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      std::get 使用枚举类作为模板参数

      std::get using enum class as template argument(std::get 使用枚举类作为模板参数)
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              • 本文介绍了std::get 使用枚举类作为模板参数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!


                我正在使用 std::tuple 并定义了一个类枚举以某种方式命名"元组的每个字段,而忘记了它们的实际索引.

                I'm using a std::tuple and defined a class enum to somehow "naming" each of the tuple's fields, forgetting about their actual indexes.


                std::tuple<A,B> tup;
                /* ... */
                std::get<0>(tup) = bleh; // was it 0, or 1?


                enum class Something {
                     MY_INDEX_NAME = 0,
                std::tuple<A,B> tup;
                /* ... */
                std::get<Something::MY_INDEX_NAME> = 0; // I don't mind the actual index...

                问题是,由于这是使用 gcc 4.5.2 编译的,我现在已经安装了 4.6.1 版本,但我的项目无法编译.此代码段重现了错误:

                The problem is that, as this compiled using gcc 4.5.2, I've now installed the 4.6.1 version, and my project failed to compile. This snippet reproduces the error:

                #include <tuple>
                #include <iostream>
                enum class Bad {
                    BAD = 0
                enum Good {
                    GOOD = 0
                int main() {
                    std::tuple<int, int> tup(1, 10);
                    std::cout << std::get<0>(tup) << std::endl;
                    std::cout << std::get<GOOD>(tup) << std::endl; // It's OK
                    std::cout << std::get<Bad::BAD>(tup) << std::endl; // NOT!

                该错误基本上是说没有与我对 std::get 的调用相匹配的重载:

                The error basically says that there's no overload that matches my call to std::get:

                test.cpp: In function ‘int main()’:
                test.cpp:16:40: error: no matching function for call to ‘get(std::tuple<int, int>&)’
                test.cpp:16:40: note: candidates are:
                /usr/include/c++/4.6/utility:133:5: note: template<unsigned int _Int, class _Tp1, class _Tp2> typename std::tuple_element<_Int, std::pair<_Tp1, _Tp2> >::type& std::get(std::pair<_Tp1, _Tp2>&)
                /usr/include/c++/4.6/utility:138:5: note: template<unsigned int _Int, class _Tp1, class _Tp2> const typename std::tuple_element<_Int, std::pair<_Tp1, _Tp2> >::type& std::get(const std::pair<_Tp1, _Tp2>&)
                /usr/include/c++/4.6/tuple:531:5: note: template<unsigned int __i, class ... _Elements> typename std::__add_ref<typename std::tuple_element<__i, std::tuple<_Elements ...> >::type>::type std::get(std::tuple<_Elements ...>&)
                /usr/include/c++/4.6/tuple:538:5: note: template<unsigned int __i, class ... _Elements> typename std::__add_c_ref<typename std::tuple_element<__i, std::tuple<_Elements ...> >::type>::type std::get(const std::tuple<_Elements ...>&)

                那么,有什么方法可以将我的枚举类用作 std::get 的模板参数?这是不是编译的东西,并在 gcc 4.6 中修复了?我可以使用简单的枚举,但我喜欢枚举类的作用域属性,所以如果可能的话,我更喜欢使用后者.

                So, is there any way I can use my enum class as a template argument to std::get? Was this something that wasn't ment to compile, and was fixed in gcc 4.6? I could use a simple enum, but I like the scoping properties of enum classes, so I'd prefer to use the latter if it's possible.


                C++11 引入的强类型枚举不能隐式转换为 int 类型的整数值,而 std::get 期望模板参数是整型.

                The strongly-typed enums introduced by C++11 cannot be implicitly converted into integral values of type say int, while std::get expects the template argument to be integral type.

                您必须使用 static_cast 来转换枚举值:

                You've to use static_cast to convert the enum values:

                std::cout <<std::get<static_cast<int>(Bad::BAD)>(tup)<< std::endl; //Ok now!


                Or you can choose to convert into underlying integral type as:

                //note that it is constexpr function
                template <typename T>
                constexpr typename std::underlying_type<T>::type integral(T value) 
                    return static_cast<typename std::underlying_type<T>::type>(value);


                std::cout <<std::get<integral(Bad::BAD)>(tup)<< std::endl; //Ok now!

                这篇关于std::get 使用枚举类作为模板参数的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!



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