/*
* Remove the provided entry from whichever list it's currently in. This
* assumes that the entry is in a list. You don't need to provide the list
* because the lists are circular, so the list's pointers will automatically
* be updated if the first or last entries are removed.
*/
private static void List_remove(List_t *entry)
{
List_t *prev = entry->prev;
List_t *next = entry->next;
prev->next = next;
next->prev = prev;
}
/*
* Append the provided entry to the end of the list. This assumes the entry
* isn't in a list already because it overwrites the linked list pointers.
*/
private static void List_push(List_t *list, List_t *entry)
{
List_t *prev = list->prev;
entry->prev = prev;
entry->next = list;
prev->next = entry;
list->prev = entry;
}
/*
* Remove and return the first entry in the list or NULL if the list is empty.
*/
private static List_t *List_pop(List_t *list)
{
List_t *back = list->prev;
if (back == list)
{
return(null);
}
List_remove(back);
return(back);
}
/*
* This array represents a linearized binary tree of bits. Every possible
* allocation larger than MIN_ALLOC has a node in this tree (and therefore a
* bit in this array).
*
* Given the index for a node, linearized binary trees allow you to traverse to
* the parent node or the child nodes just by doing simple arithmetic on the
* index:
*
* - Move to parent: index = (index - 1) / 2;
* - Move to left child: index = index * 2 + 1;
* - Move to right child: index = index * 2 + 2;
* - Move to sibling: index = ((index - 1) ^ 1) + 1;
*
* Each node in this tree can be in one of several states:
*
* - UNUSED (both children are UNUSED)
* - SPLIT (one child is UNUSED and the other child isn't)
* - USED (neither children are UNUSED)
*
* These states take two bits to store. However, it turns out we have enough
* information to distinguish between UNUSED and USED from context, so we only
* need to store SPLIT or not, which only takes a single bit.
*
* Note that we don't need to store any nodes for allocations of size MIN_ALLOC
* since we only ever care about parent nodes.
*/
//static uint8_t node_is_split[(1 << (BUCKET_COUNT - 1)) / 8];
//protected byte* node_is_split;
/*
* This is the starting address of the address range for this allocator. Every
* returned allocation will be an offset of this pointer from 0 to MAX_ALLOC.
*/
//void* base_ptr;
/*
* This is the maximum address that has ever been used by the allocator. It's
* used to know when to call "brk" to request more memory from the kernel.
*/
//void* max_ptr;
/*
* Make sure all addresses before "new_value" are valid and can be used. Memory
* is allocated in a 2gb address range but that memory is not reserved up
* front. It's only reserved when it's needed by calling this function. This
* will return false if the memory could not be reserved.
*/
//bool update_max_ptr(void* new_value)
//{
// if (new_value > max_ptr)
// {
// if (brk(new_value))
// {
// return true;
// }
// max_ptr = new_value;
// }
// return false;
//}
/*
* Initialize a list to empty. Because these are circular lists, an "empty"
* list is an entry where both links point to itself. This makes insertion
* and removal simpler because they don't need any branches.
*/
private static void List_init(List_t *list)
{
list->prev = list;
list->next = list;
}
