Solution

The correct answer is 2)(A) and (B) Only

EXPLANATION:

• (A) Pointers can be used to access array elements. -
  • This statement is CORRECT. In C, pointers can be used to navigate and access elements of an array. For example, if you have an array `int arr[] = {1, 2, 3};`, a pointer `int *ptr = arr;` can be used to access the elements like `*(ptr + 1)` for the second element.
• (B) Pointers can store the address of another pointer. -
  • This statement is CORRECT. Pointers in C can hold the address of another pointer, which is known as pointer-to-pointer. For example, `int *ptr1; int **ptr2 = &ptr1;` where `ptr2` is a pointer to `ptr1`.
• (C) Pointers are automatically dereferenced in expression. -
  • This statement is NOT CORRECT. In C, pointers are not automatically dereferenced. You must explicitly use the dereference operator `*` to access the value stored at the address contained in the pointer.
• (D) Pointers cannot be used to access structure members. -
  • This statement is NOT CORRECT. Pointers can indeed be used to access members of a structure. This is done using the arrow operator `->`. For example, if you have a structure `struct Example { int x; };` and a pointer to this structure `struct Example *ptr;`, you can access `x` using `ptr->x`.