Pointer of void array

Question

I have an array of void-Pointers and want to access the elements (inititialize them), but it do not work:

void* anyptr = ...; //a pointer to something
void* arr = (void*)malloc(sizeof(void*)*10);

int i=0;
for(i=0; i<10; i++)
   *(arr+i) = anyptr; //dont work, (arr+n) = anyptr; doesn´t work too

I guess, the reason why this won´t work is that on the left side is the result of element i. But i don´t have an idea how to do this

Why do you cast `void *` to `void *`? [You should never cast it anyway.](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — , Apr 20 '13 at 19:08
Asking "why you do that?" to somebody who doesn't know what he/she is doing almost always sounds rude and not didactic. — Rafael Eyng, Jun 03 '18 at 21:36

score 27 · Accepted Answer · answered Apr 20 '13 at 21:45

There are two ways to initialize arrays in C:

On the stack (which will handle memory for you since it will be cleaned up when your function ends)
In the heap (which will require you to handle allocation and freeing on your own).

If you would like to use the stack, you could initialize your array like this...

#define ARRAY_LENGTH 10
void *ptr;
void *arr[ARRAY_LENGTH];
for (int i = 0; i < ARRAY_LENGTH; i++) {
    arr[i] = ptr;
}

You can similarly define your array in the heap as follows...

#define ARRAY_LENGTH 10
void *ptr;
void **arr = malloc(sizeof(void *) * ARRAY_LENGTH);
for (int i = 0; i < ARRAY_LENGTH; i++) {
    arr[i] = ptr;
}
free(arr);

It is important to remember that an array (besides arrays assigned in the stack, which have some additional attributes such as length) is essentially just a pointer to the first element, and the operation arr[i] is the same as moving i*sizeof(elem) bytes away from the first element, and accessing the memory there. If you would like to get a pointer to the ith index in the array, then you would use notations such as...

void *indexPtr = arr + i;

or

void *indexPtr = &( arr[i] );

In this fashion, an array of void*'s would be of type void **, since the variable is a pointer to the first member of the array, which is a pointer. This can be a bit confusing, but just always try to keep in mind what type the elements of the array are, and creating a pointer to them. So if the array is of type int, then the array would be of type int or int[], but if you are storing pointers to integers, you would initialize an array of type int * in either of these two forms...

int **arr = malloc(sizeof(int *) * ARRAY_LENGTH);
int *arr[ARRAY_LENGTH];

Also note that you are storing pointers, so if you run the code...

int *arr[4];
for (int i = 0; i < ARRAY_LENGTH; i++) {
    arr[i] = &i;
}

Although it may seem to be that the values pointed to in the array would be as follows- [0, 1, 2, 3], but in reality it would be [4, 4, 4, 4], since what you actually have is an array of pointers all pointing to the variable i in your function, so whenever you change that, the values pointed to in the array will all be changed.
I hope this helped

I'm getting `error: invalid conversion from ‘void*’ to ‘void**’ [-fpermissive]` from the `malloc` instruction, any ideas? — Aurelien Montmejat, May 03 '22 at 08:26
Using `void **arr = (void **)malloc(sizeof(void *) * 10)` fixed it for me! — Aurelien Montmejat, May 03 '22 at 08:47

score 12 · Answer 2 · answered Apr 20 '13 at 19:06

12

You need to change this line

void* arr = (void*)malloc(sizeof(void*)*10);

to this

void** arr = malloc(sizeof(void*)*10);

answered Apr 20 '13 at 19:06

Alexey Frunze

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Could you please explain why it is declared as a pointer to a pointer? – init Aug 06 '16 at 16:24
3

@init because that's what you're allocating memory for: 10 pointers. How do you point to this memory? You point to the first pointer. Hence, it's a pointer to a pointer. You add 1 to it and you point to the next pointer and so on until the last of the 10 pointers. – Alexey Frunze Aug 06 '16 at 22:02

score 3 · Answer 3 · answered Apr 20 '13 at 19:15

3

You can't dereference a void pointer. That's the whole point of void pointers. Dereferencing a pointer provides you with access to the item that's found at the address the pointer points to. With a void pointer, however, you don't know how large the target object is (is it a 1B character or a 100B struct?). You have to cast it to a specific pointer type before dereferencing it.

Adding (or subtracting) an integer i to a pointer is then defined as adding i-times sizeof(*pointer) to the pointer's content. (You can only tell sizeof(*pointer) if your pointer has a specific type. Pointer arithmetic with void pointers makes no sense).

As for (arr+n)= anyptr;, arr+n is just an address. It's not a value you can assign something to (not an lvalue).

answered Apr 20 '13 at 19:15

Petr Skocik

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and how i can change the value on address arr+n? – 0xDEADBEEF Apr 20 '13 at 19:55
By casting it to a specific pointer type and then dereferencing it. Note, however, that ptr+n, if ptr is a non-void pointer will mean n items ahead of ptr, which doesn't necesserily mean n bytes (it does for char* pointers, but for int pointers it will most likely mean n*4 bytes ahead, and for pointers to pointers (of any type) it will mean 4*n or 8*n depending on your architecture (32 or 64 bit respectively). – Petr Skocik Apr 20 '13 at 20:05

Pointer of void array

3 Answers3