Incrementing variables using pointers

Question

I'm very new to dealing with pointers, and my C knowledge is fairly small. I'm trying to understand pointers. I wrote the following code to print a list of variables (a to f) like so:

I wrote the following code to do this:

#include <stdio.h>
int main(){
  int a,b,c,d,e,f;
  int *p;
  int i;
  a = b = c = d = f = 0;
  p = &a;
  for (i = 0; i < 5; i++){
    *p += i;
    printf("%d\n", *p);
    p++;
  }
  return 0;
}

The idea was it works through the variables and increments each by an ever-increasing number (i). I am assuming that as you initialize the variables at the same time, they'd be placed next to each other in memory. However, I get the following output:

0
1
2
3
-1218283607

If I change the for loop to only go from 0 to 3 (i < 4), it works fine, printer 0 1 2 and 3. But when I wish to print the variable f as well, it doesn't seem to set it.

As I said, I'm very new to pointers so I've probably overlooked something silly, but I've been looking through my code over and over, trying to work it out.

Thanks in advance.

Thank you all very much :) it was just my ignorance about variables in memory then. I'll read more into it, thank you all for the help. — mjp
– mjp, Commented Jan 18, 2010 at 21:45

jason · Accepted Answer · 2010-01-18 21:37:55Z

There is no guarantee that a, b, c, d, e and f will be adjacent in memory. If you want that sort of guarantee you need to use an array.

#include <stdio.h>
int main() {
    int a[6];
    int *p;
    int i;
    a[0] = a[1] = a[2] = a[3] = a[4] = a[5] = 0;
    p = &a[0];
    for (i = 0; i < 6; i++){
        *p += i;
        p++;
    }
    for(i = 0; i < 6; i++) {
        printf("%d\n", a[i]);
    }
    return 0;
}

Here int a[6] is declaring an array named a that can hold six integers. These six integers can obtained via a[0], a[1], a[2], a[3], a[4] and a[5]. You are guaranteed that a[0], a[1], a[2], a[3], a[4] and a[5] are layed out contiguously in memory. Thus the line

p = &a[0];

sets p to the address of the first element. Each increment of this pointer moves us forward one position in the array.

The second for loop shows that first for loops correctly sets a[i] to i for i in {0, 1, 2, 3, 4, 5}. If you run this program you will see

on the console.

Ignacio Vazquez-Abrams · Accepted Answer · 2010-01-18 21:32:54Z

3

You forgot to initialize e. But yes, use a packed array.

answered Jan 18, 2010 at 21:32

Ignacio Vazquez-Abrams

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Comments

John Knoeller · Accepted Answer · 2010-01-18 21:35:38Z

1

It isn't safe to assume that stack variables are arranged in memory in any particular order.

You need to use an array, a struct or possibly a union to gurantee the ordering of your ints.

union {
   int ary[6];
   struct {
      int a;
      int b;
      int c;
      int d;
      int e;
      int f;
      } s;
   } u = {0};

  p = &u.s.a;
  for (i = 0; i < 5; i++){
    *p += i;
    printf("%d\n", *p);
    p++;
  }

answered Jan 18, 2010 at 21:35

John Knoeller

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4 Comments

Alok Singhal Over a year ago

Even in the code above, there is nothing stopping a compiler from inserting padding between a and b, and b and c, and, ....

John Knoeller Over a year ago

Compilers don't insert padding arbitrarily.

Alok Singhal Over a year ago

it's about what's allowed by the standard. One could argue that the most logical thing for a compiler to do is to allocate storage for OP's a, b, ..., f sequentially, but of course, that's not required either.

John Knoeller Over a year ago

I don't buy that. Compilers may be allowed to put spaces in there, but in practice, they can only do that if they are willing to break a lot of existing code for no good reason. In fact, what actual compilers do is go beyond the spec in providing syntax for programmers to be explicit in how they want structures to be packed - and choosing backward compatible defaults otherwise.

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