Re: Which function gets specialized?
On May 13, 12:18 am, Zeppe
<z...@remove.all.this.long.comment.email.it> wrote:
desktop wrote:
I have this example:
template<class T> (1)
void f( T );
template<class T> (2)
void f( T* );
template<> (3)
void f<>(int*);
Which of (1) and (2) does (3) specialize, and why?
Is the order just a matter of when the specialization is declared?
It's simpler: (2) is a partial specialization of (1), while (3) is a
specialization of (1). Note that there is no relationship between the
specializations, they are just different implementations. So, it makes
no sense to ask if (3) specializes directly (1) or the partial
specialization (2): it's just a specialization of (1).
There's no such thing as partial specialization of a function
template. (1) and (2) are two, independent function templates.
Partial ordering of function templates (=A714.5.5.2) says that the
specialization (3) is for function template (2). This makes
sense, because the same partial ordering is used in overload
resolution, which means that in fact, (1) will never be called
with an int*.
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