How does `void_t` work

Question

I watched Walter Brown\'s talk at Cppcon14 about modern template programming (Part I, Part II) where he presented his void_t SFINAE technique.

Example:<

Answer 1

// specialized as has_member< T , void > or discarded (sfinae)
template
struct has_member> : true_type
{ };

That above specialization exists only when it is well formed, so when decltype( T::member ) is valid and not ambiguous. the specialization is so for has_member as state in the comment.

When you write has_member, it is has_member because of default template argument.

And we have specialization for has_member (so inherit from true_type) but we don't have specialization for has_member (so we use the default definition : inherit from false_type)