Function overloading with templates - applying enable_if only when alternative function exists

Question

I managed to overload insertion operator for std::ostream based on rank of object. I would like to extend below example.

When A_new.h is included, it includes A.h... in this case I am checking rank of object and I am calling f_alterantive or f_old. Works as expected :)

The problem appears when A.h is included directly, then there only one instance of f() - f_old.

If there is only one instance of f() included directly through A.h, then it should be default for all objects... (not only for rank > 1, but also for all rest) and below condition has no sense.

typename std::enable_if <(std::rank<T>::value > 1), int>::type = 0 >

I know that I need to specify condition which checks if another f() exists: something like:

//goal: typename std::enable_if <((f_alternative_exist == false) || (std::rank<T>::value > 1 && f_alternative_exist == true) ), int>::type = 0 >

I found similar question, but I am still not sure how to do it with std::enable_if

#include <iostream> #include <type_traits> #include <sstream> class Foo { }; //--- alternative header A_new.h //#include A.h template< class T, typename std::enable_if<(std::rank<T>::value == 1), int>::type = 0 > void f(std::ostream& os, const T& value) //f_alternative { os << "alternative function\n"; os << "rank == 1" << std::endl; } //---- //---- old header A.h template < class T, typename std::enable_if <(std::rank<T>::value > 1), int>::type = 0 > //goal: typename std::enable_if <((f_alternative_exist == false) || (std::rank<T>::value > 1 && f_alternative_exist == true) ), int>::type = 0 > void f(std::ostream& os, const T& value) //f_old { os << "old function\n"; os << "rank > 1" << std::endl; } template <class T> std::ostream& operator<<(std::ostream& os, const T& foo) { f<T>(os, foo); return os; } //----- int main() { Foo foo1[5]; Foo foo2[5][5]; std::cout << foo1; std::cout << foo2; return 0; }

https://coliru.stacked-crooked.com/a/7f7ef7bda1805a36

Could you tell me how can I specify std::enable_if to work also when f() is not overloaded and when it should run for all cases?

max66 · Accepted Answer · 2019-09-22 12:14:43Z

A possible f_alternative_exist is the following

template <typename T> std::false_type f_alternative_exist_f (std::ostream &, T const &, long); template <typename T> auto f_alternative_exist_f (std::ostream & os, T const & t, int) -> decltype( f(os, t), std::true_type{} ); template <typename T> using f_alternative_exist = decltype(f_alternative_exist_f(std::declval<std::ostream &>(), std::declval<T>(), 0)); template <typename T> static constexpr bool f_alternative_exist_v = f_alternative_exist<T>::value;

The old version become

template <typename T, std::enable_if_t<(std::rank<T>::value > 1) || (false == f_alternative_exist_v<T>), int> = 0 > void f (std::ostream & os, T const &) { os << "old function\n"; os << "rank > 1" << std::endl; }

The following is a full compiling C++14 example

#include <iostream> #include <type_traits> #include <sstream> class Foo { }; template <typename T> std::false_type f_alternative_exist_f (std::ostream &, T const &, long); template <typename T> auto f_alternative_exist_f (std::ostream & os, T const & t, int) -> decltype( f(os, t), std::true_type{} ); template <typename T> using f_alternative_exist = decltype(f_alternative_exist_f(std::declval<std::ostream &>(), std::declval<T>(), 0)); template <typename T> static constexpr bool f_alternative_exist_v = f_alternative_exist<T>::value; #if 0 // enable/disable template <typename T, std::enable_if_t<std::rank<T>::value == 1, int> = 0 > void f (std::ostream & os, T const &) { os << "alternative function\n"; os << "rank == 1" << std::endl; } #endif template <typename T, std::enable_if_t<(std::rank<T>::value > 1) || (false == f_alternative_exist_v<T>), int> = 0 > void f (std::ostream & os, T const &) { os << "old function\n"; os << "rank > 1" << std::endl; } template <typename T> std::ostream& operator<<(std::ostream& os, const T& foo) { f<T>(os, foo); return os; } int main () { Foo foo1[5]; Foo foo2[5][5]; std::cout << foo1; std::cout << foo2; }

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