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Pointers and Arrays in C: Array Decay, a[i] vs *(a + i), and Arrays of Strings

Understand how arrays and pointers relate in C: array-to-pointer decay, why a[i] equals *(a + i), the differences between arrays and pointers (sizeof and assignment), pointers to strings, arrays of strings, and pointers to whole arrays.

Intermediate6 min readUpdated
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Arrays and pointers are closely related in C, but they are not the same thing. Understanding the relationship explains how array parameters, strings and indexing really work.

Array decay#

In most expressions, an array's name is automatically converted ("decays") to a pointer to its first element:

C
int a[] = {5, 10, 15};
int *p = a;                  /* same as p = &a[0] */
printf("%d\n", p == &a[0]);
printf("%d %d\n", *p, p[2]);
Output
1
5 15

That is why passing an array to a function passes a pointer (Parameters).

a[i] is *(a + i)#

Indexing is defined in terms of pointer arithmetic: a[i] means exactly *(a + i). It works the same on arrays and pointers:

C
int a[] = {5, 10, 15, 20};
int *p = a + 1;              /* points to 10 */
printf("%d %d\n", a[2], *(a + 2));
printf("%d %d\n", p[0], p[2]);   /* indexes are relative to p */
Output
15 15
10 20

Where arrays and pointers differ#

Array int a[4]Pointer int *p
What it isStorage for 4 intsStorage for one address
sizeofWhole array (16 bytes)The pointer (8 bytes on 64-bit)
Can be assignedNo: a = ... is an errorYes: p = other
& givesPointer to the whole arrayAddress of the pointer variable
C
int a[4] = {0};
int *p = a;
printf("%zu %zu\n", sizeof a, sizeof p);
Output
16 8

(64-bit system.) Decay does not happen for sizeof, &, or a string literal used to initialise a char array.

Strings: array vs pointer#

C
char arr[] = "pune";          /* a modifiable array of 5 chars */
const char *ptr = "pune";     /* points to a read-only literal */

arr[0] = 'P';                 /* fine */
ptr = "delhi";                /* fine: the pointer now points elsewhere */
printf("%s %s\n", arr, ptr);
Output
Pune delhi

Arrays of strings#

An array of const char * is the usual way to store a list of fixed strings. Each element points to a literal of any length:

C
#include <stdio.h>

int main(void) {
    const char *days[] = {"Mon", "Tue", "Wednesday", "Thu", "Fri"};
    int n = sizeof days / sizeof days[0];
    for (int i = 0; i < n; i++) {
        printf("%-10s first letter %c\n", days[i], days[i][0]);
    }
    return 0;
}
Output
Mon        first letter M
Tue        first letter T
Wednesday  first letter W
Thu        first letter T
Fri        first letter F

A 2D char array, char days[5][10], instead stores the characters themselves in fixed-size rows. Use it when you need to modify the strings.

Pointer to a whole array#

int (*row)[3] is a pointer to an array of 3 ints. It is what a 2D array decays to, so stepping it moves a whole row:

C
int grid[2][3] = {{1, 2, 3}, {4, 5, 6}};
int (*row)[3] = grid;        /* points to grid[0] */
row++;                       /* now points to grid[1] */
printf("%d %d\n", (*row)[0], row[0][2]);
Output
4 6

The parentheses matter: int *row[3] would be an array of 3 pointers.

Practice#

Practice exercise

Write const char *longest(const char *words[], int n) that returns the longest word in an array of strings, and use it on {"array", "pointer", "decay", "dereference", "null"}.

Show solution
C
#include <stdio.h>
#include <string.h>

const char *longest(const char *words[], int n) {
    const char *best = words[0];
    for (int i = 1; i < n; i++) {
        if (strlen(words[i]) > strlen(best)) best = words[i];
    }
    return best;
}

int main(void) {
    const char *words[] = {"array", "pointer", "decay", "dereference", "null"};
    printf("%s\n", longest(words, 5));
    return 0;
}
Output
dereference

Key takeaway#

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