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14.8 Void Pointers

The peculiar type void *, a pointer whose target type is void, is used often in C. It represents a pointer to we-don’t-say-what. Thus,

void *numbered_slot_pointer (int);

declares a function numbered_slot_pointer that takes an integer parameter and returns a pointer, but we don’t say what type of data it points to.

The functions for dynamic memory allocation (see Dynamic Memory Allocation) use type void * to refer to blocks of memory, regardless of what sort of data the program stores in those blocks.

With type void *, you can pass the pointer around and test whether it is null. However, dereferencing it gives a void value that can’t be used (see The Void Type). To dereference the pointer, first convert it to some other pointer type.

Assignments convert void * automatically to any other pointer type, if the left operand has a pointer type; for instance,

  int *p;
  /* Converts return value to int *.  */
  p = numbered_slot_pointer (5);

Passing an argument of type void * for a parameter that has a pointer type also converts. For example, supposing the function hack is declared to require type float * for its parameter, this call to hack will convert the argument to that type.

/* Declare hack that way.
   We assume it is defined somewhere else.  */
void hack (float *);
/* Now call hack.  */
  /* Converts return value of numbered_slot_pointer
     to float * to pass it to hack.  */
  hack (numbered_slot_pointer (5));

You can also convert to another pointer type with an explicit cast (see Explicit Type Conversion), like this:

(int *) numbered_slot_pointer (5)

Here is an example which decides at run time which pointer type to convert to:

extract_int_or_double (void *ptr, bool its_an_int)
  if (its_an_int)
    handle_an_int (*(int *)ptr);
    handle_a_double (*(double *)ptr);

The expression *(int *)ptr means to convert ptr to type int *, then dereference it.

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