templates - type as returntype in c++ -

is there possibility return type returntype function , use member variable using this:

constexpr type myfunction(int a, int b){      if(a + b == 8) return int_8t;      if(a + b == 16) return int_16t;      if(a + b == 32) return int_32t;      return int_64t; }  template<int x, int y> class test{     using type = typename myfunction(x, y);     private:     type m_variable; }; 

when trying example in qt says

error: 'constexpr' not name type error: 'test' not template type class test{       ^ 

in earlier question showed me http://en.cppreference.com/w/cpp/types/conditional function, works 2 types.

you cannot normal function. however, done using template meta-programming. kind of template called type function.

#include <cstdint>  template<int bits> struct integer { /* empty */ };  // specialize bit widths want. template<> struct integer<8>  { typedef int8_t  type; }; template<> struct integer<16> { typedef int16_t type; }; template<> struct integer<32> { typedef int32_t type; }; 

it can used this.

using integer_type = integer<16>::type; integer_type number = 42; 

remember precede integer<t>::type typename keyword if t template parameter.

i leave exercise extend template accepts 2 integers parameters , returns appropriate type based on sum of two.