Role of default template arguments in the context of partial specialization

Question

I am not clear about the interaction of default template arguments in the context of partial specialization, for choosing which is the better matching template. This questions s

Answer 1

template <typename, typename Enable = bool_constant<true>>
struct cond : public bool_constant<false> {};

is identical to

template <typename, typename Enable = bool_constant<true>> struct cond;

template <typename, typename Enable>
struct cond : public bool_constant<false> {};

Later might be clearer to understand the result.

When you write cond<C>, thanks to default argument, it is equivalent to cond<C, bool_constant<true>>.

Then we try to match that to "better instantiation".

We have the choice between:

// primary template
template <typename, typename Enable>
struct cond : public bool_constant<false> {};

and partial specialization, which use SFINAE:

// specialization
template <typename T>
struct cond<T, bool_constant<(0 == T::code)>> : public bool_constant<true> {};

If 0 == T::code is ill formed, specialization is discarded and only primary template is a viable solution, so it is used.

Else if 0 == T::code evaluates to false, the specialization doesn't match and primary template is also used.
Note that using cond<C, bool_constant<false>> would use the specialization in that case.

Else, 0 == T::code evaluates to true, and then both primary and specialization are viable, but specialization is more specialized, so it is chosen.