mfontanini mfontanini - 1 year ago 195
C++ Question

std::get using enum class as template argument

I'm using a

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

So instead of doing this:

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

I did this:

enum class Something {

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

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!

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

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 ...>&)

So, is there any way I can use my enum class as a template argument to
? 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.

Answer Source

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.

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);

then use it as:

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