Question

Today when I was reading others’ code, I saw something like void *func(void* i);, what does this void* mean here for the function name and for the variable type, respectively?

In addition, when do we need to use this kind of pointer and how to use it?

Answer 1

A pointer to void is a “generic” pointer type. A void * can be converted to any other pointer type without an explicit cast. You cannot dereference a void * or do pointer arithmetic with it; you must convert it to a pointer to a complete data type first.

void * is often used in places where you need to be able to work with different pointer types in the same code. One commonly cited example is the library function qsort:

void qsort(void *base, size_t nmemb, size_t size, 
           int (*compar)(const void *, const void *));

base is the address of an array, nmemb is the number of elements in the array, size is the size of each element, and compar is a pointer to a function that compares two elements of the array. It gets called like so:

int iArr[10];
double dArr[30];
long lArr[50];
...
qsort(iArr, sizeof iArr/sizeof iArr[0], sizeof iArr[0], compareInt);
qsort(dArr, sizeof dArr/sizeof dArr[0], sizeof dArr[0], compareDouble);
qsort(lArr, sizeof lArr/sizeof lArr[0], sizeof lArr[0], compareLong);

The array expressions iArr, dArr, and lArr are implicitly converted from array types to pointer types in the function call, and each is implicitly converted from “pointer to int/double/long” to “pointer to void“.

The comparison functions would look something like:

int compareInt(const void *lhs, const void *rhs)
{
  const int *x = lhs;  // convert void * to int * by assignment
  const int *y = rhs;

  if (*x > *y) return 1;
  if (*x == *y) return 0;
  return -1;
}

By accepting void *, qsort can work with arrays of any type.

The disadvantage of using void * is that you throw type safety out the window and into oncoming traffic. There’s nothing to protect you from using the wrong comparison routine:

qsort(dArr, sizeof dArr/sizeof dArr[0], sizeof dArr[0], compareInt);

compareInt is expecting its arguments to be pointing to ints, but is actually working with doubles. There’s no way to catch this problem at compile time; you’ll just wind up with a missorted array.