c++ - Metaprogramming with std::is_same -

is possible following compiles without template specialization?

template <class t>  class { public:   #if std::is_same<t,int>   void has_int() {}   #elif std::is_same<t,char>   void has_char() {}   #endif }; a<int> a; a.has_int(); a<char> b; b.has_char(); 

yes. make function templates , conditionaly enable them using std::enable_if:

#include <type_traits>  template <class t>  class { public:    template<typename u = t>   typename std::enable_if<std::is_same<u,int>::value>::type   has_int() {}    template<typename u = t>   typename std::enable_if<std::is_same<u,char>::value>::type   has_char() {} };  int main() {     a<int> a;     a.has_int();   // ok     // a.has_char();  // error } 

the solution the other answer might not feasible if class big , has got many functions need regardless of t. can solve inheriting class used these special methods. then, can specialize base class only.

in c++14, there convenient type aliases syntax can become:

std::enable_if_t<std::is_same<u, int>::value> 

and c++17, shorter:

std::enable_if_t<std::is_same_v<u, int>>