std :: get使用枚举类作为模板参数 [英] std::get using enum class as template argument

问题描述

我使用 std :: tuple 并定义了一个类枚举，以某种方式命名每个元组的字段，忘记了他们的实际索引。

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.

而不是这样：

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

我这样做了：

enum class Something {
     MY_INDEX_NAME = 0,
     OTHER_INDEX_NAME
};

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

问题是，使用gcc 4.5.2，我现在安装了4.6.1版本，我的项目无法编译。此代码段重现错误：

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!
}

错误基本上说没有重载匹配我调用 std :: get ：

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 ...>&)

所以，有没有办法我可以使用我的枚举类作为 std :: get 的模板参数？这是不是没有编译，这是固定在gcc 4.6？我可以使用一个简单的枚举，但我喜欢枚举类的范围属性，所以我喜欢使用后者，如果可能。

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.

推荐答案

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

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.

你必须使用 static_cast 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!

或者您可以选择转换为基本积分类型：

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);
}

然后使用它：

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

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