Using a void*

Can someone explain to me how to use a void* as a function argument that can accept any type of pointer, like realloc() and free()? I'm guessing you can't dereference it directly, since that's where my error is coming in.

Correct, you can't dereference a pointer to void, nor can you use pointer arithmetic on one because it lacks size information. You'll most likely want to convert the void pointer to a char pointer, and also pass the size of the actual type so that you can do block accesses. The only downside is that you have to do the tricky size calculations manually:

Code:

---

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

void rotate_at ( void *a, int n_elem, int n, int at )
{
  char *p = a;
  char *save = malloc ( n * at );
  int inc = n * at;

  if ( save == NULL )
    return;

  /* Swap blocks */
  memcpy ( save, p, inc );
  memmove ( p, p + inc, n * ( n_elem - at ) );
  memcpy ( p + inc, save, inc );

  free ( save );
}

int main ( void )
{
  int a[4] = {1,2,3,4};
  int i;

  for ( i = 0; i < 4; i++ )
    printf ( "%d ", a[i] );
  printf ( "\n" );

  rotate_at ( a, 4, sizeof a[0], 2 );

  for ( i = 0; i < 4; i++ )
    printf ( "%d ", a[i] );
  printf ( "\n" );

  return 0;
}

---

Well, I'm trying to write a library function for myself that sets each member of an array to an argued value. The only way I can think of to do this is to copy the actual binary data of the argued value into each array member. I notice the memcpy() function you used. Should this be the focus of my research? And how would I designate how many bytes to access at a time, if I also argue the size of each element?

>I'm trying to write a library function for myself that sets each member of an array to an argued value
Basically a generic version of memset? The principle is the same. Convert the void pointer to a char pointer, copy N bytes of the value to a, then increment a by N bytes until you get to the end of the array. The function is pretty trivial; there's no real funky size stuff to deal with, and memcpy basically removes any complexity that there might be:

Code:

---

#include <stdio.h>
#include <string.h>

/*
 * Fill p with n_elem copies of x
 */
void gset ( void *p, void *x, size_t n_elem, size_t n )
{
  char *a = p;

  while ( n_elem-- > 0 ) {
    memcpy ( a, x, n );
    a += n;
  }
}

int main ( void )
{
  int a[5] = {0};
  int init = 10;
  int i;

  for ( i = 0; i < 5; i++ )
    printf ( "%d ", a[i] );
  printf ( "\n" );

  gset ( a, &init, 5, sizeof init );

  for ( i = 0; i < 5; i++ )
    printf ( "%d ", a[i] );
  printf ( "\n" );

  return 0;
}

---

>I notice the memcpy() function you used. Should this be the focus of my research?
memcpy is for copying a block of bytes that don't overlap. If you're doing something like shifting part of an array into the other part, you need to use memmove, which is designed to handle overlapping memory.