Question

I have an array as:

int x[3][5]={
        {1,2,3,4,5},
        {6,7,8,9,10},
        {11,12,13,14,15}
        };

1. What does *x refer to?
2. *(*x+2)+5 refer to “8”.How does that happen?
3. Is *(*x+2) same as *(*x)+2?

4. What if I do:

   *n=&x;
   Where is the pointer n pointing to? if it would have been only x and not an & then it would have been the base address.What for now?

Answer 1

1. *x is a dereference operation. In other words, “give me what x is pointing at”. Since this is an array (of arrays), dereferencing x will give you the first array. This is equivalent to the array access syntax of x[0].

2. *(*x+2)+5 is equivalent to x[0][2] + 5, which gives you 8. This is because:
   *x is the same as x[0] (see #1) and *(x + 2) is the same as x[2]. Once you’ve done two dereferences, you’ve gone from an array of arrays (similar to a double-pointer) to an array (single pointer) to an actual number (the third item in the first array). Then, it’s just 3 + 5 = 8.

3. *(*x+2) is equivalent to x[0][2] (see #2), which is 3 (third element in array). However, *(*x) + 2 gives you x[0][0] + 2 (first element in array plus 2), which is 1 + 2 = 3. Same answer, but very different way of getting it.