Pointer Pointer leads to Segmentation Fault in C

Question

Why can I create an array of pointers and dereference its (pointer-)elements.

int a = 1;
int* arr[1];

arr[0] = &a;

But cannot do the same with pointer to pointers:

int** arr2;
arr2[0] = &a;
--> Seg fault

Answer 1

int** arr2;

You don't initialize this pointer. Accessing it is undefined behavior and can/will result in crashes.

To initialize use something like int** arr2 = malloc(<someSize> * sizeof(*arr2)). (PS: malloc may return NULL, and you have to use free to return the memory).

Answer 2

from the first part

int a = 1;
int* arr[1];
arr[0] = &a;

So a is int. arr is what? It is int*[], an array of pointers to int. So the first and only element arr[0] is a pointer to an int and it is normal to assign an address of an int to it. The compiler is ok with that, since the operator & --- Address of --- assigns the address of the int a to a pointer to int, arr[0].

from the second part

    int** arr2;
    arr2[0] = &a;

arr2 is what? It is int**, a pointer to a pointer to int.

• arr2 is int**
• so *arr2 is int*, a pointer to an int
• and **arr is an int

arr2 is NOT what? NOT an array, so you can not write as you did arr[0].

what you can write instead

arr2 is int** so as in the first case it can get the address of a pointer to int. And you have an array of these here in the first part. So you for sure can write

    arr2 = &arr[0];

Since & will extract the address of a pointer, and arr[0] is a pointer to int. And arr2 is int**, a pointer to a pointer to int.

See the output

toninho@DSK-2009:~/projects/um$ gcc -o tst -Wall -std=c17 pp.c
toninho@DSK-2009:~/projects/um$ ./tst
*arr[0] = 1 and **arr2 = 1
toninho@DSK-2009:~/projects/um$ 

of this code

#include <stdio.h>

int main()
{
    int     a = 1;
    int*    arr[1]; // so arr is what? int*[]

    arr[0] = &a;

    int**   arr2; // arr2 is what? int**
    //arr2[0] = &a;

    arr2 = &arr[0];

    printf("*arr[0] = %d and **arr2 = %d\n",
        *arr[0], **arr2 );
    return 0;
};