When auto is used against array, why it's converted to pointer and not reference?

Question

See the below example:

int arr[10];
int *p = arr; // 1st valid choice
int (&r)[10] = arr; // 2nd valid choice

Now when we use auto against arr then, it chooses the 1st choice.

auto x = arr; // x is equivalent to *p

Is there a reason for choosing a pointer and not reference for array ?

Answer 1

Yes. In that expression, the array decays into pointer type, due to lvalue-to-rvalue conversion.

If you want array type , not pointer type, then do this:

auto & x = arr; //now it doesn't decay into pointer type!

& in the target type prevents the array from decaying into pointer type!

---

x is an array and not a pointer, can be proved as:

void f(int (&a)[10]) 
{
    std::cout << "f() is called. that means, x is an array!" << std::endl;
}
int main() {
     int arr[10];
     auto & x = arr; 
     f(x); //okay iff x is an int array of size 10, else compilation error!
}

Output:

f() is called. that means, x is an array!

Demo at ideone : http://www.ideone.com/L2Ifp

Note that f cannot be called with pointer type. It can be called with an int array of size 10. Attempting to call it with any other type, will result in compilation error.