>>::類型>{static bool const value = true;};

錯誤信息是模板參數(shù)未用于部分特化:U.

看，另一個基于 SFINAE 的用于檢測類 STL 容器的解決方案:

template結(jié)構(gòu) is_container : std::false_type {};模板結(jié)構(gòu) is_container_helper {};模板struct is_container<,std::conditional_t<錯誤的，is_container_helper<類型名稱 T::value_type，類型名稱 T::size_type,類型名稱 T::allocator_type,類型名 T::iterator,類型名 T::const_iterator,decltype(std::declval().size()),decltype(std::declval().begin()),decltype(std::declval().end()),decltype(std::declval().cbegin()),decltype(std::declval().cend())>,空白>>:公共 std::true_type {};

當(dāng)然，您可以更改要檢查的方法和類型.

如果您只想檢測 STL 容器(這意味著 std::vector、std::list 等)，您應(yīng)該執(zhí)行類似 這個.

更新.正如@Deduplicator 所指出的，容器可能不滿足 AllocatorAwareContainer 要求(例如:std::array).這就是為什么不需要檢查 T::allocator_type 的原因.但是您可以以類似的方式檢查任何/所有Container要求.>

Is it possible to write a type trait whose value is true for all common STL structures (e.g., vector, set, map, ...)?

To get started, I'd like to write a type trait that is true for a vector and false otherwise. I tried this, but it doesn't compile:

template<class T, typename Enable = void>
struct is_vector {
  static bool const value = false;
};

template<class T, class U>
struct is_vector<T, typename boost::enable_if<boost::is_same<T, std::vector<U> > >::type> {
  static bool const value = true;
};

The error message is template parameters not used in partial specialization: U.

Look, another SFINAE-based solution for detecting STL-like containers:

template<typename T, typename _ = void>
struct is_container : std::false_type {};

template<typename... Ts>
struct is_container_helper {};

template<typename T>
struct is_container<
        T,
        std::conditional_t<
            false,
            is_container_helper<
                typename T::value_type,
                typename T::size_type,
                typename T::allocator_type,
                typename T::iterator,
                typename T::const_iterator,
                decltype(std::declval<T>().size()),
                decltype(std::declval<T>().begin()),
                decltype(std::declval<T>().end()),
                decltype(std::declval<T>().cbegin()),
                decltype(std::declval<T>().cend())
                >,
            void
            >
        > : public std::true_type {};

Of course, you might change methods and types to be checked.

If you want to detect only STL containers (it means std::vector, std::list, etc) you should do something like this.

UPDATE. As @Deduplicator noted, container might not meet AllocatorAwareContainer requirements (e.g.: std::array<T, N>). That is why check on T::allocator_type is not neccessary. But you may check any/all Container requirements in a similar way.

這篇關(guān)于如何編寫類型特征“is_container"或“is_vector"?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！