c++ - How does using ellipses for SFINAE work? -

when using sfinae select constructor overloads in past, have typically used following:

template <typename t> class class { public:     template <typename u = t, typename std::enable_if<std::is_void<u>::value, int>::type=0>     class() {         std::cout << "void" << std::endl;     }      template <typename u = t, typename std::enable_if<!std::is_void<u>::value, int>::type=0>     class() {         std::cout << "not void" << std::endl;     } }; 

however, came across alternative:

template <typename u = t, typename std::enable_if<std::is_void<u>::value>::type...> class() {     std::cout << "void" << std::endl; } 

considering following illegal ...

template <typename u = t, void...> // error! class() { } 

... how alternative above using ellipses rather non-type template argument work?

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full code: http://coliru.stacked-crooked.com/a/64a1aaf13ce6099b

my previous answer wrong. sorry. i'm going fix it.

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this declaration:

template <typename u = t, void...> class() { } 

violates [temp.res]/8:

the program ill-formed, no diagnostic required, if [...] every valid specialization of variadic template requires empty template parameter pack

it's no diagnostic required, compiler chooses issue 1 anyway. either way, code ill-formed.

on other hand

template <typename u = t, std::enable_if_t<std::is_void<u>::value>...> class() { } 

doesn't violate requirement. have empty pack, don't run afoul of fact can't use void non-type template parameter. additionally, hypothetical specialization of enable_if provide type isn't void there it's not ill-formed due above limitation.