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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T07:00:33+00:00 2026-05-27T07:00:33+00:00

I am trying to declare a dynamic int array like below: int n; int

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I am trying to declare a dynamic int array like below:

int n;
int *pInt = new int[n];

Can I do this with std::auto_ptr?

I tried something like:

std::auto_ptr<int> pInt(new int[n]);

But it doesn’t compile.

I’m wondering if I could declare a dynamic array with auto_ptr construct and how. Thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-27T07:00:33+00:00Added an answer on May 27, 2026 at 7:00 am

    No, you cannot, and it will not: C++98 is very limited when it comes to arrays, and auto_ptr is a very awkward beast that often doesn’t do what you need.

    You can:

    • use std::vector<int>/std::deque<int>, or std::array<int, 10>, or

    • use C++11 and std::unique_ptr<int[]> p(new int[15]), or

    • use C++11 and std::vector<std::unique_ptr<int>> (though that feels too complicated for int).

    If the size of the array is known at compile time, use one of the static containers (array or an array-unique-pointer). If you have to modify the size at runtime, basically use vector, but for larger classes you can also use a vector of unique-pointers.

    std::unique_ptr is what std::auto_ptr wanted to be but couldn’t due to the limitations of the language.

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