请在以下 function 模板声明中解释问题（根本原因）

Question

In following code, could anyone explain why do I get following error: "error: no type named 'type' in struct std::enable_if<false, double>

According to my understanding of enable_if_t, there should not be any problem at compile time if I am not using function p. It should simply not get generated for simple types.

But when I change condition to !is_class_v<T> , it works fine for simple types but then it stops working for class types.

template<typename T>
class Smart_class
{
    public:        
        enable_if_t<is_class_v<T>, T> p(T t)
        {            
            
        };
};

void f()
{
    Smart_class<double> a;
}

Answer 1

Quoting from temp.inst/3.1 :

The implicit instantiation of a class template specialization causes the implicit instantiation of the declarations , but not of the definitions, of the non-deleted class member functions ...

This is your case, the declaration of p member function is instantiated and it refers to the non-existing return type.