Solution for Exercise 19-4

ex1904

#include <stdio.h>

int main()
{
    int numbers[10];
    int x;
    int *pn;

    pn = numbers;       /* initialize pointer */

    /* Fill array */
    for(x=0;x<10;x++)
    {
        *pn=x+1;
        pn++;
    }

    /* Display array */
    for(x=0;x<10;x++)
        printf("numbers[%d] = %d\n",
                x,numbers[x]);

    return(0);
}

Output

numbers[0] = 1
numbers[1] = 2
numbers[2] = 3
numbers[3] = 4
numbers[4] = 5
numbers[5] = 6
numbers[6] = 7
numbers[7] = 8
numbers[8] = 9
numbers[9] = 10

Notes

* Pointers are variables. You can re-initialize a pointer at any time in your code.

Exercise ex1905 →