Array name as pointer

Learn that the array name is a pointer to the first element.

Array name as pointer

In C an array name decays into the address of its first element. So you can assign an array name directly to a pointer.

int arr[3] = {10, 20, 30};
int *p = arr;  /* p holds the address of arr[0] */
printf("%d", *p);  /* prints 10 */

This is called array decay: the array info drops to a pointer. arr and p now refer to the same memory region.

Array decay in functions

When you pass an array to a function the array name decays into a pointer. The function sees the array itself, not a copy. So the function can modify the array elements.

void add_one(int *arr, int n) {
    for (int i = 0; i < n; i++) arr[i]++;
}

The array length is not stored in the pointer; you must pass the size as a separate parameter.

Exercise Access array elements via pointer

Define int arr[3] = {10, 20, 30};. Bind p to arr with int *p = arr;. With three printf calls print *p, *(p + 1) and *(p + 2) on separate lines. Output should be 10, 20, 30.

Show example solution

#include <stdio.h>

int main(void) {
    int arr[3] = {10, 20, 30};
    int *p = arr;
    printf("%d\n", *p);
    printf("%d\n", *(p + 1));
    printf("%d\n", *(p + 2));
    return 0;
}

Test yourself

int arr[3] = {10,20,30}; printf("%d", *(arr+1)); prints what?
- 10
- 20
- 30
- 11
Show answer

20 — *(arr+1) is arr[1], which is 20.

Assign array to pointer

int arr[3] = {10, 20, 30};
int *p = ____;  /* p now points to arr */

Show answer

int arr[3] = {10, 20, 30};
int *p = arr;  /* p now points to arr */

Match array-pointer concepts

arr

address of first element
arr+1

address of second element
*(arr+2)

value of third element
int *p = arr

pointer assignment

Next lesson: Pointer to array

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