Conditionally compile function if `std::optional` exists-CodePudding

I am having a mixed C 14/C 17 codebase and I want to enable a function isEmpty only if I am having std::optional at hand. So, I tried SFINAE:

template <typename T, typename int_<decltype(std::nullopt)>::type = 0>
inline bool isEmpty(const std::optional<T>& v) {
  return !v;
}

However, that doesn't work.

How can I conditionally compile isEmpty if std::optional is there?

CodePudding user response：

If your compiler is recent enough (e.g. GCC >= 9.1 or Clang >= 9.0.0), you may include header <version> and conditionally compile your function template if macro __cpp_lib_optional is defined:

#include <version> // provides, among others, __cpp_lib_optional

#ifdef __cpp_lib_optional
template <typename T>
inline bool isEmpty(const std::optional<T>& v) {
    return !v;
}
#endif

This way, isEmpty will only be compiled if you're in -std=c 17 mode (or higher).