Summary - Arrays and pointers

Combine the array-pointer relationship to write an array sum function.

Summary: Arrays and pointers

In this group you saw the close link between arrays and pointers:

An array name decays into the address of its first element.
p[i] and *(p + i) are the same.
Passing an array to a function turns it into a pointer; size is separate.
The end = arr + n pattern builds a pointer loop.
Two-dimensional arrays are walked with nested loops.

Let's combine these in a short exercise.

Exercise Summary exercise

Define int sum(int *arr, int n): sum the array with a for loop and return total. In main define int arr[3] = {10, 20, 30};. Start with int *p = arr;. Print *p, *(p + 2) and sum(arr, 3) on separate lines with printf. Output should be 10, 30, 60.

Show example solution

#include <stdio.h>

int sum(int *arr, int n) {
    int total = 0;
    for (int i = 0; i < n; i++) {
        total += arr[i];
    }
    return total;
}

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

Test yourself

What is an array name, really?
- The array length
- A copy of the array
- Pointer to the first element
- A NULL pointer
Show answer

Pointer to the first element — An array name decays into the address of the first element.
Match concepts

arr

array name = pointer
*(p+i)

pointer access
int *arr

function parameter
p < end

loop condition

Access an element via pointer

int arr[3] = {10,20,30};
int *p = arr;
printf("%d", *(p + ____));  /* prints 30 */

Show answer

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

Next lesson: Task model

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