How does int*** in this template metaprogram work?

Question

How is template metaprogramming working here (static const int value = 1 + StarCounter<\U>::value;) to print out 3 ?

#include <iostream>

template <typename T>
struct StarCounter
{
    static const int value = 0;
};

template <typename U>
struct StarCounter<U*>
{
    static const int value = 1 + StarCounter<U>::value;
};

int main()
{
    std::cout << StarCounter<int***>::value << std::endl;//How is it printing 3?
    return 0;
}

Answer 1

The first template creates a struct that will always return 0 when you call StarCounter<U>::value.

The second template specialises the first one for cases where a pointer is used. So when you call it with StarCounter<U*>::value, the second template is used, not the first and it will return StarCounter<U>::value + 1. Note that it removes the pointer at each recursion step.

So the call to StarCounter<int***>::value will expend to:

StarCounter<int***>::value // second template is used due to pointer
1 + StarCounter<int**>::value // second template is used due to pointer
1 + 1 + StarCounter<int*>::value // second template is used due to pointer
1 + 1 + 1 + StarCounter<int>::value // no pointer here, so first template is used
1 + 1 + 1 + 0
3

Answer 2

StarCounter<int>::value

equals 0, because it's matched with first instantiation of the template, where value is explicitly defined.

StarCounter<int*>::value = 1 + StarCounter<int>::value

equals 1, because StarCounter<int*> is matched with StarCounter<U*>. Yes, StarCounter<T> can also be considered as a match, but StarCounter<U*> is more specific and that's why this one is preferred.

Similarly,

StarCounter<int**>::value = 1 + StarCounter<int*>::value

equals 2 and

StarCounter<int***>::value = 1 + StarCounter<int**>::value

equals 3.