c++ - Function template specialization

Question

While reading this, I'm confused by the following examples:

// Example 2: Explicit specialization 
// 
template<class T> // (a) a base template 
void f( T );

template<class T> // (b) a second base template, overloads (a) 
void f( T* );     //     (function templates can't be partially 
                  //     specialized; they overload instead)

template<>        // (c) explicit specialization of (b) 
void f<>(int*);

// ...

int *p; 
f( p );           // calls (c)

Here, (c) is an explicit specialization of (b).

// Example 3: The Dimov/Abrahams Example 
// 
template<class T> // (a) same old base template as before 
void f( T );

template<>        // (c) explicit specialization, this time of (a)
void f<>(int*);

template<class T> // (b) a second base template, overloads (a) 
void f( T* );

// ...

int *p; 
f( p );           // calls (b)! overload resolution ignores 
                  // specializations and operates on the base 
                  // function templates only

Here (c) is an explicit specialization of (a). Why is that? Is this because of the ordering of the declaration?

Answer 1

Yes, it's because of the ordering of the declaration. When the compiler encounters (c) in the second set, the only defined template to specialize is (a).

This is why you must be careful in ordering your template specializations.

The C++ Programming Language goes into quite some detail about this (Section 13.5.1). I highly recommend it.