Assign values to a void array? - c

Question

Here's my code :
#include <stdio.h>
#include <stdlib.h>
void *create_array(int size) {
double *v;
v = (double*)malloc(size * sizeof(double));
v[0] = (double)size;
return (void*)v;
}
int main() {
void *a;
double b = 5;
a = create_array(10);
a[1] = b;
return 0;
}
I'm trying to assign the value stored in b to (*a[1]). I get the following error :
"main.c:19:9: error: incomplete type 'void' is not assignable
a[1]=b;"
~~~~^

Answer 1

Main Answer
void is an incomplete type and cannot be used in expressions, including assignments. You can, however, have pointers to this type, void *. void * is a type that means “this points to something, but we are not saying what at this point.”
In order to use a void *, you must say what it points to, by converting it to a different kind of pointer, such as double *, which says “this points to a double.”
C, unlike C++, will do this conversion automatically in assignments. If we declare a with double *a; and then we assign it with a = create_array(10);, then the void * that is returned from create_array is automatically converted to a double * for storing in a. After that, we can write a[1] = b; to store b in the double at a[1].
Supplement
Your create_array contains a mix of double * and void *. It is better to either write a routine that is designed for a specific type:
double *create_array(int NumberOfElements)
{
double *v = malloc(NumberOfElements * sizeof *v);
v[0] = NumberOfElements; // (Yuck, this is ugly.)
return v;
}
or to write a routine that handles any type:
void *create_array(int NumberOfElements, size_t SizeOfElement)
{
void *v = malloc(NumberOfElements * SizeOfElement);
// Cannot store size easily since we do not know what element type is.
return v;
}
As noted, that business about storing the number of elements in the first element is messy. It is better to handle the size separately, even if that requires writing more code.
Also, avoid calling something just a “size,” because this can be confusing: Is the size of an array the number of elements in it or the number of bytes? The name or accompanying comments should always make the unit of measurement clear.

Answer 2

neither pointers nor variables can be void type, just function's return
change to:
#include <stdio.h>
#include <stdlib.h>
double *create_array(int size){
double *v;
v=(double*)malloc(size*sizeof(double));
v[0]=(double)size;
return (double*)v;
}
int main() {
double *a;
double b;
b=5;
a=create_array(10);
a[1]=b;
return 0;
}

Answer 3

You can't dereference a void *, but you can use a cast:
#include <stdio.h>
#include <stdlib.h>
void *
create_array(int size)
{
double *v = malloc(size * sizeof *v);
if( v == NULL ){
perror("malloc");
exit(EXIT_FAILURE);
}
v[0] = (double)size;
return v;
}
int
main(void)
{
void *a;
double b = 5.0;
double *c;
c = a = create_array(10);
((double *)a)[1] = b;
printf("%f %f\n", c[0], c[1]);
return 0;
}