Question

Is there an elegant way to specialize a template based on one of its template parameters?

Ie.

template<int N> struct Junk {     static int foo() {         // stuff         return Junk<N - 1>::foo();     } };  // compile error: template argument '(size * 5)' involves template parameter(s) template<int N> struct Junk<N*5> {     static int foo() {         // stuff         return N;     } };  template<> struct Junk<0> {     static int foo() {         // stuff         return 0;     } }; 

Ie. I am trying to specialize a template based on the parameter being divisible by 5. The only way I can seem to do it is like below:

template<int N> struct JunkDivisibleBy5 {     static int foo() {         // stuff         return N;     } };  template<int N> struct Junk {     static int foo() {         // stuff         if ((N - 1) % 5 == 0 && N != 1)             return JunkDivisibleBy5<N - 1>::foo();         else             return Junk<N - 1>::foo();     } };   template<> struct Junk<0> {     static int foo() {         // stuff         return 0;     } }; 

But this is significantly less elegant, and also necessitates instantiation of all templates even if the template argument shouldn’t require it.

Answer 1

How’s this:

#include <iostream> using namespace std;  template < typename T, T N, T D > struct fraction {     typedef T value_type;     static const value_type num = N;     static const value_type denom = D;     static const bool is_div = (num % denom == 0); };  template< typename T, T N, T D, bool P > struct do_if {     static void op() { cout << N << ' NOT divisible by ' << D << endl; } };  template< typename T, T N, T D > struct do_if< T, N, D, true > {     static void op() { cout << N << ' divisible by ' << D << endl; } };  template < int N > void foo() {     typedef fraction< int, N, 5 > f;     do_if< typename f::value_type, f::num, f::denom, f::is_div >::op(); }  int main() {     foo< -5 >();     foo< -1 >();     foo< 0 >();     foo< 1 >();     foo< 5 >();     foo< 10000005 >();     return 0; }