Using enable_if with struct specialization

问题

I am attempting to define a template that will will specify a storage type given another type T. I'd like to use enable_if to catch all the arithmetic types. Below is my attempt at this which complains the template is redeclared with 2 parameters. I tried adding a 2nd dummy parm to the primary template but get a different error. How can this be done?

#include <string>
#include <type_traits>
template <typename T> struct storage_type; // want compile error if no match
// template <typename T, typename T2=void> struct storage_type; // no joy
template <> struct storage_type<const char *> { typedef std::string type; };
template <> struct storage_type<std::string> { typedef std::string type; };
template <typename T, typename std::enable_if<std::is_arithmetic<T>::value>::type* = nullptr> 
    struct storage_type { typedef double type; };

// Use the storage_type template to allocate storage
template<typename T>
class MyStorage {
  public:
  typename storage_type<T>::type storage;
};

MyStorage<std::string> s;  // uses std::string
MyStorage<const char *> s2; // uses std::string
MyStorage<float> f;  // uses 'double'

回答1:

You can do this by adding a second parameter to the primary template, then specialising to match it; you were on the right track, but didn't do it correctly.

#include <string>
#include <type_traits>
// template <typename T> struct storage_type;                // Don't use this one.
template <typename T, typename T2=void> struct storage_type; // Use this one instead.
template <> struct storage_type<const char *> { typedef std::string type; };
template <> struct storage_type<std::string> { typedef std::string type; };

// This is a partial specialisation, not a separate template.
template <typename T> 
struct storage_type<T, typename std::enable_if<std::is_arithmetic<T>::value>::type> {
    typedef double type;
};

// Use the storage_type template to allocate storage
template<typename T>
class MyStorage {
  public:
  typename storage_type<T>::type storage;
};

MyStorage<std::string> s;  // uses std::string
MyStorage<const char *> s2; // uses std::string
MyStorage<float> f;  // uses 'double'

// -----

struct S {};

//MyStorage<S> breaker; // Error if uncommented.

And voila.

来源：https://stackoverflow.com/questions/42678338/using-enable-if-with-struct-specialization

标签

c++

templates

enable-if