How to declare a pointer to : array of void pointers? in c

Question

I'm little bit confused about how to declare a pointer to array or void*.

let say :

void* myarr[20];

how to declare a pointer to "myarr?"

Related, if not the same (in the sense of how to declare a pointer to an array): stackoverflow.com/q/9779450/694576 — alk
– alk, Commented Nov 30, 2014 at 14:01

Vlad from Moscow · Accepted Answer · 2014-11-30 09:48:41Z

3

I think you mean a pointer to the first element of an array of pointers to void.

It is simply to do if you will use the following general approach. Let's assume that you have an array of some type T

T myarr[20];

then the definition of the pointer to the first element of the array will look like

T *ptr = myarr;

Now all what you need is substitute T for you particulat type and you will get

void * myarr[20];

void * *ptr = myarr;

If you mean indeed a pointer to the array then approach is the following

T myarr[20];
T ( *ptr )[20] = &myarr;

Or if to substitute T for void * you will get

void * myarr[20];
void * ( *ptr )[20] = &myarr;

answered Nov 30, 2014 at 9:48

Vlad from Moscow

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tohava · Accepted Answer · 2014-11-30 09:31:47Z

3

typedef void *myarr_t[20];
myarr_t *ptr_to_myarr;

answered Nov 30, 2014 at 9:31

tohava

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2 Comments

user2637293 Over a year ago

why you chooe typdef?

tohava Over a year ago

Because declaring a "pointer to an array of 20 void pointers" directly would result in complicated syntax. When I use the typedef, it means I get the type of myarr and put it as myarr_t, afterwards I can just say "give me a pointer to type of myarr".

D3Hunter · Accepted Answer · 2014-11-30 09:50:56Z

1

void* (*myarr_ptr)[20] = myarr_ptr;

here is my test code:

#include <stdio.h>

int main()

{
    int* myarr[20];
    int * (*myarr_ptr)[20] = myarr;
    printf("%p %p\n", myarr, *myarr_ptr);

    return 0;
}

$ ./a.out 
0x7fff8bd39dd0 0x7fff8bd39dd0

answered Nov 30, 2014 at 9:50

D3Hunter

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Pranit Kothari · Accepted Answer · 2014-11-30 09:32:48Z

-1

void** ptrToArrary = myarr_t; //pointing to base address of array.

answered Nov 30, 2014 at 9:32

Pranit Kothari

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6 Comments

user2637293 Over a year ago

why not void* prtToArray = myarr_t; ?

Pranit Kothari Over a year ago

To hold address of normal variable you need pointer, to hold address of pointer you need pointer to pointer.

user2637293 Over a year ago

So what i wrote to you (void* prtToArray = myarr_t;) it illegal?

Pranit Kothari Over a year ago

name of array represents pointer to element zero. arr[10] then arr is arr = &arr[0]

Pranit Kothari Over a year ago

@user3729959 It would have been legal if you write void prtToArray = myarr_t[0];

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