C pointer 練習

From JOS Lab1:
http://www.cs.sunysb.edu/~porter/courses/cse506/f11/lab1.html

1. Read 5.1 (Pointers and Addresses) through 5.5 (Character Pointers and Functions) in K&R. Then download the code for pointers.c, run it, and make sure you understand where all of the printed values come from. In particular, make sure you understand where the pointer addresses in lines 1 and 6 come from, how all the values in lines 2 through 4 get there, and why the values printed in line 5 are seemingly corrupted.


2. There are other references on pointers in C, though not as strongly recommended. A tutorial by Ted Jensen that cites K&R heavily is available in the course readings.

3. Here are a few specific points you read about in K&R Chapter 5 that are worth remembering for the following exercise and for future labs.

• If int *p = (int*)100, then (int)p + 1 and (int)(p + 1) are different numbers: the first is 101 but the second is 104. When adding an integer to a pointer, as in the second case, the integer is implicitly multiplied by the size of the object the pointer points to.
• p[i] is defined to be the same as *(p+i), referring to the i'th object in the memory pointed to by p. The above rule for addition helps this definition work when the objects are larger than one byte.
• &p[i] is the same as (p+i), yielding the address of the i'th object in the memory pointed to by p.

pointers.c
-----------

#include <stdio.h>
#include <stdlib.h>

void
f(void)
{
    int a[4];
    int *b = malloc(16);
    int *c;
    int i;

    printf("1: a = %p, b = %p, c = %p\n", a, b, c);

    c = a;
    for (i = 0; i < 4; i++)
 a[i] = 100 + i;
    c[0] = 200;
    printf("2: a[0] = %d, a[1] = %d, a[2] = %d, a[3] = %d\n",
    a[0], a[1], a[2], a[3]);

    c[1] = 300;
    *(c + 2) = 301;
    3[c] = 302;
    printf("3: a[0] = %d, a[1] = %d, a[2] = %d, a[3] = %d\n",
    a[0], a[1], a[2], a[3]);

    c = c + 1;
    *c = 400;
    printf("4: a[0] = %d, a[1] = %d, a[2] = %d, a[3] = %d\n",
    a[0], a[1], a[2], a[3]);

    c = (int *) ((char *) c + 1);
    *c = 500;
    printf("5: a[0] = %d, a[1] = %d, a[2] = %d, a[3] = %d\n",
    a[0], a[1], a[2], a[3]);

    b = (int *) a + 1;
    c = (int *) ((char *) a + 1);
    printf("6: a = %p, b = %p, c = %p\n", a, b, c);
}

int
main(int ac, char **av)
{
    f();
    return 0;
}