c++ - How to restrict typenames in template -

have function can accept 4 different types. , implementations similar.

template< typename t > void foo( std::string &&name, t value ){     //... } 

typename t must 1 of 4 predefined types. other typenames not acceptable, , should restricted @ compile-time.
possible write in c++?

<type_traits> let generalize logic class template. how works take template parameter t , parameter pack of ts , start checking if t same head of list of types. if match found, inherit std::true_type , done. if no match found, pop head , recursively instantiate same template tail of ts. eventually, if no match found @ all, parameter pack size drop 0 , compiler instantiate base template class inherit std::false_type. please check video better , in dept explanation mr. walter e. brown.

template<class t, class...> struct is_any_of: std::false_type{}; template<class t, class head, class... tail>  struct is_any_of<t, head, tail...>: std::conditional_t<   std::is_same<t, head>::value,    std::true_type,   is_any_of<t, tail...>> {}; 

now can sfinae over, using enable_if in english wording.

#include <type_traits> #include <string>  template<   class t,   class = std::enable_if_t<is_any_of<t, int, float, unsigned, double>::value> > void foo(std::string &&str, t value) {}   int main() {   foo(std::string{"hello"}, 3);   foo(std::string{"world"}, '0'); //compile-time error } 

sfanie language feature, tool, that's used or abused achieve ask for,

the standard library component std::enable_if allows creating substitution failure in order enable or disable particular overloads based on condition evaluated @ compile time. fyi http://en.cppreference.com/w/cpp/language/sfinae.

note std::conditional_t<> , std::enable_if_t<> shorthanded std::conditional<>::type , std::enable_if<>::type respectively. simple replace these in code, should put typename keyword before enable_if then.