CooCox CoOS User's Guide

while allocate the memory. To all the memory blocks, whether they are free or
have been allocated, CooCox CoOS will link them through a doubly linked list.
Thus, when an allocated memory releases, there is no need to find the
insertion point of the memory from the beginning of the free list. You only need
to find the former free block in the doubly linked list and then insert the
memory into the free list, which has greatly improved the speed of memory
releasing.
In CooCox CoOS, since all the memory is 4-byte alignment (if the space that
needs to allocate is not 4-byte alignment, force it to be), the last two bits of the
previous and next memory address that saved in the head of the memory are all
invalid. Therefore, CooCox CoOS determines whether the memory is free
through the least significant bit: bit 0 refers to the free block, otherwise refers
to the allocated ones. If the memory list points to the free block, get the list
address directly. If it points to the allocated one, decrease it by one.
Code 3 The head of allocated memory block
typedef struct UsedMemBlk
{
void* nextMB;
void* preMB;
}UMB,*P_UMB;
Code 4 The head of free memory block
typedef struct FreeMemBlk
{
struct FreeMemBlk* nextFMB;
struct UsedMemBlk* nextUMB;
struct UsedMemBlk* preUMB;
}FMB,*P_FMB;
For memory block 1 and 2, the memory block itself preserves the address of
the previous free memory block when released, so it is very easy to plug back to
the free list. You only need to decide whether to merge them according to if the
memory block’s next block is free.
For memory block 3 and 4, its previous memory block is not a free memory
block when released. It is essential to get the previous free memory address
through two-way list when plug it back to the free list.
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