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• Type alias for template member function - 3 Updates

Type alias for template member function

Alvin <Alvin@invalid.invalid>: Apr 23 09:51PM +0200 Is there a way to write this more elegantly, introducing an alias for "std::tuple_element<type_nr, tuple_t>::type"?       #include <tuple>   template<typename... types> struct S { using tuple_t = std::tuple<types...>;   template<size_t type_nr> void f(typename std::tuple_element<type_nr, tuple_t>::type v1, typename std::tuple_element<type_nr, tuple_t>::type v2, typename std::tuple_element<type_nr, tuple_t>::type v3, typename std::tuple_element<type_nr, tuple_t>::type v4) { } };

"Alf P. Steinbach" <alf.p.steinbach+usenet@gmail.com>: Apr 23 10:41PM +0200 On 23-Apr-17 9:51 PM, Alvin wrote: > typename std::tuple_element<type_nr, tuple_t>::type v3, > typename std::tuple_element<type_nr, tuple_t>::type v4) { } > };   One possibility:     template< class... Types > struct S { template< int i > using Arg = std::tuple_element_t<i, std::tuple<Types...>>;   template< int i > void f( Arg<i> v1, Arg<i> v2, Arg<i> v3, Arg<i> v4 ) {} };     Cheers & hth.,   - Alf

Alvin <Alvin@invalid.invalid>: Apr 23 11:08PM +0200 On 2017-04-23 22:41, Alf P. Steinbach wrote:   > template< int i > > void f( Arg<i> v1, Arg<i> v2, Arg<i> v3, Arg<i> v4 ) {} > };   That is better.   Thanks!

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