Pointer to array

Learn to access array elements with pointer arithmetic.

Pointer arithmetic with arrays

Adding an integer to a pointer jumps to the next element. p + 1 moves forward by the size of the pointer's type; 4 bytes for int.

int arr[3] = {10, 20, 30};
int *p = arr;
printf("%d", *(p + 2));  /* 30 */
printf("%d", p[2]);       /* 30: same thing */

p[i] and *(p + i) mean exactly the same thing. Index notation is more readable, pointer notation more flexible.

p[i] and *(p+i) are the same

In C the p[i] notation is actually shorthand for *(p + i). They are fully equivalent; the compiler produces the same machine code.

int arr[4] = {5, 10, 15, 20};
int *p = arr;
printf("%d", p[0]);      /* 5 */
printf("%d", *(p + 1));   /* 10 */
printf("%d", p[2]);      /* 15 */

Which form you choose is a matter of style. Index notation is easier for beginners.

Exercise Mix both notations

Define int arr[4] = {5, 10, 15, 20};. Start with int *p = arr;. With four printf calls print p[0], *(p + 1), p[2] and *(p + 3) on separate lines. Output should be 5, 10, 15, 20.

Show example solution

#include <stdio.h>

int main(void) {
    int arr[4] = {5, 10, 15, 20};
    int *p = arr;
    printf("%d\n", p[0]);
    printf("%d\n", *(p + 1));
    printf("%d\n", p[2]);
    printf("%d\n", *(p + 3));
    return 0;
}

Test yourself

Access the third element

int arr[5] = {1,2,3,4,5};
int *p = arr;
printf("%d", *(p + ____));  /* prints 3 */

Show answer

int arr[5] = {1,2,3,4,5};
int *p = arr;
printf("%d", *(p + 2));  /* prints 3 */

Match access forms

p[0]

first element
p[1]

second element
*(p+2)

third element
p+3

fourth address
p = arr; p[2] is equal to which?
- *(p + 1)
- *(p + 3)
- *(p + 2)
- *p
Show answer

*(p + 2) — p[2] and *(p + 2) are the same.

Next lesson: Passing arrays to functions

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