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jeremybenn |
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xml:id="manual.ext.allocator.bitmap" xreflabel="bitmap_allocator">
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The bitmap_allocator
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ISO C++
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allocator
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As this name suggests, this allocator uses a bit-map to keep track
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of the used and unused memory locations for its book-keeping
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purposes.
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This allocator will make use of 1 single bit to keep track of
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whether it has been allocated or not. A bit 1 indicates free,
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while 0 indicates allocated. This has been done so that you can
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easily check a collection of bits for a free block. This kind of
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Bitmapped strategy works best for single object allocations, and
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with the STL type parameterized allocators, we do not need to
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choose any size for the block which will be represented by a
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single bit. This will be the size of the parameter around which
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the allocator has been parameterized. Thus, close to optimal
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performance will result. Hence, this should be used for node based
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containers which call the allocate function with an argument of 1.
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The bitmapped allocator's internal pool is exponentially growing.
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Meaning that internally, the blocks acquired from the Free List
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Store will double every time the bitmapped allocator runs out of
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memory.
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The macro __GTHREADS decides whether to use
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Mutex Protection around every allocation/deallocation. The state
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of the macro is picked up automatically from the gthr abstraction
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layer.
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The Free List Store (referred to as FLS for the remaining part of this
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document) is the Global memory pool that is shared by all instances of
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the bitmapped allocator instantiated for any type. This maintains a
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sorted order of all free memory blocks given back to it by the
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bitmapped allocator, and is also responsible for giving memory to the
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bitmapped allocator when it asks for more.
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Internally, there is a Free List threshold which indicates the
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Maximum number of free lists that the FLS can hold internally
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(cache). Currently, this value is set at 64. So, if there are
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more than 64 free lists coming in, then some of them will be given
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back to the OS using operator delete so that at any given time the
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Free List's size does not exceed 64 entries. This is done because
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a Binary Search is used to locate an entry in a free list when a
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request for memory comes along. Thus, the run-time complexity of
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the search would go up given an increasing size, for 64 entries
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however, lg(64) == 6 comparisons are enough to locate the correct
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free list if it exists.
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Suppose the free list size has reached its threshold, then the
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largest block from among those in the list and the new block will
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be selected and given back to the OS. This is done because it
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reduces external fragmentation, and allows the OS to use the
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larger blocks later in an orderly fashion, possibly merging them
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later. Also, on some systems, large blocks are obtained via calls
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to mmap, so giving them back to free system resources becomes most
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important.
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The function _S_should_i_give decides the policy that determines
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whether the current block of memory should be given to the
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allocator for the request that it has made. That's because we may
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not always have exact fits for the memory size that the allocator
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requests. We do this mainly to prevent external fragmentation at
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the cost of a little internal fragmentation. Now, the value of
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this internal fragmentation has to be decided by this function. I
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can see 3 possibilities right now. Please add more as and when you
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find better strategies.
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Equal size check. Return true only when the 2 blocks are of equal
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size.
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Difference Threshold: Return true only when the _block_size is
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greater than or equal to the _required_size, and if the _BS is > _RS
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by a difference of less than some THRESHOLD value, then return true,
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else return false.
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Percentage Threshold. Return true only when the _block_size is
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greater than or equal to the _required_size, and if the _BS is > _RS
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by a percentage of less than some THRESHOLD value, then return true,
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else return false.
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Currently, (3) is being used with a value of 36% Maximum wastage per
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Super Block.
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A super block is the block of memory acquired from the FLS from
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which the bitmap allocator carves out memory for single objects
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and satisfies the user's requests. These super blocks come in
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sizes that are powers of 2 and multiples of 32
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(_Bits_Per_Block). Yes both at the same time! That's because the
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next super block acquired will be 2 times the previous one, and
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also all super blocks have to be multiples of the _Bits_Per_Block
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value.
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How does it interact with the free list store?
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The super block is contained in the FLS, and the FLS is responsible for
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getting / returning Super Bocks to and from the OS using operator new
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as defined by the C++ standard.
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Each Super Block will be of some size that is a multiple of the
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number of Bits Per Block. Typically, this value is chosen as
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Bits_Per_Byte x sizeof(size_t). On an x86 system, this gives the
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figure 8 x 4 = 32. Thus, each Super Block will be of size 32
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x Some_Value. This Some_Value is sizeof(value_type). For now, let
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it be called 'K'. Thus, finally, Super Block size is 32 x K bytes.
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This value of 32 has been chosen because each size_t has 32-bits
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and Maximum use of these can be made with such a figure.
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Consider a block of size 64 ints. In memory, it would look like this:
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(assume a 32-bit system where, size_t is a 32-bit entity).
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Bitmap Allocator Memory Map
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268
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0
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4294967295
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4294967295
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Data -> Space for 64 ints
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The first Column(268) represents the size of the Block in bytes as
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seen by the Bitmap Allocator. Internally, a global free list is
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used to keep track of the free blocks used and given back by the
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bitmap allocator. It is this Free List Store that is responsible
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for writing and managing this information. Actually the number of
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bytes allocated in this case would be: 4 + 4 + (4x2) + (64x4) =
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272 bytes, but the first 4 bytes are an addition by the Free List
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Store, so the Bitmap Allocator sees only 268 bytes. These first 4
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bytes about which the bitmapped allocator is not aware hold the
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value 268.
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What do the remaining values represent?
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The 2nd 4 in the expression is the sizeof(size_t) because the
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Bitmapped Allocator maintains a used count for each Super Block,
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which is initially set to 0 (as indicated in the diagram). This is
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incremented every time a block is removed from this super block
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(allocated), and decremented whenever it is given back. So, when
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the used count falls to 0, the whole super block will be given
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back to the Free List Store.
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The value 4294967295 represents the integer corresponding to the bit
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representation of all bits set: 11111111111111111111111111111111.
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The 3rd 4x2 is size of the bitmap itself, which is the size of 32-bits
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x 2,
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which is 8-bytes, or 2 x sizeof(size_t).
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Maximum Wasted Percentage
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This has nothing to do with the algorithm per-se,
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only with some vales that must be chosen correctly to ensure that the
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allocator performs well in a real word scenario, and maintains a good
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balance between the memory consumption and the allocation/deallocation
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speed.
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The formula for calculating the maximum wastage as a percentage:
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(32 x k + 1) / (2 x (32 x k + 1 + 32 x c)) x 100.
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where k is the constant overhead per node (e.g., for list, it is
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8 bytes, and for map it is 12 bytes) and c is the size of the
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base type on which the map/list is instantiated. Thus, suppose the
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type1 is int and type2 is double, they are related by the relation
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sizeof(double) == 2*sizeof(int). Thus, all types must have this
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double size relation for this formula to work properly.
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Plugging-in: For List: k = 8 and c = 4 (int and double), we get:
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33.376%
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For map/multimap: k = 12, and c = 4 (int and double), we get: 37.524%
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Thus, knowing these values, and based on the sizeof(value_type), we may
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create a function that returns the Max_Wastage_Percentage for us to use.
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The allocate function is specialized for single object allocation
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ONLY. Thus, ONLY if n == 1, will the bitmap_allocator's
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specialized algorithm be used. Otherwise, the request is satisfied
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directly by calling operator new.
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Suppose n == 1, then the allocator does the following:
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Checks to see whether a free block exists somewhere in a region
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of memory close to the last satisfied request. If so, then that
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block is marked as allocated in the bit map and given to the
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user. If not, then (2) is executed.
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Is there a free block anywhere after the current block right
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up to the end of the memory that we have? If so, that block is
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found, and the same procedure is applied as above, and
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returned to the user. If not, then (3) is executed.
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Is there any block in whatever region of memory that we own
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free? This is done by checking
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The use count for each super block, and if that fails then
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The individual bit-maps for each super block.
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Note: Here we are never touching any of the memory that the
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user will be given, and we are confining all memory accesses
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to a small region of memory! This helps reduce cache
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misses. If this succeeds then we apply the same procedure on
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that bit-map as (1), and return that block of memory to the
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user. However, if this process fails, then we resort to (4).
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This process involves Refilling the internal exponentially
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growing memory pool. The said effect is achieved by calling
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_S_refill_pool which does the following:
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Gets more memory from the Global Free List of the Required
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size.
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Adjusts the size for the next call to itself.
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Writes the appropriate headers in the bit-maps.
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Sets the use count for that super-block just allocated to 0
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(zero).
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All of the above accounts to maintaining the basic invariant
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for the allocator. If the invariant is maintained, we are
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sure that all is well. Now, the same process is applied on
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the newly acquired free blocks, which are dispatched
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accordingly.
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Thus, you can clearly see that the allocate function is nothing but a
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combination of the next-fit and first-fit algorithm optimized ONLY for
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single object allocations.
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The deallocate function again is specialized for single objects ONLY.
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For all n belonging to > 1, the operator delete is called without
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further ado, and the deallocate function returns.
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However for n == 1, a series of steps are performed:
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We first need to locate that super-block which holds the memory
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location given to us by the user. For that purpose, we maintain
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a static variable _S_last_dealloc_index, which holds the index
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into the vector of block pairs which indicates the index of the
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last super-block from which memory was freed. We use this
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strategy in the hope that the user will deallocate memory in a
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region close to what he/she deallocated the last time around. If
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the check for belongs_to succeeds, then we determine the bit-map
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for the given pointer, and locate the index into that bit-map,
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and mark that bit as free by setting it.
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If the _S_last_dealloc_index does not point to the memory block
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that we're looking for, then we do a linear search on the block
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stored in the vector of Block Pairs. This vector in code is
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called _S_mem_blocks. When the corresponding super-block is
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found, we apply the same procedure as we did for (1) to mark the
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block as free in the bit-map.
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405 |
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406 |
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407 |
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408 |
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Now, whenever a block is freed, the use count of that particular
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409 |
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super block goes down by 1. When this use count hits 0, we remove
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410 |
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that super block from the list of all valid super blocks stored in
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411 |
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the vector. While doing this, we also make sure that the basic
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412 |
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invariant is maintained by making sure that _S_last_request and
|
413 |
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_S_last_dealloc_index point to valid locations within the vector.
|
414 |
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415 |
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416 |
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417 |
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418 |
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420 |
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421 |
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422 |
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423 |
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Q1) The "Data Layout" section is
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424 |
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cryptic. I have no idea of what you are trying to say. Layout of what?
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425 |
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The free-list? Each bitmap? The Super Block?
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426 |
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|
427 |
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|
428 |
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The layout of a Super Block of a given
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429 |
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size. In the example, a super block of size 32 x 1 is taken. The
|
430 |
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general formula for calculating the size of a super block is
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431 |
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32 x sizeof(value_type) x 2^n, where n ranges from 0 to 32 for 32-bit
|
432 |
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systems.
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433 |
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434 |
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435 |
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436 |
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437 |
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438 |
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439 |
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And since I just mentioned the
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440 |
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term `each bitmap', what in the world is meant by it? What does each
|
441 |
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bitmap manage? How does it relate to the super block? Is the Super
|
442 |
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Block a bitmap as well?
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443 |
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444 |
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|
445 |
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Each bitmap is part of a Super Block which is made up of 3 parts
|
446 |
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as I have mentioned earlier. Re-iterating, 1. The use count,
|
447 |
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2. The bit-map for that Super Block. 3. The actual memory that
|
448 |
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will be eventually given to the user. Each bitmap is a multiple
|
449 |
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of 32 in size. If there are 32 x (2^3) blocks of single objects
|
450 |
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to be given, there will be '32 x (2^3)' bits present. Each 32
|
451 |
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bits managing the allocated / free status for 32 blocks. Since
|
452 |
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each size_t contains 32-bits, one size_t can manage up to 32
|
453 |
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blocks' status. Each bit-map is made up of a number of size_t,
|
454 |
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whose exact number for a super-block of a given size I have just
|
455 |
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mentioned.
|
456 |
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|
457 |
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458 |
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|
459 |
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460 |
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|
461 |
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|
462 |
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How do the allocate and deallocate functions work in regard to
|
463 |
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bitmaps?
|
464 |
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|
465 |
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|
466 |
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The allocate and deallocate functions manipulate the bitmaps and
|
467 |
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have nothing to do with the memory that is given to the user. As
|
468 |
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I have earlier mentioned, a 1 in the bitmap's bit field
|
469 |
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indicates free, while a 0 indicates allocated. This lets us
|
470 |
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check 32 bits at a time to check whether there is at lease one
|
471 |
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free block in those 32 blocks by testing for equality with
|
472 |
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(0). Now, the allocate function will given a memory block find
|
473 |
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the corresponding bit in the bitmap, and will reset it (i.e.,
|
474 |
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make it re-set (0)). And when the deallocate function is called,
|
475 |
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it will again set that bit after locating it to indicate that
|
476 |
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that particular block corresponding to this bit in the bit-map
|
477 |
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is not being used by anyone, and may be used to satisfy future
|
478 |
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requests.
|
479 |
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|
480 |
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|
481 |
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e.g.: Consider a bit-map of 64-bits as represented below:
|
482 |
|
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1111111111111111111111111111111111111111111111111111111111111111
|
483 |
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|
484 |
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|
485 |
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|
486 |
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Now, when the first request for allocation of a single object
|
487 |
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comes along, the first block in address order is returned. And
|
488 |
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since the bit-maps in the reverse order to that of the address
|
489 |
|
|
order, the last bit (LSB if the bit-map is considered as a
|
490 |
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binary word of 64-bits) is re-set to 0.
|
491 |
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|
492 |
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|
493 |
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|
494 |
|
|
The bit-map now looks like this:
|
495 |
|
|
1111111111111111111111111111111111111111111111111111111111111110
|
496 |
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|
497 |
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|
498 |
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|
499 |
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|
500 |
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|
501 |
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|
502 |
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|
503 |
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Another issue would be whether to keep the all bitmaps in a
|
504 |
|
|
separate area in memory, or to keep them near the actual blocks
|
505 |
|
|
that will be given out or allocated for the client. After some
|
506 |
|
|
testing, I've decided to keep these bitmaps close to the actual
|
507 |
|
|
blocks. This will help in 2 ways.
|
508 |
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|
509 |
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|
510 |
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|
511 |
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|
Constant time access for the bitmap themselves, since no kind of
|
512 |
|
|
look up will be needed to find the correct bitmap list or its
|
513 |
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|
equivalent.
|
514 |
|
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And also this would preserve the cache as far as possible.
|
515 |
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|
516 |
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|
517 |
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|
518 |
|
|
So in effect, this kind of an allocator might prove beneficial from a
|
519 |
|
|
purely cache point of view. But this allocator has been made to try and
|
520 |
|
|
roll out the defects of the node_allocator, wherein the nodes get
|
521 |
|
|
skewed about in memory, if they are not returned in the exact reverse
|
522 |
|
|
order or in the same order in which they were allocated. Also, the
|
523 |
|
|
new_allocator's book keeping overhead is too much for small objects and
|
524 |
|
|
single object allocations, though it preserves the locality of blocks
|
525 |
|
|
very well when they are returned back to the allocator.
|
526 |
|
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|
527 |
|
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|
528 |
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|
529 |
|
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|
530 |
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|
531 |
|
|
|
532 |
|
|
Expected overhead per block would be 1 bit in memory. Also, once
|
533 |
|
|
the address of the free list has been found, the cost for
|
534 |
|
|
allocation/deallocation would be negligible, and is supposed to be
|
535 |
|
|
constant time. For these very reasons, it is very important to
|
536 |
|
|
minimize the linear time costs, which include finding a free list
|
537 |
|
|
with a free block while allocating, and finding the corresponding
|
538 |
|
|
free list for a block while deallocating. Therefore, I have
|
539 |
|
|
decided that the growth of the internal pool for this allocator
|
540 |
|
|
will be exponential as compared to linear for
|
541 |
|
|
node_allocator. There, linear time works well, because we are
|
542 |
|
|
mainly concerned with speed of allocation/deallocation and memory
|
543 |
|
|
consumption, whereas here, the allocation/deallocation part does
|
544 |
|
|
have some linear/logarithmic complexity components in it. Thus, to
|
545 |
|
|
try and minimize them would be a good thing to do at the cost of a
|
546 |
|
|
little bit of memory.
|
547 |
|
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|
548 |
|
|
|
549 |
|
|
|
550 |
|
|
Another thing to be noted is the pool size will double every time
|
551 |
|
|
the internal pool gets exhausted, and all the free blocks have
|
552 |
|
|
been given away. The initial size of the pool would be
|
553 |
|
|
sizeof(size_t) x 8 which is the number of bits in an integer,
|
554 |
|
|
which can fit exactly in a CPU register. Hence, the term given is
|
555 |
|
|
exponential growth of the internal pool.
|
556 |
|
|
|
557 |
|
|
|
558 |
|
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|
559 |
|
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|
560 |
|
|
|
561 |
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|
|