C++ single template specialisation with multiple template parameters -

hallo!

i specialise 1 of 2 template types. e.g. template <typename a, typename b> class x should have special implementation single function x<float, sometype>::somefunc().

sample code:

main.h:

#include <iostream>  template <typename f, typename i> class b { public:     void somefunc()     {         std::cout << "normal" << std::endl;     };      void somefuncnotspecial()     {         std::cout << "normal" << std::endl;     }; };  template <typename i> void b<float, i>::somefunc(); 

main.cpp:

#include <iostream> #include "main.h"  using namespace std;  template <typename i> void b<float, i>::somefunc() {     cout << "special" << endl; }  int main(int argc, char *argv[]) {     b<int, int> b1;     b1.somefunc();     b1.somefuncnotspecial();      b<float, int> b2;     b2.somefunc();     b2.somefuncnotspecial(); } 

compilation fails class b. true, not possible in c++ in way? best workaround?

[edit]

template <float, typename i> void b<float, i>::somefunc(); leads main.h:26: error: ‘float’ not valid type template constant parameter

template <typename i> void b<float, i>::somefunc(); leads main.h:27: error: invalid use of incomplete type ‘class b’

and i'm using gcc.

[edit]

i don't want specialise whole class, there other functions don't have specialisation.

you have provide partial specialization of class template b:

template <typename i> class b<float, i> { public:     void somefunc(); };  template <typename i> void b<float, i>::somefunc() {     ... } 

you can define somefunc inside specialization.

however, if want specialize function, , not class e. g.

template <typename f, typename i> void somefunc(f f, i) { somefuncimpl::act(f, i); }  template <typename f, typename i> struct somefuncimpl { static void act(f f, i) { ... } };  // partial specialization template <typename i> struct somefuncimpl<float, i> { static void act(float f, i) { ... } }; 

but can't specialize function template without trick.