Consider the following code snippet. What is the output

Solution

In the code, ptr initially points to the first element of the array arr. The line ptr += 2; moves the pointer two positions ahead, meaning ptr now points to the third element of the array, which is 3. Therefore, *ptr dereferences the pointer to print the value 3. Why Other Options are Wrong: a) 1 is the first element, but the pointer has moved past it. b) 2 is the second element, but ptr skips over it. d) 4 is the fourth element, but ptr stops at the third. e) 5 is the fifth element, beyond where ptr is pointing.