Apprendre Void | Pointers

We've previously encountered void when we looked at functions that don't return anything or just display text on the screen.

Keep in mind that we can't declare a variable of the void type. However, we can create a void* pointer that points to a memory location.

Note

If you've worked through the pointer exercises in the section, you'll know that all pointers occupy 8 bytes. A void* pointer is no different.

A void* pointer isn't tied to any specific data type. This flexibility allows us to store the address of any data type in it.

However, there's a catch: we can't dereference a void* pointer.

Note

This limitation stems from the same reason we can't declare variables of the void type.

Perform an explicit type conversion!

Main.c


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#include <stdio.h>

int main()
{
    char c = 'F';
    int i = 100;
    double d = 3.15;

    void* pV;

    pV = &c;
    printf("%c \n", *((char*)pV));

    pV = &i;
    printf("%d \n", *((int*)pV));

    pV = &d;
    printf("%.2f \n", *((double*)pV));

    return 0;
}

Congratulations on completing the basic course in C programming! If you're eager to dive deeper into this fascinating and logical programming language, I highly recommend delving into topics like macros, sorting algorithms, and data structures. Additionally, familiarizing yourself with one of the Linux distributions is invaluable for any seasoned C programmer.

Best of luck!

Tout était clair ?

Merci pour vos commentaires !

Section 6. Chapitre 6

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We've previously encountered void when we looked at functions that don't return anything or just display text on the screen.

Keep in mind that we can't declare a variable of the void type. However, we can create a void* pointer that points to a memory location.

Note

If you've worked through the pointer exercises in the section, you'll know that all pointers occupy 8 bytes. A void* pointer is no different.

A void* pointer isn't tied to any specific data type. This flexibility allows us to store the address of any data type in it.

However, there's a catch: we can't dereference a void* pointer.

Note

This limitation stems from the same reason we can't declare variables of the void type.

Perform an explicit type conversion!

Main.c


              123456789101112131415161718192021
            
#include <stdio.h>

int main()
{
    char c = 'F';
    int i = 100;
    double d = 3.15;

    void* pV;

    pV = &c;
    printf("%c \n", *((char*)pV));

    pV = &i;
    printf("%d \n", *((int*)pV));

    pV = &d;
    printf("%.2f \n", *((double*)pV));

    return 0;
}

Best of luck!

Tout était clair ?

Merci pour vos commentaires !

Section 6. Chapitre 6

Void

Perform an explicit type conversion!

Awesome!

Void

Perform an explicit type conversion!