std::enable_if : parameter vs template parameter
c++, c++11, enable-if, g++, templates
Solution
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();
}
};
Problem
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.