Computer Systems Organization

Start Lecture #8

Some Basic Pointer Examples

The next few sections present some simple examples using pointers. Several were written in class with the students.

Value Versus Address

#include <stdio.h>
int x, *p;
int main () {
  p = &x;
  x = 12;
  printf ("p = %d\n", p);
  p++;
  printf ("p = %d\n", p);
  printf ("*p = %d\n", *p);
}

The example on the right illustrates well the difference between a variable, in this case x, and its address &x. The first value printed is the address of x. This is not 12. Instead it is (probably) some large number.

A compiler warning occurs since in (modern) C, integers (%d) and pointers (&x) are not the same type.

Incrementing p does not increment x. Instead, the result is that p points to the next integer after x. In this program there is no further integer after x, so the result is unpredictable. If, instead of x, we had p point to A[7] for some integer array A, then the last line would have printed the value of A[8] and the penultimate line would have printed the address of A[8].

include <stdio.h>
int mystrlen (char *s);
int main () {
  char stg[] = "hello";
  printf ("The string %s has %d characters\n",
          stg, mystrlen(stg));
}
int mystrlen (char *s) {
  int i;
  for (i = 0; s[i] != '\0'; i++);
  return i;
}
#include <stdio.h>
int mystrlen (char s[]);
int main () {}   // unchanged
int mystrlen (char s[]) {
  int i = 0;
  while (*s++ != '\0') i++;
  return i;
}

String Length with Arrays and Pointers

On the right we show two versions of the string length function: The first version uses array notation for the string; the second uses pointer notation. The main() program is identical in the two versions so is shown only once. Note how very close the two string length functions are; another illustration of the similarity of arrays and pointers in C.

In the second version we encounter a common C idiom *s++. First note that the precedence of the operators is such that *s++ is the same as *(s++). That is, we are moving (incrementing) the pointer and examining what it now points at. We are not incrementing a part of the string. Specifically, we are not executing (*s)++;

Note that the two declarations

    int mystrlen (char *s)
    int mystrlen (char s[])
  

Changing any of them to the other form does not change the meaning of the program.

#include <stdio.h>
int changeltox (char*);
int main () {
  char stg[] = "hello";
  changeltox (stg);
  printf ("The string is now %s\n", stg);
}
int changeltox (char *s) {
  while (*s != '\0') {
    if (*s == 'l')
      *s = 'x';
    s++;
  }
}

Simple Substitution

The program on the right simply loops through the input string and replaces each occurence of l with x.

The while loop and increment of s could have been combined into a for loop. This version is written in pointer style.

Homework: Rewrite changeltox() to use array style and a for loop.