Concept:

Understanding *p++:

This is a combination of dereferencing and post-increment. First, the value pointed to by `p` is used, and then `p` is incremented. This means that the initial value pointed to by `p` (which is `arr[0]` or `1`) is used in the expression, and then `p` is incremented to point to the next element of the array (`arr[1]`).

Understanding *(p + 1):

This expression adds 1 to the pointer `p`, but does not change the value of `p`. Instead, it returns the address of the element that is one position ahead of the current element pointed to by `p`. Therefore, `p + 1` points to `arr[2]` after the increment operation in Line 1.

Explanation:

Let's break down the code step by step:

Step 1:

• Initially, `arr` is an array with elements `{1, 2, 3, 4, 5}` and `p` is a pointer pointing to the first element of the array, i.e., `arr[0]`.

Line 1:

• `printf("%d", *p++);`
• Here, `*p` gives the value at `p`, which is `1` (the first element of the array). Then `p` is incremented to point to the next element of the array (`arr[1]`). Therefore, the output of this line is `1`.

Step 2:

• After Line 1, `p` now points to `arr[1]` (which is `2`).

Line 2:

• `printf("%d", *(p + 1));`
• Here, `p + 1` points to `arr[2]` (since `p` is currently pointing to `arr[1]`). Therefore, `*(p + 1)` gives the value at `arr[2]`, which is `3`. Therefore, the output of this line is `3`.

Hence the correct answer is:
The output of the above code is 1, 3.