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.

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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.

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#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 declaration: ...