Question

int a[10][5];
for (int i = 0; i < 10; i++)
{
    for (int j = 0; j < 10; j++)
    {
        cout << i[j[a]];
        cout << j[i[a]];
    }
}  

Edit:assume the values are already initialized to the array and is this cout valid then?

please explain the i[j[a]]; part only regardless of the program I’m concerned about that statement only!

Answer 1

C arrays have a strange quirk that allows them to be accessed through the “opposite” direction. This is deeply rooted in the pointer arithmetic of arrays. For example, a[1] is equivalent to *(a + 1). Likewise, 1[a] is equivalent to *(1 + a). Due to the commutative nature of addition, this works out quite nicely. More details can be found here.

With that knowledge in tact, the expression i[j[a]] can be broken down into two different parts. j[a] is equivalent to *(j + a) which would return another array due to the multi-dimensional nature of the array you have, for example purposes we’ll call this returned array p. Then you have the statement i[p] which would be equivalent to *(i + p). Bringing it back all together would show you that i[j[a]] is equivalent to *(i + *(j + a)). Indeed, this means that i[j[a]] is just an obfuscated way of writing a[j][i].