Why void_t doesnt work in SFINAE but enable_if does

Question

I was trying to understand how SFINAE works and I was experimenting with this code

#include <type_traits>

struct One { 
  using x = int; 
};
struct Two { 
  using y = int; 
};

template <typename T, std::void_t<typename T::x>* = nullptr>
void func() {}
template <typename T, std::void_t<typename T::y>* = nullptr>
void func() {}

/*template <typename T, std::enable_if_t<std::is_same_v<typename T::x, typename T::x>>* = nullptr>
void func() {}
template <typename T, std::enable_if_t<std::is_same_v<typename T::y, typename T::y>>* = nullptr>
void func() {} */



int main() {
  func<One>();
  func<Two>();
}

The commented code works but the first doesn't. The compiler gives me errors saying that there is a redefinition and that template argument deduction failed. Could someone explain why this happens? The two void_ts should be independent right? Since one line checks for x and the other for y. How can I fix?

Darill

Note that this works with the make_void trick. – Teofilateosinte

Answer 1

This seems to be related to CWG issue #1980 (credits to T.C. for correcting me).

As a workaround you can define void_t as:

template<typename... Ts> struct make_void { typedef void type;};
template<typename... Ts> using void_t = typename make_void<Ts...>::type;

(from cppreference)

live example on wandbox

Teofilateosinte

Whoa this problem has still not been fixed in the new installations of C++17 on gcc? – Darill

Apparently not. :( I do see a difference on g++ 7.1.0 between the standard void_t and the above void_t. – Rudolfrudolfo

@Whoami: I am surprised as well. If my answer is indeed correct, this is quite sad – Teofilateosinte

@VittorioRomeo they could have at least used the make_void trick in the library.. – Darill

@VittorioRomeo sorry but could you confirm to me once that they have not used the make_void trick in the library? I am not sure where to check – Darill

@Whoami: the sources for libstdc++ and libc++ are publicly available, there are some mirrors on GitHub. – Teofilateosinte

@Whoami: added links in my answer. – Teofilateosinte

std::void_t is specified as template <class...> using void_t = void; – Dryclean

Yeah, those definitely aren't compiler bugs. LEWG issue? – Wivestad

This is not 1558. This has to do with declaration matching, i.e., whether the two func definitions actually declare distinct function templates or the same one. Closer to 2037. – Statolith

@T.C.: interesting, all the hints seemed to point to 1558 and the make_void workaround. I tried my best, apologies for the possible misinformation. Would like to know why this is not related to 1558 though. – Teofilateosinte

"all the hints seemed to point to 1558"? Not to me. First, you don't get anything like "ambiguous overload", which would be an indication that SFINAE's not kicking in; instead, you get "redefinition". Second, if you remove one of the two funcs, then the remaining one gets called - or yield an error - appropriately. That's also not consistent with a SFINAE problem. Finally, just having the two definitions of func, without any calling, is sufficient to trigger the error. Once again, that's not consistent with a SFINAE problem. – Statolith