c++ - std::conditional compile-time branch evaluation -

compiling this:

template < class t, class y, class ...args > struct issame {     static constexpr bool value = std::conditional<                                       sizeof...( args ),                                       typename std::conditional<                                           std::is_same< t, y >::value,                                           issame< y, args... >, // error!                                           std::false_type >::type,                                       std::is_same< t, y > >::type::value; };  int main() {     qdebug() << issame< double, int >::value;     return exit_success; } 

gives me compiler error:

error: wrong number of template arguments (1, should 2 or more) 

the issue issame< double, int > has empty args parameter pack, issame< y, args... > becomes issame< y > not match signature.

but question is: why branch being evaluated @ all? sizeof...( args ) false, inner std:conditional should not evaluated. isn't runtime piece of code, compiler knows sizeof..( args ) never true given template types.

if you're curious, it's supposed variadic version of std::is_same, not works...

you have error because type has correct when used template parameter.
may use template specialization solve issue, like:

#include <type_traits>  template <typename ... ts> struct are_same;  template <> struct are_same<> : std::true_type {}; template <typename t> struct are_same<t> : std::true_type {};  template <typename t1, typename t2, typename... ts> struct are_same<t1, t2, ts...> :     std::conditional<         std::is_same<t1, t2>::value,         are_same<t2, ts...>,         std::false_type     >::type {};  static_assert(are_same<char, char, char>::value, "all type should identical"); static_assert(!are_same<char, char, int>::value, "all type should not identical");