Typecast void pointer to int and string

Question

What I'm trying to do is store values in a array of Element struczs (that has a member void* data)

typedef struct {
void* data
} Element;

I have malloc'd the array to have 3 elements

Element* array = malloc(3 * sizeof(Element));

I know you can do it like this if data is a int but how do you do it if data is void *

array[0].data = 65;

I've tried typecasting

((*int)(array[0].data)) = 65;

but I get the message expected expression before 'int'

My second question is how would I go about storing a string in the Element struct (I can't make another member that is a string it has to be using the void pointer)

Can I use strcopy or sscanf ?

Sample main below in case I'm doing something wrong

#include "test.h"
int main (void)
{  
    Element* array = malloc(3 * sizeof(Element));
    ((*int)(array[0].data)) = 65;
    ((*int)(array[1].data)) = 64;
    ((*int)(array[2].data)) = 66;
}

EDIT

Sorry I should have added that the data will be added using a loop (the iterations is based on user input, so arrayOfElements could have 5,8,20 ect elements), based on user3386109 answer

Element* arrayOfElements = malloc(3 * sizeof(Element));
arrayOfElements[0].data = malloc( sizeof(int) );
int *ptr = arrayOfElements[0].data;
*ptr = 65;

work perfectly but if I want to make arrayOfElements[1].data = 67

Element* arrayOfElements = malloc(3 * sizeof(Element));
 arrayOfElements[1].data = malloc( sizeof(int) );
 int *ptr = arrayOfElements[1].data;
 *ptr = 76;

I can't reuse *ptr even if I do free(ptr) , is there anyway to do this if you don't know how many elements arrayOfElements will have?

Something is fundamentally wrong with your program design. If you use void* for any purpose, you must have a plan for how to know what types it points to, or your program will not doing anything useful. You store misc types through a void* then what? How will you know what each such pointer points at? — Lundin
Oh I do have a way to track what type it is, what is pictured is only a small segment. But I do understand your concerns — CrispyCashew

user3386109 user3386109 · Accepted Answer · 2016-10-20T07:44:06

The malloc in your question creates an array of three structs. Each struct has an uninitialized pointer called data. Before you use that pointer you must initialize it with another call to malloc.

To store an int:

array[0].data = malloc( sizeof(int) );
int *ptr = array[0].data;
*ptr = 65;

To store a string:

array[0].data = malloc( strlen(str) + 1 );
strcpy( array[0].data, str );

Here's a complete example that shows how to initialize the array, print the elements of the array, and then free the array. Note that ptr is just a temporary variable that's used to avoid the nasty syntax associated with casting and dereferencing a void *. (A good compiler will optimize out the ptr variable.)

int main( void )
{
    int *ptr;
    char *str = "hello";

    Element *array = malloc(3 * sizeof(Element) );

    array[0].data = malloc( sizeof(int) );
    ptr = array[0].data;
    *ptr = 65;

    array[1].data = malloc( strlen(str) +  1 );
    strcpy( array[1].data, str );

    array[2].data = malloc( sizeof(int) );
    ptr = array[2].data;
    *ptr = 76;

    ptr = array[0].data;
    printf( "%d\n", *ptr );
    str = array[1].data;
    printf( "%s\n", str );
    ptr = array[2].data;
    printf( "%d\n", *ptr );

    for ( int i = 0; i < 3; i++ )
        free( array[i].data );
    free( array );
}

Typecast void pointer to int and string

2 Answers