There is a std::remove_pointer type trait helper. If you give it a pointer type T*, its type member type is T. Otherwise, the type member type is just the template type unchanged.

But what if you want to remove all pointers? For example, remove_all_pointers<int*const*volatile*>::type should be int.

You can define this as a recursive operation. In pseudo-code:

template<typename T>
auto remove_all_pointers
{
    if (std::is_pointer_v<T>) {
        return remove_all_pointers<
            std::remove_pointer_t<T>
        >;
    } else {
        return T;
    }
}

One way to express conditional evaluation in template metaprogramming is to use std::conditional<a, b, c>::type, which is b if a is true and is c if a is false.

Therefore, your first attempt might be to write it as a one-liner built out of std::conditional.

template<typename T>
using remove_all_pointers_t =
    std::conditional_t<
        std::is_pointer_v<T>,
        remove_all_pointers_t<
            std::remove_pointer_t<T>&gt,;
        T>;

Okay, this doesn’t work because of the recursive reference to remove_all_pointers_t before it has completed its declaration. We can sidestep this by using a struct.

template<typename T>
struct remove_all_pointers
{
  using type = std::conditional_t<
        std::is_pointer_v<T>,
        typename remove_all_pointers<
            std::remove_pointer_t<T>>::type,
        T>;
};

This compiles, but you get an error when you try to use it:

using test = remove_all_pointers<int*const*volatile*>::type;

// gcc
In instantiation of 'struct remove_all_pointers<int>':
    recursively required from 'struct remove_all_pointers<int* const* volatile>'
    required from 'struct remove_all_pointers<int* const* volatile*>'
    required from here
error: invalid use of incomplete type 'struct remove_all_pointers<int>'
    |   using type = std::conditional_t<
    |         ^~~~
note: definition of 'struct remove_all_pointers<int>' is not complete until the closing brace
    | struct remove_all_pointers
    |        ^~~~~~~~~~~~~~~~~~~

// clang
error: no type named 'type' in 'remove_all_pointers<int>'
    |         typename remove_all_pointers<
    |         ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    |             std::remove_pointer_t<T>>::type,
    |             ~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
note: in instantiation of template class 'remove_all_pointers<int>' requested here
    |         typename remove_all_pointers<
    |                  ^
note: in instantiation of template class 'remove_all_pointers<int *const>' requested here
note: in instantiation of template class 'remove_all_pointers<int *const *volatile>' requested here
note: in instantiation of template class 'remove_all_pointers<int *const *volatile *>' requested here
   | using test = remove_all_pointers<int*const*volatile*>::type;
   |              ^

// msvc
error C2146: syntax error: missing '>' before identifier 'type'
note: the template instantiation context (the oldest one first) is
note: see reference to class template instantiation 'remove_all_pointers<int *const *volatile *>' being compiled
note: see reference to class template instantiation 'remove_all_pointers<int *const *volatile >' being compiled
note: see reference to class template instantiation 'remove_all_pointers<int *const >' being compiled
note: see reference to class template instantiation 'remove_all_pointers<int>' being compiled

Okay, maybe we were too ambitious.

All the error messages show that the template was able to recurse and strip away pointers, but then it ran into a problem when it reached the base case. Let’s look at that base case:

struct remove_all_pointers<int>
{
  using type = std::conditional_t<
        std::is_pointer_v<int>,
        remove_all_pointers<
            std::remove_pointer_t<int>>::type,
        int>;
};

After substituting std::remove_pointer_t<int> = int, we get

struct remove_all_pointers<int>
{
  using type = std::conditional_t<
        std::is_pointer_v<int>,
        remove_all_pointers<int>::type,
        int>;
};

Now we see the problem. The definition of remove_all_pointers<int>::type is dependent on itself.

The catch here is that std::conditional is not a short-circuiting operator. How can it be? It’s a template!

In order to instantiate a template, the compiler first evaluates the template parameters, and then it looks at the template expansion that results. The compiler doesn’t “look ahead” and say, “Oh, I can tell that the template expansion never uses its second parameter, so I will skip the evaluation of the second parameter.”¹

One way to solve this problem is to move the expansion of the two parameters to a partial specialization. That way, only the pointer cases invoke the template recursively.

template<typename T,
    bool = std::is_pointer_v<T>>
struct remove_all_pointers;

template<typename T>
struct remove_all_pointers<T, false>
{
    using type = T;
};

template<typename T>
struct remove_all_pointers<T, true>
{
    using type = typename remove_all_pointers<
        std::remove_pointer_t<T>>::type;
};

We add a hidden second template parameter which defaults to std::is_pointer_v<T>. We then partially specialize the template on that second template parameter: If it’s false (T is not a pointer), then the type is T itself, which provides our base case (no longer accidentally referring to itself). If it’s true (T is a pointer), then the type is calculated recursively after stripping away one layer of indirection.

template<typename T>
using remove_all_pointers_t =
    typename remove_all_pointers<T>::type;

static_assert(std::is_same_v<
    remove_all_pointers_t<int*const*volatile*>,
    int>);

As a small tuning step, we can fold the base case into the initial definition, so that only the recursive case is a partial specialization.

template<typename T,
    bool = std::is_pointer_v<T>>
struct remove_all_pointers
{                         
    using type = T;       
};                        

template<typename T>
struct remove_all_pointers<T, true>
{
    using type = typename remove_all_pointers<
        std::remove_pointer_t<T>>::type;
};

Are we done?

No, not yet.

We’ll continue next time.

¹ Indeed, the “I evaluate all the parameters even if they aren’t used” behavior is one of the things that SFINAE relies on!