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Message: [QUOTE="Gennaro Prota, post: 3649170"] That's the info you should have given :-) About your original question, you *can* have one function parameter and enable_if on it (its type), but you'll usually face another issue: something like foo< ... >:: is a so-called "non-deduced context". A "classical" example is: template< typename T > struct identity { typedef T type ; } ; template< typename T > void f( typename identity< T >::type t ) { } int main() { f( 1 ) ; // can't deduce } So, even if you had simply struct Foo { template< typename T > void Func( typename enable_if_c< true, T >::type ) { } } ; it wouldn't work for your p (FWIW, f.Func< char * >( p ) would, instead). On the contrary, your second template gives a way to deduce an argument for T; and, after all arguments are deduced, they are substituted in non-deduced contexts (later, SFINAE applies). [/QUOTE]