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I'm building some input checker that needs to have specific functions for integer and/or double (for example 'isPrime' should only be available for integers).

If I'm using enable_if as a parameter it's working perfectly :

template <class T>
class check
{
public:
   template< class U = T>
   inline static U readVal(typename std::enable_if<std::is_same<U, int>::value >::type* = 0)
   {
      return BuffCheck.getInt();
   }

   template< class U = T>
   inline static U readVal(typename std::enable_if<std::is_same<U, double>::value >::type* = 0)
   {
      return BuffCheck.getDouble();
   }   
};

but if I'm using it as a template paramater (as demonstrated on http://en.cppreference.com/w/cpp/types/enable_if )

template <class T>
class check
{
public:
   template< class U = T, class = typename std::enable_if<std::is_same<U, int>::value>::type >
   inline static U readVal()
   {
      return BuffCheck.getInt();
   }

   template< class U = T, class = typename std::enable_if<std::is_same<U, double>::value>::type >
   inline static U readVal()
   {
      return BuffCheck.getDouble();
   }
};

then I have the following error :

error: ‘template<class T> template<class U, class> static U check::readVal()’ cannot be overloaded
error: with ‘template<class T> template<class U, class> static U check::readVal()’

I can't figure out what is wrong in the second version.

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1 Answer

Default template arguments are not part of the signature of a template (so both definitions try to define the same template twice). Their parameter types are part of the signature, however. So you can do

template <class T>
class check
{
public:
   template< class U = T, 
             typename std::enable_if<std::is_same<U, int>::value, int>::type = 0>
   inline static U readVal()
   {
      return BuffCheck.getInt();
   }

   template< class U = T, 
             typename std::enable_if<std::is_same<U, double>::value, int>::type = 0>
   inline static U readVal()
   {
      return BuffCheck.getDouble();
   }
};

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