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C Structs

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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.

main.

copy

              
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.

Tehtävä

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

Ratkaisu

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book
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.

main.

copy

              
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.

Tehtävä

Swipe to start coding

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

Ratkaisu

Switch to desktopVaihda työpöytään todellista harjoitusta vartenJatka siitä, missä olet käyttämällä jotakin alla olevista vaihtoehdoista
Oliko kaikki selvää?

Miten voimme parantaa sitä?

Kiitos palautteestasi!

close

Awesome!

Completion rate improved to 4.17

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