While going through the various rules on list-initialization, I found this in dcl.init.list#3.6:
Otherwise, if T is a specialization of
std::initializer_list<E>, the object is constructed as described below.
On the other hand, in the synopsis of std::initializer_list, in support.initlist, I found the following statement:
If an explicit specialization or partial specialization of
initializer_listis declared, the program is ill-formed.
These appear to be contradictory statements, so what am I misunderstanding?
"A template specialization" has two distinct meanings:
"Explicit (full) specialization" or "partial specialization" - a language construct that changes the meaning of a template for some combination of template parameters.
Something that was generated from a template by substituting template arguments into it.
In other words, if you specify template arguments for a template, the resulting type/function/variable/... is a specialization of that template. E.g. std::vector<int> is a specialization of std::vector.
Looks like the first passage you quote uses (2).
So "if T is a specialization of std::initializer_list<E>" roughly means "if there exists such E that std::is_same_v<T, std::initializer_list<E>>", or "if T is a std::initializer_list<E>".