How do I convert from void * back to int

Question

if I have

int a= 5;
long b= 10;
int count0 = 2;
void ** args0;
args0 = (void **)malloc(count0 * sizeof(void *));
args0[0] = (void *)&a;
args0[1] = (void *)&b;

how can I convert from args[0] and args0[1] back to int and long? for example

int c=(something im missing)args0[0]
long d=(something im missing)args1[0]

Answer 1

Assuming that your &a0 and &b0 are supposed to be &a and &b, and that you mean args0[1] for setting up long d, you have stored a pointer to a in args0[0] and a pointer to b in args0[1]. This means you need to convert them to the correct pointer types.

int c = *((int *)args0[0]);
int d = *((long *)args0[1]);

Answer 2

Try this:

 int c =  *( (int *)  args0[0]);

 long d = *( (long *) args0[1]);

Answer 3

You need to tell it that the void* should be interpreted as an int* or long* when you dereference.

int a = 5;
long b = 10;
void *args[2];
args[0] = &a;
args[1] = &b;

int c = *(int*)args[0];
long d = *(long*)args[1];

Answer 4

To literally answer your question, you'd write

int c = *((int *)args0[0]);
long d *((long *)args[1]);

What might concern me about your code is that you have allocated space for the pointers to your locations, but you haven't allocated memory for the values themselves. If you expect to persist these locations beyond the local scope, you have to do something like:

int *al = malloc(sizeof(int));
long *bl = malloc(sizeof(long));
*al = a;
*bl = b;
void **args0 = malloc(2 * sizeof(void *));
args0[0] = al;
args0[1] = bl;

Answer 5

While others have answered your question, I will make a comment about the last three lines in the first part of your code snippet:

args0 = (void **)malloc(count0 * sizeof(void *));
args0[0] = (void *)&a;
args0[1] = (void *)&b;

The about is better written as:

args0 = malloc(count0 * sizeof *args0);
args0[0] = &a;
args0[1] = &b;

The malloc() call is easier to read this way, and less error-prone. You don't need a cast in the last two statements since C guarantees conversions to and from an object pointer and a void pointer.

Answer 6

sizeof(void *)

won't work. The compiler doesn't know the size. void is by definition not defined so the size is unknown.