bitmap_allocator
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## bitmap_allocator

### Design

As this name suggests, this allocator uses a bit-map to keep track of the used and unused memory locations for it's book-keeping purposes.

This allocator will make use of 1 single bit to keep track of whether it has been allocated or not. A bit 1 indicates free, while 0 indicates allocated. This has been done so that you can easily check a collection of bits for a free block. This kind of Bitmapped strategy works best for single object allocations, and with the STL type parameterized allocators, we do not need to choose any size for the block which will be represented by a single bit. This will be the size of the parameter around which the allocator has been parameterized. Thus, close to optimal performance will result. Hence, this should be used for node based containers which call the allocate function with an argument of 1.

The bitmapped allocator's internal pool is exponentially growing. Meaning that internally, the blocks acquired from the Free List Store will double every time the bitmapped allocator runs out of memory.

The macro `__GTHREADS` decides whether to use Mutex Protection around every allocation/deallocation. The state of the macro is picked up automatically from the gthr abstraction layer.

### Implementation

#### Free List Store

The Free List Store (referred to as FLS for the remaining part of this document) is the Global memory pool that is shared by all instances of the bitmapped allocator instantiated for any type. This maintains a sorted order of all free memory blocks given back to it by the bitmapped allocator, and is also responsible for giving memory to the bitmapped allocator when it asks for more.

Internally, there is a Free List threshold which indicates the Maximum number of free lists that the FLS can hold internally (cache). Currently, this value is set at 64. So, if there are more than 64 free lists coming in, then some of them will be given back to the OS using operator delete so that at any given time the Free List's size does not exceed 64 entries. This is done because a Binary Search is used to locate an entry in a free list when a request for memory comes along. Thus, the run-time complexity of the search would go up given an increasing size, for 64 entries however, lg(64) == 6 comparisons are enough to locate the correct free list if it exists.

Suppose the free list size has reached it's threshold, then the largest block from among those in the list and the new block will be selected and given back to the OS. This is done because it reduces external fragmentation, and allows the OS to use the larger blocks later in an orderly fashion, possibly merging them later. Also, on some systems, large blocks are obtained via calls to mmap, so giving them back to free system resources becomes most important.

The function _S_should_i_give decides the policy that determines whether the current block of memory should be given to the allocator for the request that it has made. That's because we may not always have exact fits for the memory size that the allocator requests. We do this mainly to prevent external fragmentation at the cost of a little internal fragmentation. Now, the value of this internal fragmentation has to be decided by this function. I can see 3 possibilities right now. Please add more as and when you find better strategies.

1. Equal size check. Return true only when the 2 blocks are of equal size.

2. Difference Threshold: Return true only when the _block_size is greater than or equal to the _required_size, and if the _BS is > _RS by a difference of less than some THRESHOLD value, then return true, else return false.

3. Percentage Threshold. Return true only when the _block_size is greater than or equal to the _required_size, and if the _BS is > _RS by a percentage of less than some THRESHOLD value, then return true, else return false.

Currently, (3) is being used with a value of 36% Maximum wastage per Super Block.

#### Super Block

A super block is the block of memory acquired from the FLS from which the bitmap allocator carves out memory for single objects and satisfies the user's requests. These super blocks come in sizes that are powers of 2 and multiples of 32 (_Bits_Per_Block). Yes both at the same time! That's because the next super block acquired will be 2 times the previous one, and also all super blocks have to be multiples of the _Bits_Per_Block value.

How does it interact with the free list store?

The super block is contained in the FLS, and the FLS is responsible for getting / returning Super Bocks to and from the OS using operator new as defined by the C++ standard.

#### Super Block Data Layout

Each Super Block will be of some size that is a multiple of the number of Bits Per Block. Typically, this value is chosen as Bits_Per_Byte x sizeof(size_t). On an x86 system, this gives the figure 8 x 4 = 32. Thus, each Super Block will be of size 32 x Some_Value. This Some_Value is sizeof(value_type). For now, let it be called 'K'. Thus, finally, Super Block size is 32 x K bytes.

This value of 32 has been chosen because each size_t has 32-bits and Maximum use of these can be made with such a figure.

Consider a block of size 64 ints. In memory, it would look like this: (assume a 32-bit system where, size_t is a 32-bit entity).

Table 33.1. Bitmap Allocator Memory Map

268 | 0 | 4294967295 | 4294967295 | Data -> Space for 64 ints

The first Column(268) represents the size of the Block in bytes as seen by the Bitmap Allocator. Internally, a global free list is used to keep track of the free blocks used and given back by the bitmap allocator. It is this Free List Store that is responsible for writing and managing this information. Actually the number of bytes allocated in this case would be: 4 + 4 + (4x2) + (64x4) = 272 bytes, but the first 4 bytes are an addition by the Free List Store, so the Bitmap Allocator sees only 268 bytes. These first 4 bytes about which the bitmapped allocator is not aware hold the value 268.

What do the remaining values represent?

The 2nd 4 in the expression is the sizeof(size_t) because the Bitmapped Allocator maintains a used count for each Super Block, which is initially set to 0 (as indicated in the diagram). This is incremented every time a block is removed from this super block (allocated), and decremented whenever it is given back. So, when the used count falls to 0, the whole super block will be given back to the Free List Store.

The value 4294967295 represents the integer corresponding to the bit representation of all bits set: 11111111111111111111111111111111.

The 3rd 4x2 is size of the bitmap itself, which is the size of 32-bits x 2, which is 8-bytes, or 2 x sizeof(size_t).

#### Maximum Wasted Percentage

This has nothing to do with the algorithm per-se, only with some vales that must be chosen correctly to ensure that the allocator performs well in a real word scenario, and maintains a good balance between the memory consumption and the allocation/deallocation speed.

The formula for calculating the maximum wastage as a percentage:

(32 x k + 1) / (2 x (32 x k + 1 + 32 x c)) x 100.

where k is the constant overhead per node (e.g., for list, it is 8 bytes, and for map it is 12 bytes) and c is the size of the base type on which the map/list is instantiated. Thus, suppose the type1 is int and type2 is double, they are related by the relation sizeof(double) == 2*sizeof(int). Thus, all types must have this double size relation for this formula to work properly.

Plugging-in: For List: k = 8 and c = 4 (int and double), we get: 33.376%

For map/multimap: k = 12, and c = 4 (int and double), we get: 37.524%

Thus, knowing these values, and based on the sizeof(value_type), we may create a function that returns the Max_Wastage_Percentage for us to use.

#### `allocate`

The allocate function is specialized for single object allocation ONLY. Thus, ONLY if n == 1, will the bitmap_allocator's specialized algorithm be used. Otherwise, the request is satisfied directly by calling operator new.

Suppose n == 1, then the allocator does the following:

1. Checks to see whether a free block exists somewhere in a region of memory close to the last satisfied request. If so, then that block is marked as allocated in the bit map and given to the user. If not, then (2) is executed.

2. Is there a free block anywhere after the current block right up to the end of the memory that we have? If so, that block is found, and the same procedure is applied as above, and returned to the user. If not, then (3) is executed.

3. Is there any block in whatever region of memory that we own free? This is done by checking

   - The use count for each super block, and if that fails then

   - The individual bit-maps for each super block.

   Note: Here we are never touching any of the memory that the user will be given, and we are confining all memory accesses to a small region of memory! This helps reduce cache misses. If this succeeds then we apply the same procedure on that bit-map as (1), and return that block of memory to the user. However, if this process fails, then we resort to (4).

4. This process involves Refilling the internal exponentially growing memory pool. The said effect is achieved by calling _S_refill_pool which does the following:

   - Gets more memory from the Global Free List of the Required size.

   - Adjusts the size for the next call to itself.

   - Writes the appropriate headers in the bit-maps.

   - Sets the use count for that super-block just allocated to 0 (zero).

   - All of the above accounts to maintaining the basic invariant for the allocator. If the invariant is maintained, we are sure that all is well. Now, the same process is applied on the newly acquired free blocks, which are dispatched accordingly.

Thus, you can clearly see that the allocate function is nothing but a combination of the next-fit and first-fit algorithm optimized ONLY for single object allocations.

#### `deallocate`

The deallocate function again is specialized for single objects ONLY. For all n belonging to > 1, the operator delete is called without further ado, and the deallocate function returns.

However for n == 1, a series of steps are performed:

1. We first need to locate that super-block which holds the memory location given to us by the user. For that purpose, we maintain a static variable _S_last_dealloc_index, which holds the index into the vector of block pairs which indicates the index of the last super-block from which memory was freed. We use this strategy in the hope that the user will deallocate memory in a region close to what he/she deallocated the last time around. If the check for belongs_to succeeds, then we determine the bit-map for the given pointer, and locate the index into that bit-map, and mark that bit as free by setting it.

2. If the _S_last_dealloc_index does not point to the memory block that we're looking for, then we do a linear search on the block stored in the vector of Block Pairs. This vector in code is called _S_mem_blocks. When the corresponding super-block is found, we apply the same procedure as we did for (1) to mark the block as free in the bit-map.

Now, whenever a block is freed, the use count of that particular super block goes down by 1. When this use count hits 0, we remove that super block from the list of all valid super blocks stored in the vector. While doing this, we also make sure that the basic invariant is maintained by making sure that _S_last_request and _S_last_dealloc_index point to valid locations within the vector.

#### Questions

##### 1

Q1) The "Data Layout" section is cryptic. I have no idea of what you are trying to say. Layout of what? The free-list? Each bitmap? The Super Block?

The layout of a Super Block of a given size. In the example, a super block of size 32 x 1 is taken. The general formula for calculating the size of a super block is 32 x sizeof(value_type) x 2^n, where n ranges from 0 to 32 for 32-bit systems.

##### 2

And since I just mentioned the term `each bitmap', what in the world is meant by it? What does each bitmap manage? How does it relate to the super block? Is the Super Block a bitmap as well?

Each bitmap is part of a Super Block which is made up of 3 parts as I have mentioned earlier. Re-iterating, 1. The use count, 2. The bit-map for that Super Block. 3. The actual memory that will be eventually given to the user. Each bitmap is a multiple of 32 in size. If there are 32 x (2^3) blocks of single objects to be given, there will be '32 x (2^3)' bits present. Each 32 bits managing the allocated / free status for 32 blocks. Since each size_t contains 32-bits, one size_t can manage up to 32 blocks' status. Each bit-map is made up of a number of size_t, whose exact number for a super-block of a given size I have just mentioned.

##### 3

How do the allocate and deallocate functions work in regard to bitmaps?

The allocate and deallocate functions manipulate the bitmaps and have nothing to do with the memory that is given to the user. As I have earlier mentioned, a 1 in the bitmap's bit field indicates free, while a 0 indicates allocated. This lets us check 32 bits at a time to check whether there is at lease one free block in those 32 blocks by testing for equality with (0). Now, the allocate function will given a memory block find the corresponding bit in the bitmap, and will reset it (i.e., make it re-set (0)). And when the deallocate function is called, it will again set that bit after locating it to indicate that that particular block corresponding to this bit in the bit-map is not being used by anyone, and may be used to satisfy future requests.

e.g.: Consider a bit-map of 64-bits as represented below: 1111111111111111111111111111111111111111111111111111111111111111

Now, when the first request for allocation of a single object comes along, the first block in address order is returned. And since the bit-maps in the reverse order to that of the address order, the last bit (LSB if the bit-map is considered as a binary word of 64-bits) is re-set to 0.

The bit-map now looks like this: 1111111111111111111111111111111111111111111111111111111111111110

#### Locality

Another issue would be whether to keep the all bitmaps in a separate area in memory, or to keep them near the actual blocks that will be given out or allocated for the client. After some testing, I've decided to keep these bitmaps close to the actual blocks. This will help in 2 ways.

1. Constant time access for the bitmap themselves, since no kind of look up will be needed to find the correct bitmap list or it's equivalent.

2. And also this would preserve the cache as far as possible.

So in effect, this kind of an allocator might prove beneficial from a purely cache point of view. But this allocator has been made to try and roll out the defects of the node_allocator, wherein the nodes get skewed about in memory, if they are not returned in the exact reverse order or in the same order in which they were allocated. Also, the new_allocator's book keeping overhead is too much for small objects and single object allocations, though it preserves the locality of blocks very well when they are returned back to the allocator.

#### Overhead and Grow Policy

Expected overhead per block would be 1 bit in memory. Also, once the address of the free list has been found, the cost for allocation/deallocation would be negligible, and is supposed to be constant time. For these very reasons, it is very important to minimize the linear time costs, which include finding a free list with a free block while allocating, and finding the corresponding free list for a block while deallocating. Therefore, I have decided that the growth of the internal pool for this allocator will be exponential as compared to linear for node_allocator. There, linear time works well, because we are mainly concerned with speed of allocation/deallocation and memory consumption, whereas here, the allocation/deallocation part does have some linear/logarithmic complexity components in it. Thus, to try and minimize them would be a good thing to do at the cost of a little bit of memory.

Another thing to be noted is the pool size will double every time the internal pool gets exhausted, and all the free blocks have been given away. The initial size of the pool would be sizeof(size_t) x 8 which is the number of bits in an integer, which can fit exactly in a CPU register. Hence, the term given is exponential growth of the internal pool.

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