How to transmit pointer to dynamically allocated array as a function parameter

Question

I want to allocate an array inside of a function and to be able to use the pointer to that array outside of it as well. I don't know what's wrong with my code

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

void alloc_array (int **v, int n)
{
    *v=(int *)calloc(n,sizeof(int));
    printf ("the address of the array is %p",v);
}

void display_pointer (int *v)
{
    printf ("\n%p",v);
}

int main()
{
    int *v;
    alloc_array(&v,3);
    display_pointer(v);

    return 0;
}

I would expect to get the same address in both printfs, but that is not the case. What is my mistake?

`printf ("the address of the array is %p",v);` ---> `printf ("the address of the array is %p",*v);` Recall what the double pointer is pointing to, and what its value is. — StoryTeller - Unslander Monica, Nov 28 '17 at 18:38
@StoryTeller thank you. I understand now. I actually feel pretty stupid for not realising this. — Andrei, Nov 28 '17 at 18:41

score 3 · Accepted Answer · answered Nov 28 '17 at 18:39

3

void alloc_array (int **v, int n)
{
    *v = calloc(n,sizeof(int));
    printf("the address of the array is %p", *v);
    //                                   ----^
}

Note the additional star in my printf call.

Also, don't cast the return value of malloc / calloc.

answered Nov 28 '17 at 18:39

Jonathon Reinhart

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You are right. My bad for asking such an obvious thing, I didn't pay enough attention. – Andrei Nov 28 '17 at 18:42
It's an easy mistake to make when you are dealing with multiple levels of indirection. – Jonathon Reinhart Nov 28 '17 at 18:43
Could you please explain why the return value shouldn't be casted? I was taught that we should do this because those functions return a type void pointer – Andrei Nov 28 '17 at 18:43
In C, `void*` can be freely assigned to pointers of other types without cast and vice-versa. – Jonathon Reinhart Nov 28 '17 at 18:44
https://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc – Jonathon Reinhart Nov 28 '17 at 18:45
I understand. So casting it is just superfluous, or is it bad practice as well? edit: thank you for the link, I am going to look into that – Andrei Nov 28 '17 at 18:46

score 1 · Answer 2 · answered Nov 28 '17 at 18:42

v in alloc_array contains address of a pointer variable. That variable is vin main().

You don't care about it. Because it won't change. But it the content of v in main would.

You should print *v but why?

Because the content of v in alloc_array contains the address of the memory that you allocated.

But display_pointer doesn't need this. because here you pass the pointer variable itself and the local copy of it in the called function will contain the address of the memory that you allocated.

score 1 · Answer 3 · answered Nov 28 '17 at 18:49

In this code snippet

int *v;
alloc_array(&v,3);
display_pointer(v);

the function alloc_array accepts the address of the variable (pointer) v and this address is outputted in the function

printf ("the address of the array is %p",v);

On the other hand the function display_pointer accepts the value stored in the variable v and this value is outputted within the function

printf ("\n%p",v);

Add one more statement in the function alloc_array and you will see the difference

printf ("the address of the original pointer is %p", ( void * )v);
printf ("the address of the first element of the allocated array is %p", ( void * )*v);

How to transmit pointer to dynamically allocated array as a function parameter

3 Answers3