Question

I want to be able to specialize based on a constant c style string. The problem is that when I call my templatized function the type is const char[N] where ‘N’ is the size of the string +1 (null character). How can I specialize for all c-style strings?

The following code displays the problem. You can see the specialization for const char [15] gets matched for “const char [15]” but for “const char[5]” it goes to Generic.

Is there any way to do this?

template <typename T>
struct Test {
  static const char* type() { return "Generic"; }
};

template <>
struct Test<const char*> {
  static const char* type() { return "const char*"; }
};

template <>
struct Test<const char[]> {
  static const char* type() { return "const char[]"; }
};

template <>
struct Test<const char[15]> {
  static const char* type() { return "const char[15]"; }
};

template <>
struct Test<char*> {
  static const char* type() { return "char*"; }
};

template <>
struct Test<char[]> {
  static const char* type() { return "char[]"; }
};

template <typename T>
void PrintType(const T& expected) {
  std::cerr << expected << " type " << Test<T>::type() << std::endl;
}
int main(int argc, char* argv[]) {
  const char* tmp = "const char*";
  PrintType(tmp);
  PrintType("const char[]");
  PrintType("const char[15]");
  PrintType("const char[5]");
}

output when run in Windows 7 – VS 2008

const char* type const char*
const char[] type Generic
const char[15] type const char[15]
const char[5] type Generic

Answer 1

A specialization for any array of chars is:

template< std::size_t N >
struct Test< const char[N] > { ... };

However you can no longer return a char* from type(), unless you write more metafunctions to turn a non-type template parameter into its textual representation.