Tricky pointer question

Question

I\'m having trouble with a past exam question on pointers in c which I found from this link, http://www.cl.cam.ac.uk/teaching/exams/pastpapers/y2007p3q4.pdf

The question

Answer 1

The expression is parsed as &((*pps)[1]); pps is being treated as a pointer to an array, you're accessing the first element of that pointed-to array, and then taking the address of that element.

Answer 2

pps is a pointer to pointer to short,

which means that *pps is a pointer to short (or array of shorts),

(*pps)[1] is just like *(*pps + 1) [pointers arithmetic],

and &(*(*pps + 1)) is the address of *(*pps+1),

or, in other words - (*pps+1) (which is a pointer to short).

Answer 3

pps is a pointer to a pointer. It is dereferencing pps. So now you have a pointer. As arrays are just pointers you are then using pps as an array.

It is then same as:

short ps[2] = {0x0001,0x034c};
short **pps = &ps;

so the result is: 0x034c

Answer 4

The [] operator takes precedence over the & operator. So the code is dereferencing pps to get to the first element of an array of short*. Since this element is also a pointer, we may treat it as an array and look up the element one position to the right of what it points to, wth [1]. Finally, we take the address of that element.

It might be useful to note that &p[i] is the same as p + i - it gives you a pointer to the element i positions to the right of where p points to.

The intermediate values are:

pps == 0x1c
*pps == 0x18
&(*pps)[1] == *pps + 1 == 0x1A

(the +1 adds two bytes, since it is used on a short*)