An In-Depth Understanding of Memory and Pointers in C/

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Memory Representation Using Char Arrays

Memory Representation Using Char Arrays

Get introduced to strings in C from a memory point of view.

We'll cover the following...

Introduction

In C, we represent strings using char arrays. Therefore, everything we discussed in the “Pointers And Arrays” chapter holds for this chapter.

However, a few things are specific to character arrays, which we’ll now discuss.

In this chapter, we’ll refer to character arrays as strings.

First code with char arrays

Let’s use the ARRAY_SIZE macro with a character array and see if we can deduce its number of elements.

Moreover, let’s also print the string’s size in bytes.

C
#include <stdio.h>
#define ARRAY_SIZE(arr) (sizeof(arr) / sizeof(arr[0]))
int main()
{
    char str[32];
    printf("sizeof(str) = %u\n", sizeof(str));
    printf("ARRAY_SIZE(str) = %u\n", ARRAY_SIZE(str));
    return 0;
}

The output is as follows:

sizeof(str) = 32
ARRAY_SIZE(str) = 32
  • sizeof: The size of the array is 32 bytes. The string has 32 elements, and each element is a char that occupies 1 byte. 1 * 32 = 32 bytes.
  • ARRAY_SIZE: The number of elements in the array is 32 since each is 1 byte.

Previously, we noticed a difference between the total size and number of elements for integer arrays. It happened because the size of int was 4 bytes. The size of char is 1 byte, so the total size and the number of elements are equal.

In conclusion, sizeof and ARRAY_SIZE are equivalent for character arrays. In this chapter, we’ll use sizeof to determine the number of elements in a char array. We stress again that for all other data types, we have to use ARRAY_SIZE.

Memory drawing

Let’s assume the following string ...