Why function template cannot be partially specialized?

Question

I know the language specification forbids partial specialization of function template.

I would like to know the rationale why it forbids it? Are they not useful?

template<typename T, typename U> void f() {}   //allowed!
template<> void f<int, char>()            {}   //allowed!
template<typename T> void f<char, T>()    {}   //not allowed!
template<typename T> void f<T, int>()     {}   //not allowed!

Answer 1

AFAIK that's changed in C++0x.

I guess it was just an oversight (considering that you can always get the partial specialization effect with more verbose code, by placing the function as a static member of a class).

You might look up the relevant DR (Defect Report), if there is one.

EDIT: checking this, I find that others have also believed that, but no-one is able to find any such support in the draft standard. This SO thread seems to indicate that partial specialization of function templates is not supported in C++0x.

EDIT 2: just an example of what I meant by "placing the function as a static member of a class":

#include <iostream>
using namespace std;

// template<typename T, typename U> void f() {}   //allowed!
// template<> void f<int, char>()            {}   //allowed!
// template<typename T> void f<char, T>()    {}   //not allowed!
// template<typename T> void f<T, int>()     {}   //not allowed!

void say( char const s[] ) { std::cout << s << std::endl; }

namespace detail {
    template< class T, class U >
    struct F {
        static void impl() { say( "1. primary template" ); }
    };

    template<>
    struct F<int, char> {
        static void impl() { say( "2. <int, char> explicit specialization" ); }
    };

    template< class T >
    struct F< char, T > {
        static void impl() { say( "3. <char, T> partial specialization" ); }
    };

    template< class T >
    struct F< T, int > {
        static void impl() { say( "4. <T, int> partial specialization" ); }
    };
}  // namespace detail

template< class T, class U >
void f() { detail::F<T, U>::impl(); }    

int main() {
    f<char const*, double>();       // 1
    f<int, char>();                 // 2
    f<char, double>();              // 3
    f<double, int>();               // 4
}

Answer 2

Well, you really can't do partial function/method specialization however you can do overloading.

template <typename T, typename U>
T fun(U pObj){...}

// acts like partial specialization <T, int> AFAIK 
// (based on Modern C++ Design by Alexandrescu)
template <typename T>
T fun(int pObj){...} 

It is the way but I do not know if it satisfy you.

Answer 3

In general, it's not recommended to specialize function templates at all, because of troubles with overloading. Here's a good article from the C/C++ Users Journal: http://www.gotw.ca/publications/mill17.htm

And it contains an honest answer to your question:

For one thing, you can't partially specialize them -- pretty much just because the language says you can't.

Answer 4

Since you can partially specialize classes, you can use a functor:

#include <iostream>

template < typename dtype , int k > struct fun
{
 int operator()()
 {
  return k ;
 }
} ;

template < typename dtype > struct fun < dtype , 0 >
{
 int operator()()
 {
  return 42 ;
 }
} ;

int main ( int argc , char * argv[] )
{
 std::cout << fun<float,5>()() << std::endl ;
 std::cout << fun<float,0>()() << std::endl ;
}