Malloc and Void Pointers

Question

I'm studying this malloc function and I could use some help:

static void *malloc(int size)
  {
        void *p;

        if (size < 0)
                 error("Malloc error");
        if (!malloc_ptr)
                 malloc_ptr = free_mem_ptr;

        malloc_ptr = (malloc_ptr + 3) & ~3;     /* Align */

        p = (void *)malloc_ptr;
        malloc_ptr += size;

        if (free_mem_end_ptr && malloc_ptr >= free_mem_end_ptr)
                 error("Out of memory");

        malloc_count++;
        return p;
 }

I know that the malloc func allocates memory space for any type, if there is enough memory, but the lines i don't understand are:

p = (void *)malloc_ptr;
malloc_ptr += size;

How can it point to any data type like that? I just can't understand that void pointer or its location.

NOTE: malloc_ptr is an unsigned long

Answer 1

The reason it returns a void pointer is because it has no idea what you are allocating space for in the malloc call. All it knows is the amount of space you requested. It is up to you or your compiler to decide what will fill the memory. The void pointer's location is typically implemented as a linked list to maintain integrity and know what values of memory are free which is surprisingly kept track of in the free function.

Answer 2

This is the implementation of malloc, so it is allowed to do things that would not be legitimate in a regular program. Specifically, it is making use of the implementation-defined conversion from unsigned long to void *. Program initialization sets malloc_ptr to the numeric address of a large block of unallocated memory. Then, when you ask for an allocation, malloc makes a pointer out of the current value of malloc_ptr and increases malloc_ptr by the number of bytes you asked for. That way, the next time you call malloc it will return a new pointer.

This is about the simplest possible implementation of malloc. Most notably, it appears not to ever reuse freed memory.