How to create dynamic sized array of NULL pointers?

Question

I want to have dynamic size of array of pointers, how can I do it with malloc? without malloc it looks like this:

typedef struct Nodes {
    UINT_64 Datasequence; 
    UINT_32 Data; 
    struct Nodes *next;
} Node;

Node *head_ri_interface[2][50];

which will give me an array of [2][50] of null pointers.

I have tried this:

void NULLPointerArrayInit(int MAX_LIST_HEAD) {
    int i = 0;

    for (i = 0; i < 2; i++) {
        head_ri_interface[i] = (Node *)malloc(sizeof(Node *) *  MAX_LIST_HEAD);
    }
}

but I see that I get that the struct I am pointing at are not NULL. what am I doing wrong?

But `head_ri_interface[i]` is an array of 50 pointers to your structure? Are you *sure* you want an array of arrays? — Some programmer dude, Jul 29 '18 at 12:24
yes , what i want is in 2D array of dynamic size (50 as an example) of pointers to head of link list that strart from point to NULL — Omer Anisfeld, Jul 29 '18 at 12:26
[How do I work with dynamic multi-dimensional arrays in C? - Stack Overflow](https://stackoverflow.com/questions/917783/how-do-i-work-with-dynamic-multi-dimensional-arrays-in-c) — user202729, Jul 29 '18 at 12:30
So your *real* declaration is something like `Node ** head_ri_interface[2];`? Please try to create a [Minimal, Complete, and Verifiable Example](http://stackoverflow.com/help/mcve) to show us. — Some programmer dude, Jul 29 '18 at 12:31
And perhaps you would be interested in learning about [`calloc`](http://en.cppreference.com/w/c/memory/calloc)? — Some programmer dude, Jul 29 '18 at 12:33

score 1 · Accepted Answer · answered Jul 29 '18 at 12:42

1

void ** 2Darray = malloc(2*sizeof(void*));

int i;
for(i = 0; i < 2; i++){
    2Darray[i] = malloc(MAX_LIST_HEAD*sizeof(void*));
    memset(2Darray[i], 0, MAX_LIST_HEAD*sizeof(void*));
}

So what i did here: I allocated memory for 2 pointer types. Than pointed each of this two pointers to MAX_LIST_HEAD times size of pointer type memory (you can assume it's an array of pointers size - MAX_LIST_HEAD); After that I wrote zeros to all pointers, this means they are pointing to NULL.

(If pointer is pointing to 0 it means it is pointing to NULL. But when pointer is pointing to NULL it does not always mean it is pointing to 0)

answered Jul 29 '18 at 12:42

Data Chanturia

149
9

You should consider using `calloc` or initializing the array of pointers with a loop. – chqrlie Jul 29 '18 at 12:43
@chqrlie Yes I forgot, In `calloc` it would have been better visible what I did here. Thanks. But I think it works either. – Data Chanturia Jul 29 '18 at 12:45
The type `void**` is inconsistent with the allocation method. The OP wants to allocate a pseudo-2D array of `Node*`. – chqrlie Jul 29 '18 at 12:52
1

thank's i used this answer just with this minor change of alyways first dimension is size of 2 meands : ~ Node * head_ri_interface[2] ; void NULLPointerArrayInit(int MAX_LIST_HEAD){ int i = 0 ; for(i = 0; i < 2; i++) { head_ri_interface[i] = (Node *) CALLOC (MAX_LIST_HEAD, sizeof(ACL_Node *)); } } ~ – Omer Anisfeld Jul 29 '18 at 14:08

How to create dynamic sized array of NULL pointers?

1 Answers1