C++ Partial Specialization ( Function Pointers )
问题 Can any one please tell, whether below is legal c++ or not ? template < typename s , s & (*fn) ( s * ) > class c {}; // partial specialization template < typename s , s & (*fn) ( s * ) > class c < s*, s* & (*fn)(s**) {}; g++ ( 4.2.4) error: a function call cannot appear in a constant-expression error: template argument 2 is invalid Although it does work for explicit specialization int & func ( int * ) { return 0; } template <> class c < int , func> class c {}; 回答1: I think you mean template <