bitmap_allocator

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.

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.

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

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).


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).

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

    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:

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.

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:

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.