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<int,5> test (const std::array<int,5>& a, const std::array<int,5>& b)
{
std::array<int,5> 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])