std::is_compound
From cppreference.com
                    
                                        
                    
                    
                                                            
                    |   Defined in header <type_traits>
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|   template< class T > struct is_compound;  | 
(since C++11) | |
If T is a compound type (that is, array, function, object pointer, function pointer, member object pointer, member function pointer, reference, class, union, or enumeration, including any cv-qualified variants), provides the member constant value equal true. For any other type, value is false.
Contents | 
Inherited from std::integral_constant
Member constants
|    value [static]  | 
   true if  T is a compound type , false otherwise   (public static member constant)  | 
Member functions
|    operator bool  | 
   converts the object to bool, returns value   (public member function)  | 
Member types
| Type | Definition | 
| value_type | bool | 
| type | std::integral_constant<bool, value> | 
[edit] Notes
Compound types are the types that are constructed from fundamental types. Any C++ type is either fundamental or compound.
[edit] Possible implementation
template< class T > struct is_compound : std::integral_constant<bool, !std::is_fundamental<T>::value> {};  | 
[edit] Example
#include <iostream> #include <type_traits> int main() { class cls {}; std::cout << (std::is_compound<cls>::value ? "T is compound" : "T is not a compound") << '\n'; std::cout << (std::is_compound<int>::value ? "T is compound" : "T is not a compound") << '\n'; }
Output:
T is compound T is not a compound
[edit] See also
|    (C++11)  | 
   checks if a type is fundamental type   (class template)  | 
|    (C++11)  | 
   checks if a type is scalar type   (class template)  | 
|    (C++11)  | 
   checks if a type is object type   (class template)  | 
|    (C++11)  | 
   checks if a type is an array type   (class template)  |