Hi all, I am trying to enable only two version of a function. The caller may choose one of the two versions by specifying a template parameter. I've got two versions for the problem, both of them failing the same way. -------------------------- Version A): integral template parameter -------------------------- #include <iostream> #include #include using namespace std; using namespace boost; template enable_if_c f(T t) { cout << 1 << endl; } template enable_if_c f(T t) { cout << 2 << endl; } int _tmain(int argc, _TCHAR* argv[]) { f<1>(10); // ERROR f<2>(5.5); // ERROR f<3>(7); // ERROR return 0; --- The error message on all 3 lines is: error C2668: 'f' : ambiguous call to overloaded function The compiler's "explanation" for the first one: (for the other two it is much the same except the type) could be 'boost::enable_if_c f<1,int>(int)' with [ B=false, T=void ] or 'boost::enable_if_c f<1,int>(int)' with [ B=true, T=void ] while trying to match the argument list '(int)' -------------------------- Version B) : type template parameter -------------------------- #include <iostream> #include #include using namespace std; using namespace boost; struct A1 {}; struct A2 {}; template enable_if< is_same > g(T t) { cout << 1 << endl; } template enable_if< is_same > g(T t) { cout << 2 << endl; } int _tmain(int argc, _TCHAR* argv[]) { g<A1>(10); // ERROR g<A2>(5.5); // ERROR g<int>(7); // ERROR return 0; } ---- Error messages: error C2668: 'g' : ambiguous call to overloaded function (for all three of them) "Explanation" sample: could be 'boost::enable_if g(int)' with [ Cond=boost::is_same, T=void ] or 'boost::enable_if g(int)' with [ Cond=boost::is_same, T=void ] while trying to match the argument list '(int)' ----------------- enable_if should actually prevent the compiler from trying to compile the 'false' version, shouldn't it? Thx, Agoston