1 <sect1 id="manual.ext.allocator.bitmap" xreflabel="mt allocator">
2 <?dbhtml filename="bitmap_allocator.html"?>
15 <title>bitmap_allocator</title>
20 <sect2 id="allocator.bitmap.design" xreflabel="allocator.bitmap.design">
24 As this name suggests, this allocator uses a bit-map to keep track
25 of the used and unused memory locations for it's book-keeping
29 This allocator will make use of 1 single bit to keep track of
30 whether it has been allocated or not. A bit 1 indicates free,
31 while 0 indicates allocated. This has been done so that you can
32 easily check a collection of bits for a free block. This kind of
33 Bitmapped strategy works best for single object allocations, and
34 with the STL type parameterized allocators, we do not need to
35 choose any size for the block which will be represented by a
36 single bit. This will be the size of the parameter around which
37 the allocator has been parameterized. Thus, close to optimal
38 performance will result. Hence, this should be used for node based
39 containers which call the allocate function with an argument of 1.
43 The bitmapped allocator's internal pool is exponentially growing.
44 Meaning that internally, the blocks acquired from the Free List
45 Store will double every time the bitmapped allocator runs out of
50 The macro <literal>__GTHREADS</literal> decides whether to use
51 Mutex Protection around every allocation/deallocation. The state
52 of the macro is picked up automatically from the gthr abstraction
58 <sect2 id="allocator.bitmap.impl" xreflabel="allocator.bitmap.impl">
59 <title>Implementation</title>
61 <sect3 id="bitmap.impl.free_list_store" xreflabel="Free List Store">
62 <title>Free List Store</title>
65 The Free List Store (referred to as FLS for the remaining part of this
66 document) is the Global memory pool that is shared by all instances of
67 the bitmapped allocator instantiated for any type. This maintains a
68 sorted order of all free memory blocks given back to it by the
69 bitmapped allocator, and is also responsible for giving memory to the
70 bitmapped allocator when it asks for more.
73 Internally, there is a Free List threshold which indicates the
74 Maximum number of free lists that the FLS can hold internally
75 (cache). Currently, this value is set at 64. So, if there are
76 more than 64 free lists coming in, then some of them will be given
77 back to the OS using operator delete so that at any given time the
78 Free List's size does not exceed 64 entries. This is done because
79 a Binary Search is used to locate an entry in a free list when a
80 request for memory comes along. Thus, the run-time complexity of
81 the search would go up given an increasing size, for 64 entries
82 however, lg(64) == 6 comparisons are enough to locate the correct
83 free list if it exists.
86 Suppose the free list size has reached it's threshold, then the
87 largest block from among those in the list and the new block will
88 be selected and given back to the OS. This is done because it
89 reduces external fragmentation, and allows the OS to use the
90 larger blocks later in an orderly fashion, possibly merging them
91 later. Also, on some systems, large blocks are obtained via calls
92 to mmap, so giving them back to free system resources becomes most
96 The function _S_should_i_give decides the policy that determines
97 whether the current block of memory should be given to the
98 allocator for the request that it has made. That's because we may
99 not always have exact fits for the memory size that the allocator
100 requests. We do this mainly to prevent external fragmentation at
101 the cost of a little internal fragmentation. Now, the value of
102 this internal fragmentation has to be decided by this function. I
103 can see 3 possibilities right now. Please add more as and when you
104 find better strategies.
108 <listitem><para>Equal size check. Return true only when the 2 blocks are of equal
109 size.</para></listitem>
110 <listitem><para>Difference Threshold: Return true only when the _block_size is
111 greater than or equal to the _required_size, and if the _BS is > _RS
112 by a difference of less than some THRESHOLD value, then return true,
113 else return false. </para></listitem>
114 <listitem><para>Percentage Threshold. Return true only when the _block_size is
115 greater than or equal to the _required_size, and if the _BS is > _RS
116 by a percentage of less than some THRESHOLD value, then return true,
117 else return false.</para></listitem>
121 Currently, (3) is being used with a value of 36% Maximum wastage per
126 <sect3 id="bitmap.impl.super_block" xreflabel="Super Block">
127 <title>Super Block</title>
130 A super block is the block of memory acquired from the FLS from
131 which the bitmap allocator carves out memory for single objects
132 and satisfies the user's requests. These super blocks come in
133 sizes that are powers of 2 and multiples of 32
134 (_Bits_Per_Block). Yes both at the same time! That's because the
135 next super block acquired will be 2 times the previous one, and
136 also all super blocks have to be multiples of the _Bits_Per_Block
140 How does it interact with the free list store?
143 The super block is contained in the FLS, and the FLS is responsible for
144 getting / returning Super Bocks to and from the OS using operator new
145 as defined by the C++ standard.
149 <sect3 id="bitmap.impl.super_block_data" xreflabel="Super Block Data">
150 <title>Super Block Data Layout</title>
152 Each Super Block will be of some size that is a multiple of the
153 number of Bits Per Block. Typically, this value is chosen as
154 Bits_Per_Byte x sizeof(size_t). On an x86 system, this gives the
155 figure 8 x 4 = 32. Thus, each Super Block will be of size 32
156 x Some_Value. This Some_Value is sizeof(value_type). For now, let
157 it be called 'K'. Thus, finally, Super Block size is 32 x K bytes.
160 This value of 32 has been chosen because each size_t has 32-bits
161 and Maximum use of these can be made with such a figure.
164 Consider a block of size 64 ints. In memory, it would look like this:
165 (assume a 32-bit system where, size_t is a 32-bit entity).
169 <title>Bitmap Allocator Memory Map</title>
170 <tgroup cols='5' align='left' colsep='1' rowsep='1'>
171 <colspec colname='c1'></colspec>
172 <colspec colname='c2'></colspec>
173 <colspec colname='c3'></colspec>
174 <colspec colname='c4'></colspec>
175 <colspec colname='c5'></colspec>
181 <entry>4294967295</entry>
182 <entry>4294967295</entry>
183 <entry>Data -> Space for 64 ints</entry>
190 The first Column(268) represents the size of the Block in bytes as
191 seen by the Bitmap Allocator. Internally, a global free list is
192 used to keep track of the free blocks used and given back by the
193 bitmap allocator. It is this Free List Store that is responsible
194 for writing and managing this information. Actually the number of
195 bytes allocated in this case would be: 4 + 4 + (4x2) + (64x4) =
196 272 bytes, but the first 4 bytes are an addition by the Free List
197 Store, so the Bitmap Allocator sees only 268 bytes. These first 4
198 bytes about which the bitmapped allocator is not aware hold the
203 What do the remaining values represent?</para>
205 The 2nd 4 in the expression is the sizeof(size_t) because the
206 Bitmapped Allocator maintains a used count for each Super Block,
207 which is initially set to 0 (as indicated in the diagram). This is
208 incremented every time a block is removed from this super block
209 (allocated), and decremented whenever it is given back. So, when
210 the used count falls to 0, the whole super block will be given
211 back to the Free List Store.
214 The value 4294967295 represents the integer corresponding to the bit
215 representation of all bits set: 11111111111111111111111111111111.
218 The 3rd 4x2 is size of the bitmap itself, which is the size of 32-bits
220 which is 8-bytes, or 2 x sizeof(size_t).
224 <sect3 id="bitmap.impl.max_wasted" xreflabel="Max Wasted Percentage">
225 <title>Maximum Wasted Percentage</title>
228 This has nothing to do with the algorithm per-se,
229 only with some vales that must be chosen correctly to ensure that the
230 allocator performs well in a real word scenario, and maintains a good
231 balance between the memory consumption and the allocation/deallocation
235 The formula for calculating the maximum wastage as a percentage:
239 (32 x k + 1) / (2 x (32 x k + 1 + 32 x c)) x 100.
243 Where, k => The constant overhead per node. eg. for list, it is
244 8 bytes, and for map it is 12 bytes. c => The size of the
245 base type on which the map/list is instantiated. Thus, suppose the
246 type1 is int and type2 is double, they are related by the relation
247 sizeof(double) == 2*sizeof(int). Thus, all types must have this
248 double size relation for this formula to work properly.
251 Plugging-in: For List: k = 8 and c = 4 (int and double), we get:
256 For map/multimap: k = 12, and c = 4 (int and double), we get: 37.524%
259 Thus, knowing these values, and based on the sizeof(value_type), we may
260 create a function that returns the Max_Wastage_Percentage for us to use.
265 <sect3 id="bitmap.impl.allocate" xreflabel="Allocate">
266 <title><function>allocate</function></title>
269 The allocate function is specialized for single object allocation
270 ONLY. Thus, ONLY if n == 1, will the bitmap_allocator's
271 specialized algorithm be used. Otherwise, the request is satisfied
272 directly by calling operator new.
275 Suppose n == 1, then the allocator does the following:
280 Checks to see whether a free block exists somewhere in a region
281 of memory close to the last satisfied request. If so, then that
282 block is marked as allocated in the bit map and given to the
283 user. If not, then (2) is executed.
288 Is there a free block anywhere after the current block right
289 up to the end of the memory that we have? If so, that block is
290 found, and the same procedure is applied as above, and
291 returned to the user. If not, then (3) is executed.
296 Is there any block in whatever region of memory that we own
297 free? This is done by checking
302 The use count for each super block, and if that fails then
307 The individual bit-maps for each super block.
313 Note: Here we are never touching any of the memory that the
314 user will be given, and we are confining all memory accesses
315 to a small region of memory! This helps reduce cache
316 misses. If this succeeds then we apply the same procedure on
317 that bit-map as (1), and return that block of memory to the
318 user. However, if this process fails, then we resort to (4).
323 This process involves Refilling the internal exponentially
324 growing memory pool. The said effect is achieved by calling
325 _S_refill_pool which does the following:
330 Gets more memory from the Global Free List of the Required
336 Adjusts the size for the next call to itself.
341 Writes the appropriate headers in the bit-maps.
346 Sets the use count for that super-block just allocated to 0
352 All of the above accounts to maintaining the basic invariant
353 for the allocator. If the invariant is maintained, we are
354 sure that all is well. Now, the same process is applied on
355 the newly acquired free blocks, which are dispatched
364 Thus, you can clearly see that the allocate function is nothing but a
365 combination of the next-fit and first-fit algorithm optimized ONLY for
366 single object allocations.
371 <sect3 id="bitmap.impl.deallocate" xreflabel="Deallocate">
372 <title><function>deallocate</function></title>
374 The deallocate function again is specialized for single objects ONLY.
375 For all n belonging to > 1, the operator delete is called without
376 further ado, and the deallocate function returns.
379 However for n == 1, a series of steps are performed:
384 We first need to locate that super-block which holds the memory
385 location given to us by the user. For that purpose, we maintain
386 a static variable _S_last_dealloc_index, which holds the index
387 into the vector of block pairs which indicates the index of the
388 last super-block from which memory was freed. We use this
389 strategy in the hope that the user will deallocate memory in a
390 region close to what he/she deallocated the last time around. If
391 the check for belongs_to succeeds, then we determine the bit-map
392 for the given pointer, and locate the index into that bit-map,
393 and mark that bit as free by setting it.
396 If the _S_last_dealloc_index does not point to the memory block
397 that we're looking for, then we do a linear search on the block
398 stored in the vector of Block Pairs. This vector in code is
399 called _S_mem_blocks. When the corresponding super-block is
400 found, we apply the same procedure as we did for (1) to mark the
401 block as free in the bit-map.
406 Now, whenever a block is freed, the use count of that particular
407 super block goes down by 1. When this use count hits 0, we remove
408 that super block from the list of all valid super blocks stored in
409 the vector. While doing this, we also make sure that the basic
410 invariant is maintained by making sure that _S_last_request and
411 _S_last_dealloc_index point to valid locations within the vector.
415 <sect3 id="bitmap.impl.questions" xreflabel="Questions">
416 <title>Questions</title>
418 <sect4 id="bitmap.impl.question.1" xreflabel="Question 1">
421 Q1) The "Data Layout" section is
422 cryptic. I have no idea of what you are trying to say. Layout of what?
423 The free-list? Each bitmap? The Super Block?
426 The layout of a Super Block of a given
427 size. In the example, a super block of size 32 x 1 is taken. The
428 general formula for calculating the size of a super block is
429 32 x sizeof(value_type) x 2^n, where n ranges from 0 to 32 for 32-bit
434 <sect4 id="bitmap.impl.question.2" xreflabel="Question 2">
437 And since I just mentioned the
438 term `each bitmap', what in the world is meant by it? What does each
439 bitmap manage? How does it relate to the super block? Is the Super
440 Block a bitmap as well?
443 Each bitmap is part of a Super Block which is made up of 3 parts
444 as I have mentioned earlier. Re-iterating, 1. The use count,
445 2. The bit-map for that Super Block. 3. The actual memory that
446 will be eventually given to the user. Each bitmap is a multiple
447 of 32 in size. If there are 32 x (2^3) blocks of single objects
448 to be given, there will be '32 x (2^3)' bits present. Each 32
449 bits managing the allocated / free status for 32 blocks. Since
450 each size_t contains 32-bits, one size_t can manage up to 32
451 blocks' status. Each bit-map is made up of a number of size_t,
452 whose exact number for a super-block of a given size I have just
457 <sect4 id="bitmap.impl.question.3" xreflabel="Question 3">
460 How do the allocate and deallocate functions work in regard to
464 The allocate and deallocate functions manipulate the bitmaps and
465 have nothing to do with the memory that is given to the user. As
466 I have earlier mentioned, a 1 in the bitmap's bit field
467 indicates free, while a 0 indicates allocated. This lets us
468 check 32 bits at a time to check whether there is at lease one
469 free block in those 32 blocks by testing for equality with
470 (0). Now, the allocate function will given a memory block find
471 the corresponding bit in the bitmap, and will reset it (i.e.,
472 make it re-set (0)). And when the deallocate function is called,
473 it will again set that bit after locating it to indicate that
474 that particular block corresponding to this bit in the bit-map
475 is not being used by anyone, and may be used to satisfy future
479 e.g.: Consider a bit-map of 64-bits as represented below:
480 1111111111111111111111111111111111111111111111111111111111111111
484 Now, when the first request for allocation of a single object
485 comes along, the first block in address order is returned. And
486 since the bit-maps in the reverse order to that of the address
487 order, the last bit (LSB if the bit-map is considered as a
488 binary word of 64-bits) is re-set to 0.
492 The bit-map now looks like this:
493 1111111111111111111111111111111111111111111111111111111111111110
498 <sect3 id="bitmap.impl.locality" xreflabel="Locality">
499 <title>Locality</title>
501 Another issue would be whether to keep the all bitmaps in a
502 separate area in memory, or to keep them near the actual blocks
503 that will be given out or allocated for the client. After some
504 testing, I've decided to keep these bitmaps close to the actual
505 blocks. This will help in 2 ways.
509 <listitem><para>Constant time access for the bitmap themselves, since no kind of
510 look up will be needed to find the correct bitmap list or it's
511 equivalent.</para></listitem>
512 <listitem><para>And also this would preserve the cache as far as possible.</para></listitem>
516 So in effect, this kind of an allocator might prove beneficial from a
517 purely cache point of view. But this allocator has been made to try and
518 roll out the defects of the node_allocator, wherein the nodes get
519 skewed about in memory, if they are not returned in the exact reverse
520 order or in the same order in which they were allocated. Also, the
521 new_allocator's book keeping overhead is too much for small objects and
522 single object allocations, though it preserves the locality of blocks
523 very well when they are returned back to the allocator.
527 <sect3 id="bitmap.impl.grow_policy" xreflabel="Grow Policy">
528 <title>Overhead and Grow Policy</title>
530 Expected overhead per block would be 1 bit in memory. Also, once
531 the address of the free list has been found, the cost for
532 allocation/deallocation would be negligible, and is supposed to be
533 constant time. For these very reasons, it is very important to
534 minimize the linear time costs, which include finding a free list
535 with a free block while allocating, and finding the corresponding
536 free list for a block while deallocating. Therefore, I have
537 decided that the growth of the internal pool for this allocator
538 will be exponential as compared to linear for
539 node_allocator. There, linear time works well, because we are
540 mainly concerned with speed of allocation/deallocation and memory
541 consumption, whereas here, the allocation/deallocation part does
542 have some linear/logarithmic complexity components in it. Thus, to
543 try and minimize them would be a good thing to do at the cost of a
544 little bit of memory.
548 Another thing to be noted is the pool size will double every time
549 the internal pool gets exhausted, and all the free blocks have
550 been given away. The initial size of the pool would be
551 sizeof(size_t) x 8 which is the number of bits in an integer,
552 which can fit exactly in a CPU register. Hence, the term given is
553 exponential growth of the internal pool.
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