standard c++11 way to remove all pointers of a type

Question

Is there a way to do this with some c++11 or at most a boost library?

#include <iostream>
#include <typeinfo>
using namespace std;

template <typename T> class remove_all_pointers{
public:
    typedef T type;
};

template <typename T> class remove_all_pointers<T*>{
public:
    typedef typename remove_all_pointers<T>::type type;
};

int main(){
    //correctly prints 'i' on gcc
    cout<<typeid(remove_all_pointers<int****>::type).name()<<endl;
}

Your `remove_all_pointers` template does remove all pointers. What else do you need? — n. m. could be an AI, Mar 24 '12 at 14:05
Is the question here "is there something in the standard library that does the same job as this code I wrote"? — zwol, Mar 24 '12 at 17:08
@downvoters I cannot see the problem in looking for a standard feature to achieve something. Even if you can write this yourself in 3 lines, why not see if there's a standard facility for the purpose? And what the OP needs it for is pretty much irrelevant for answering the question. — Christian Rau, Mar 24 '12 at 17:36
Plus, it's not like this is some functionality that's so esoteric that one wonders why someone could possibly want it at all. The C++11 standard and Boost already have `remove_pointer`. All he's asking for is `remove_all_pointer`, much like it has `remove_all_extent` for removing all array qualifiers. He wants the base type; why is that so questionable a thing to want? — Nicol Bolas, Mar 24 '12 at 17:42

score 7 · Answer 1 · answered Mar 24 '12 at 18:52

That doesn't quite work for all pointer types. You need to account for different cv-qualifiers as well:

template <typename T> class remove_all_pointers<T* const>{
public:
    typedef typename remove_all_pointers<T>::type type;
};

template <typename T> class remove_all_pointers<T* volatile>{
public:
    typedef typename remove_all_pointers<T>::type type;
};

template <typename T> class remove_all_pointers<T* const volatile >{
public:
    typedef typename remove_all_pointers<T>::type type;
};

Good point. You could use std:remove_cv and dispatch to a routine that does what the poster has. — emsr, Mar 25 '12 at 00:46

klaus triendl · Answer 2 · 2021-05-14T13:35:00.953

Since C++17 you can create a readable, simple and cv-qualifier aware meta function.

Use it like:

int main()
{
    remove_all_pointers_t<int* const* volatile* const volatile*> v = 42;
    return 0;
}

C++20

#include <type_traits>

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

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

C++17

In C++17 std::type_identity isn't available yet and std::identity isn't available anymore, hence you need to create your own 'identity' meta function:

#include <type_traits>

// your custom 'identity' meta function
template <typename T>
struct identity
{
    using type = T;
};

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

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

score 2 · Accepted Answer · answered Mar 24 '12 at 17:10

2

Neither Boost nor C++11 features such a trait template. But your code should work.

answered Mar 24 '12 at 17:10

Nicol Bolas

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standard c++11 way to remove all pointers of a type

3 Answers3

C++20

C++17

Linked