How to get integral type from a template type array

Question

Im trying to write a scoped pointer class which calls delete once it is destroyed. I realized that i would need to check whether my pointer is pointing to an array or not, so i could call the right delete. Taking inspiration from std::unique_ptr i used type_traits to check if the template argument, which holds the type pointer, is an array or not:

template <typename type, bool _Dx = std::is_array<type>::value>
    class scoped_ptr {
    private:
        type* m_ptr;
    //...
    };

template <typename type>
    class scoped_ptr<type, true> {};

But if my template argument type is "int[]" the codes invalidates because i cant have an "int[]* m_ptr" How can i solve this problem? How can i pass int[] argument and have "int* m_ptr"

NathanOliver · Accepted Answer · 2019-11-12T22:03:34.060

1

What you want is std::remove_extent. If you give it an array it gives you the element type, otherwise it just gives you the type you gave it. That would look like

template <typename type, bool _Dx = std::is_array<type>::value>
class scoped_ptr {
private:
    std::remove_extent_t<type>* m_ptr;
//...
};

Also note that _Dx is an illegal name. All names that start with an underscore and are followed with a capital letter are reserved for the implementation.

edited Nov 12 '19 at 22:03

answered Nov 12 '19 at 21:42

NathanOliver

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Wow thanks, that was really helpfull, can i alias that type with a using/typedef? – Denomycor Nov 12 '19 at 21:57
@Denomycor Yep. You can do `using BaseType = std::conditional_t<_Dx, std::remove_extent_t, type>; ...; BaseType* m_ptr` – NathanOliver Nov 12 '19 at 21:59
@Denomycor Please note that I have updated my answer. The `std::conditional_t` was not needed. – NathanOliver Nov 12 '19 at 22:05

How to get integral type from a template type array

1 Answers1