[Courses] Re: [C] Beginner's Lesson 12: another answer to Exercise 1 (Simple Pointers)

[Lorne Gutz]

On Monday 25 November 2002 15:40, Morgon Kanter wrote:
> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
> Your code won't work:
> >   for (i = 0; i < SIZE; i++)
> >   {
> >     array[*(ptr + 1)] = 0;     /* use pointer arithmetic to get to  */
> >     printf("%3d", *(ptr + 1)); /* the next element in the array     */
> >   }
>
> rather, try:
> array[(*ptr + i)] = 0;
> printf("%3d", (*ptr + i));


This code only works because each element of the array has been
initilized to the index.   ie    array[4] = 4.

Change the initilization so that each element is array[i] = 1200 +i,
and see what happens.

Look at what you have here.  first ptr is set equal to the address of
array[0].  Now that element just happens to contain a zero so when
you go *ptr you have a zero, and array[*(ptr +i )] is equal to array[0];
but if you make the change I suggested above *(ptr +i) will equal
(1200 + i) ,  array[ (1200 +i)] is an array overrun and you will
probably get a Segmentation fault.

When you have set ptr = array[0], all you need to do is ptr++ and
you will be pointing at array[1].

cheers
Lorne