How to specialize a template for 2 different values? - c++

Question

Given the code below, I was wondering if it is possible to have a specialization for a set of values. In my example, I want to create an specialization for N=3 or N=4 to use an array of known size. Is it possible to avoid code duplication in this case?
template <typename T, unsigned int N>
class A
{
public:
T* data;
};
template <typename T>
class A<T, 3>
{
public:
T data[3];
};
template <typename T>
class A<T, 4>
{
public:
T data[4];
};
int main()
{
A<int, 1> u;
std::cout << sizeof(u.data) << std::endl; // Size of pointer
A<int, 3> v;
std::cout << sizeof(v.data) << std::endl; // Size of data
A<int, 4> w;
std::cout << sizeof(w.data) << std::endl; // Size of data
return 0;
}

Answer 1

You can use std::enable_if by introducing a default void template parameter in the general case.
template <typename T, unsigned int N, typename = void>
class A
{
public:
T* data;
};
template <typename T, unsigned int N>
class A<T, N, typename std::enable_if<N == 3 || N == 4>::type>
{
public:
T data[N];
};
live wandbox example
If N == 3 || N == 4 is true, then typename std::enable_if<N == 3 || N == 4>::type is well-formed and evaluates to void. The specialization is then chosen.
If N == 3 || N == 4 is false, then typename std::enable_if<N == 3 || N == 4>::type is ill-formed and the specialization is "SFINAEd away". The general case is then chosen.