Course Content

C Structs

1. Introduction to Structs

What are Structs Structs Application Defining and Declaring Structs Creating Your First Struct Accessing Struct Fields

2. Pointers and Structs

Pointers Brief Overview Pointers to Structs Pointers Inside Structs Dynamically Allocating Structs Passing Structs to the Functions

3. Structs and Memory

Understanding Memory Layout of Structs Alignment and Padding in Structs Unions Arrays with Structs

4. Advanced Structs Usage

Nested Structs Struct with Arrays and Other Structs Anonymous Struct and Union Complex Data Structure

5. Implementing Data Structures

Introduction to Linked List Basic Concept and Structure Building a Linked List Travesting and Displaying Linked List Completion Overview

Understanding Memory Layout of Structs

Memory structures in the C programming language play a key role in understanding how data is stored and accessed in memory. When a structure is defined in C, the compiler determines how to place its members in memory based on alignment and padding rules.

Here's an overview of how basic memory allocation for structures works in C:

main


              123456789101112131415
            
#include <stdio.h>

// simple struct 
struct Test {
    char x; // 1 byte
    int y; // 4 bytes
};

int main() {
    struct Test example;
    printf("Size of struct Point: %zu\n", sizeof(example));
    printf("Address of p.x (char): %p\n", &example.x);
    printf("Address of p.y (int): %p\n", &example.y);
    return 0;
}

As expected, such a structure should occupy 5 bytes: 1 byte for int x, 4 bytes for int y, but this will be 8 bytes.

Why is the size of the structure much larger than we expected?

This occurs because the compiler can insert padding between members to ensure that each member begins at an address that is a multiple of its size.

Task

Create a structure with four char fields;
Initialize the structure and display its size in bytes;
Display the address of each field.

Task

Create a structure with four char fields;
Initialize the structure and display its size in bytes;
Display the address of each field.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Everything was clear?

Section 3. Chapter 1

Understanding Memory Layout of Structs

Here's an overview of how basic memory allocation for structures works in C:

main


              123456789101112131415
            
#include <stdio.h>

// simple struct 
struct Test {
    char x; // 1 byte
    int y; // 4 bytes
};

int main() {
    struct Test example;
    printf("Size of struct Point: %zu\n", sizeof(example));
    printf("Address of p.x (char): %p\n", &example.x);
    printf("Address of p.y (int): %p\n", &example.y);
    return 0;
}

As expected, such a structure should occupy 5 bytes: 1 byte for int x, 4 bytes for int y, but this will be 8 bytes.

Why is the size of the structure much larger than we expected?

This occurs because the compiler can insert padding between members to ensure that each member begins at an address that is a multiple of its size.

Task

Create a structure with four char fields;
Initialize the structure and display its size in bytes;
Display the address of each field.

Task

Create a structure with four char fields;
Initialize the structure and display its size in bytes;
Display the address of each field.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Everything was clear?

Section 3. Chapter 1

Understanding Memory Layout of Structs

Here's an overview of how basic memory allocation for structures works in C:

main


              123456789101112131415
            
#include <stdio.h>

// simple struct 
struct Test {
    char x; // 1 byte
    int y; // 4 bytes
};

int main() {
    struct Test example;
    printf("Size of struct Point: %zu\n", sizeof(example));
    printf("Address of p.x (char): %p\n", &example.x);
    printf("Address of p.y (int): %p\n", &example.y);
    return 0;
}

As expected, such a structure should occupy 5 bytes: 1 byte for int x, 4 bytes for int y, but this will be 8 bytes.

Why is the size of the structure much larger than we expected?

This occurs because the compiler can insert padding between members to ensure that each member begins at an address that is a multiple of its size.

Task

Create a structure with four char fields;
Initialize the structure and display its size in bytes;
Display the address of each field.

Task

Create a structure with four char fields;
Initialize the structure and display its size in bytes;
Display the address of each field.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Everything was clear?

Here's an overview of how basic memory allocation for structures works in C:

main


              123456789101112131415
            
#include <stdio.h>

// simple struct 
struct Test {
    char x; // 1 byte
    int y; // 4 bytes
};

int main() {
    struct Test example;
    printf("Size of struct Point: %zu\n", sizeof(example));
    printf("Address of p.x (char): %p\n", &example.x);
    printf("Address of p.y (int): %p\n", &example.y);
    return 0;
}

As expected, such a structure should occupy 5 bytes: 1 byte for int x, 4 bytes for int y, but this will be 8 bytes.

Why is the size of the structure much larger than we expected?

This occurs because the compiler can insert padding between members to ensure that each member begins at an address that is a multiple of its size.

Task

Create a structure with four char fields;
Initialize the structure and display its size in bytes;
Display the address of each field.

Switch to desktop for real-world practiceContinue from where you are using one of the options below

Section 3. Chapter 1

Switch to desktop for real-world practiceContinue from where you are using one of the options below