Can you please help me understand, why is the answer 5?

Question

Code:
Int mat [2][2]={{1,2}, {3,5}};
cout <<1[mat[1]];

Answer 1

That's because what actually happens behind an array subscription is simple pointer arithmetic. For example, the elements of one-dimensional array `int a[2]` are accessible as

a[0], a[1]

also, through the pointer arithmetic, it will be

*(a + 0), *(a + 1)

Since addition has commutative property, the following are also valid operations

*(0 + a), *(1 + a)

so are 0[a], 1[a].

That means arrays with higher dimensionality also follow the same rule. Here is the illustration of other variants for accessing individual elements of `mat` array.

int main()
{
int mat [2][2] = { {1,2}, {3,5} };

cout << mat[1][1] << endl; // regular access using bracket
cout << *(mat[1] + 1) << endl; // access using partial pointer arithmetic
cout << *(*(mat + 1) + 1) << endl; // access using full pointer arithmetic

cout << 1[mat [1]] << endl; // access using commutative property of pointer arithmetic
cout << *(1 + mat[1]) << endl;
cout << *(1 + *(mat + 1)) << endl;

cout << 1[mat][1] << endl; // access using the same property
cout << *(1 + 1[mat]) << endl;
cout << *(1 + *(1 + mat)) << endl;
}

Live demo: http://cpp.sh/9drls

Answer 2

array index start from zero....u given index 1 so this wll print 5..