std::enable_if_t to separate string from non string function parameters

Question

I need to print all kind of string data within double quotation, and others without double quotation.

This are my functions to check if the argument is a string.

template<class T>
struct is_str : std::integral_constant<bool, false> {};
template<> struct is_str<char*> : std::integral_constant<bool, true> {};
template<> struct is_str<wchar_t*> : std::integral_constant<bool, true> {};
template<> struct is_str<const char*> : std::integral_constant<bool, true> {};
template<> struct is_str<const wchar_t*> : std::integral_constant<bool, true> {};
template<> struct is_str<std::string> : std::integral_constant<bool, true> {};
template<> struct is_str<const std::string> : std::integral_constant<bool, true> {};
template<> struct is_str<std::wstring> : std::integral_constant<bool, true> {};
template<> struct is_str<const std::wstring> : std::integral_constant<bool, true> {};

And in the printer function, I use the above functions like this

template<typename T>
std::enable_if_t<is_str<T>::value>, std::ostream&>
    operator<<(std::ostream& xx, const T& ar)
{
    xx << "\"" << ar << "\"";
    return xx;
}

template<typename T>
std::enable_if_t<!is_str<T>::value>, std::ostream&>
    operator<<(std::ostream& xx, const T& ar)
{
    xx << ar;
    return xx;
}

It doesn't compile with the error

unrecognized template declaration/definition

Can someone please tell me how to fix this or better way to handle this situation?

Answer 1

Graham Best is right but there is another (bigger, IMHO) problem.

You're not defining operator<<; you're redefining it calling itself.

As far I know, it's impossible (you redefine a function, the compiler doesn't know which version call) and when you write (inside the redefined operator)

xx << "\"" << ar << "\"";

which version of operator<<() should be used? The new redefined (causing a loop recursion) or the old one?

I think that the best way to exit from this problem is avoid to redefine operator<<() and define a simple function; by example, the following print()

template<typename T>
std::enable_if_t<is_str<T>::value, std::string> print (T const & ar)
 {
   std::ostringstream oss;

   oss << "\"" << ar << "\"";

   return oss.str();
 }

template<typename T>
std::enable_if_t<!is_str<T>::value, T const &> print (T const & ar)
 { return ar; }

and use it in this way

std::cout << print(std::string{"abc"}) << std::endl; // add quotes
std::cout << print(1) << std::endl;                  // no quotes

En passant: there are a couple of classes that can be useful to write a more compact code: std::true_type, defined as std::integral_constant<bool, true>, and std::false_type, defined as std::integral_constant<bool, false>.

So you can define is_str as follows

template<class T>
struct is_str : std::false_type {};

template<> struct is_str<char*> : std::true_type {};
template<> struct is_str<wchar_t*> : std::true_type {};
// ...

OBSERVE also that if you write

std::cout << print("abc") << std::endl; // add quotes

the print() doesn't add the quotes.

This is because "abc" isn't a char const *; it's a const char[4].

So, if you want add the quotes to char[N] and wchar_t[N] too, you should add the following specializations

template<std::size_t N> struct is_str<char[N]> : std::true_type {};
template<std::size_t N> struct is_str<wchar_t[N]> : std::true_type {};