Use decltype to declare the return type

Question

I've got the following code snippet:

int odd[] = { 1,3,5,7,9 };
int even[] = { 0,2,4,6,8 };

// returns a pointer to an array of five int elements
decltype(odd) *arrPtr(int i) {
    return (i % 2) ? &odd : &even; // returns a pointer to the array
}

int main()
{
    int *res1 = arrPtr(3);
    decltype(odd) *res2 = arrPtr(3);
    auto res3 = arrPtr(3);
}

For the first line I get the following error message:

int (*arrPtr(int i))[5]

returns a pointer to an array of five int elements

Error: a value of type "int (*)[5]" cannot be used to initialize an entity of type "int *"

Why can't int* be used to initialize my return value from arrayPtr()? I would have assumed that the compiler is doing an explicit conversion.

Furthermore, what would one use as return type (best practice)?

Answer 1

It seems odd is declared as an array of int:

int odd[5];

The type of odd is int[5] and taking the address (or adding a * to decltype(odd)) of this array yields an int(*)[5]. It seems, you actually want to use the decayed type of odd:

decltype(+odd) arrPtr(int i) {
    return i % 2? odd: even;
}

int main() {
   int*           res1 = arrPtr(3);
   decltype(+odd) res2 = arrPtr(3);
   auto           res3 = arrPtr(3)
}

Note this use of unary + which forces decay of the array to become a pointer to the first element of the array.

If you really mean to return a pointer to the array, you'd need to use a suitable type when capturing the result (assuming the original implementation):

int (*res1)[5]      = arrPtr(3);
decltype(odd)* res2 = arrPtr(3);
auto res3           = arrPtr(3);

Note, that this is a pointer to an array of 5 int, i.e., you'd access an element of it using

(*res1)[3];

or

res1[0][3];

Answer 2

int (&)[5] can decay to int *, it is not the case for int (*)[5].

You may change your code to

decltype(odd)& arrPtr(int i) {
    return (i % 2) ? odd : even; // returns the reference to the array
}

and then

int main()
{
    int *res1 = arrPtr(3);
    decltype(odd)& res2 = arrPtr(3); // int (&)[5]
    auto res3 = arrPtr(3); // int*
    const auto& res4 = arrPtr(3); // const int (&)[5]
}