问题描述

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.

所以不要这样做:

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

我是这样做的:

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，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.

推荐答案

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!

或者您可以选择转换为底层整数类型为:

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);
}

然后将其用作:

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！