Error returning an array from a consteval function

Question

I'm trying to create a constant function that takes in a size int as a template parameter and returns an array that size with each index filled with its respective power of 2.

 template <int sz>
consteval int* conflagtab() {
    int ar[sz] = {};
    for (unsigned int i = 0; i < sz - 1; i++) {
        ar[i] = conflag(i);
    }
    return ar;
}

However when I am trying to use it I get the error:

call to consteval function "conflagtab<sz>() [with sz=31]" did not produce a valid constant expression.

I tried putting it in global namespace to see if that had a change but it had no difference. I'm expecting it to just return a pointer to the array.

Unrelated: Your for loop leaves the last element untouched (maybe it's intended) — Fareanor
– Fareanor, Commented Jun 27, 2024 at 9:14
You could return std::array by value. Or you could accept a pointer to the array as a parameter, and fill the existing array with powers of two. — HolyBlackCat
– HolyBlackCat, Commented Jun 27, 2024 at 9:18
how is the title related to the question or error message? "invalid use of interpreter storage" isnt mentioned elsewhere — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented Jun 27, 2024 at 9:32

Caleth · Accepted Answer · 2024-06-27 11:54:02Z

6

You can't return a raw array from a consteval function, especially by pointer. Your example, if it were not consteval, would be a dangling pointer. Finishing the constant evaluation and keeping a value that new[] created is not permitted, you have to delete[] somewhere later.

You can return a std::array:

template <int sz>
consteval std::array<int, sz> conflagtab() {
    std::array<int, sz> ar;
    for (unsigned int i = 0; i < sz - 1; i++) {
        ar[i] = conflag(i);
    }
    return ar;
}

edited Jun 27, 2024 at 11:54

answered Jun 27, 2024 at 9:21

Caleth

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