C++ correct way to return pointer to array from function

Question

I am fairly new to C++ and have been avoiding pointers. From what I\'ve read online I cannot return an array but I can return a pointer to it. I made a small code to test it

Answer 1

New answer to new question:

You cannot return pointer to automatic variable (int c[5]) from the function. Automatic variable ends its lifetime with return enclosing block (function in this case) - so you are returning pointer to not existing array.

Either make your variable dynamic:

int* test (int a[5], int b[5]) {
    int* c = new int[5];
    for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
    return c;
}

Or change your implementation to use std::array:

std::array test (const std::array& a, const std::array& b) 
{
   std::array c;
   for (int i = 0; i < 5; i++) c[i] = a[i]+b[i];
   return c;
}

In case your compiler does not provide std::array you can replace it with simple struct containing an array:

struct array_int_5 { 
   int data[5];
   int& operator [](int i) { return data[i]; } 
   int operator const [](int i) { return data[i]; } 
};

---

Old answer to old question:

Your code is correct, and ... hmm, well, ... useless. Since arrays can be assigned to pointers without extra function (note that you are already using this in your function):

int arr[5] = {1, 2, 3, 4, 5};
//int* pArr = test(arr);
int* pArr = arr;

Morever signature of your function:

int* test (int in[5])

Is equivalent to:

int* test (int* in)

So you see it makes no sense.

However this signature takes an array, not pointer:

int* test (int (&in)[5])