c++ - Why function template cannot be partially specialized? -
i know language specification forbids partial specialization of function template.
i know rationale why forbids it? 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!
afaik that's changed in c++0x.
i guess oversight (considering can partial specialization effect more verbose code, placing function static member of class).
you might relevant dr (defect report), if there one.
edit: checking this, find others have believed that, no-one able find such support in draft standard. this thread seems indicate partial specialization of function templates not supported in c++0x.
edit 2: example of meant "placing function static member of 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 }
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