C [C] byte address range.

[Xenon02]

Hello !

I am struggeling with understanding how this works :

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int rows = 2;
    int columns = 5;
    int **matrix = (int **) malloc(rows * sizeof(int *));
    matrix[0] = (int *) malloc(rows * columns * sizeof(int));
    for (int i = 1; i < rows; i++)
    {
        matrix[i] = matrix[0] + i * columns;
        //printf("%p", *matrix[i]);
    }
     for (int i = 0; i < rows; i++)
    {
      
        printf("Indeks : %d, Address of (matrix) : %p\n", i, matrix+i);
        printf("Indeks : %d, Address of (*matrix) : %p\n", i, *(matrix+i));
    }
    
    return 0;
}

The results are like here :

Indeks : 0, Address of (matrix) : 0x55eff083e2a0
Indeks : 0, Address of (*matrix) : 0x55eff083e2c0
Indeks : 1, Address of (matrix) : 0x55eff083e2a8
Indeks : 1, Address of (*matrix) : 0x55eff083e2d4

From tutorial like here :



It shows that the addres of matrix should change every 4 bytes but it changes every 8 bytes in my code and it uses malloc. But for the *matrix if changes every 4 bytes and it is also malloc "array", so why in one malloc it changes every 8 bytes and in other 4 bytes they are both int and int value takes 4 bytes.

So how does he know that when I write matrix + 1 he know that I want to move 4 bytes to another number which is located in the picture 604 this is anotherr number. Like how does he convert it into 8 bytes or 4 bytes when both mallocs are pointers (I heard that pointer takes 8 bytes but (int*) malloc is also a pointer but it takes 4 bytes not 8)

Or another example with chars.

#include <stdio.h>

int main()
{
 
    char* arr[3] = { "geek", "Geeks", "Geeksfor" };
 
    for (int i = 0; i < 3; i++) {
        printf("Indeks : %d, Address of (arr) : %p\n",i , *(arr+i));
        printf("Indeks : %d, Address of (&arr) : %p\n",i , &arr[i]);
    }
 
    return 0;
}

The results :

Indeks : 0, Address of (arr) : 0x55ae7f302008
Indeks : 0, Address of (&arr) : 0x7fff60e25ef0
Indeks : 1, Address of (arr) : 0x55ae7f30200d
Indeks : 1, Address of (&arr) : 0x7fff60e25ef8
Indeks : 2, Address of (arr) : 0x55ae7f302013
Indeks : 2, Address of (&arr) : 0x7fff60e25f00

Now here also 4 bytes difference or more when looking at address arr (char has only 1 byte when there are 4 letters it takes 4 bytes logical). But the &arr has 8 bytes. Why ?

It doesn't make sense when I write &arr + 1 knowing that char is 1 byte then why it goes every 8 bytes ?
Same goes with the 1st example I am struggeling to understand when I type *matrix + 1 it will go 4 bytes but when I type matrix + 1 it will go 8 bytes how does he know when both malloc addreses are pointers which means they should be 8 bytes adresses

 

[Xenon02]

PS.
For the second example when I write size of(arr) it will say it has 8 bytes although for arr[0] it takes 4 bytes because there are 4 addresses for 4 letters. But it says it's size is 8 bytes what ??? Even though if I write (arr + 1) it will go 1 byte (for example arr[0] is 55ae7f302008 so are[1] 55ae7f302009) and so it is not 8 byte. If it was 8 byte size it would go 8 bytes. If I write arr +1 it would go 8 byte and not 1. Not only &arr is a pointer to the address *arr but also *arr is a pointer to a value. Because char arr is a array of pointers.
Two pointers but one is 8 byte and second is one byte but the systems says it is 8 byte.

I don't get it.