Question

I was following template class restriction, but ran into errors in gcc:

error: multiple types in one declaration

error: declaration does not declare anything

It compiles if I remove the enable_if block. Can anybody explain if I am missing something?

template<class A, class B, class C, class D>                                                                               
typename std::enable_if<                                                  
std::is_base_of<baseofA, A>::value &&                      
std::is_base_of<baseofB, B>::value &&      
std::is_base_of<baseofC, C>::value &&            
std::is_base_of<baseofD, D>::value>::type       
class library {
    //whatever
};

Answer 1

You’re not using enable_if correctly. static_assert would be more appropriate in this case.

template<class A, class B, class C, class D>       
class library {
  static_assert(
    std::is_base_of<baseofA, A>::value &&                      
    std::is_base_of<baseofB, B>::value &&      
    std::is_base_of<baseofC, C>::value &&            
    std::is_base_of<baseofD, D>::value, 
    "template argument A must derive from baseofA and so on ..." );
      //whatever
};

If you want to use enable_if instead you must create a dummy template parameter that depends on the enabled type for it to work as you desire.

template<class A, 
         class B, 
         class C, 
         class D, 
         class _ = 
           typename std::enable_if<                                                  
             std::is_base_of<baseofA, A>::value &&                      
             std::is_base_of<baseofB, B>::value &&      
             std::is_base_of<baseofC, C>::value &&            
             std::is_base_of<baseofD, D>::value>
           ::type>
class library {
    //whatever
};

But IMO, the static_assert method is better because you can provide a descriptive error message instead of the compiler complaining about failing to find a type named type in the latter case.