1 <chapter xmlns="http://docbook.org/ns/docbook" version="5.0"
2 xml:id="manual.ext.allocator.bitmap" xreflabel="bitmap_allocator">
3 <?dbhtml filename="bitmap_allocator.html"?>
5 <info><title>The bitmap_allocator</title>
7 <keyword>ISO C++</keyword>
8 <keyword>allocator</keyword>
17 <section xml:id="allocator.bitmap.design"><info><title>Design</title></info>
21 As this name suggests, this allocator uses a bit-map to keep track
22 of the used and unused memory locations for its book-keeping
26 This allocator will make use of 1 single bit to keep track of
27 whether it has been allocated or not. A bit 1 indicates free,
28 while 0 indicates allocated. This has been done so that you can
29 easily check a collection of bits for a free block. This kind of
30 Bitmapped strategy works best for single object allocations, and
31 with the STL type parameterized allocators, we do not need to
32 choose any size for the block which will be represented by a
33 single bit. This will be the size of the parameter around which
34 the allocator has been parameterized. Thus, close to optimal
35 performance will result. Hence, this should be used for node based
36 containers which call the allocate function with an argument of 1.
40 The bitmapped allocator's internal pool is exponentially growing.
41 Meaning that internally, the blocks acquired from the Free List
42 Store will double every time the bitmapped allocator runs out of
47 The macro <literal>__GTHREADS</literal> decides whether to use
48 Mutex Protection around every allocation/deallocation. The state
49 of the macro is picked up automatically from the gthr abstraction
55 <section xml:id="allocator.bitmap.impl"><info><title>Implementation</title></info>
56 <?dbhtml filename="bitmap_allocator_impl.html"?>
59 <section xml:id="bitmap.impl.free_list_store" xreflabel="Free List Store"><info><title>Free List Store</title></info>
63 The Free List Store (referred to as FLS for the remaining part of this
64 document) is the Global memory pool that is shared by all instances of
65 the bitmapped allocator instantiated for any type. This maintains a
66 sorted order of all free memory blocks given back to it by the
67 bitmapped allocator, and is also responsible for giving memory to the
68 bitmapped allocator when it asks for more.
71 Internally, there is a Free List threshold which indicates the
72 Maximum number of free lists that the FLS can hold internally
73 (cache). Currently, this value is set at 64. So, if there are
74 more than 64 free lists coming in, then some of them will be given
75 back to the OS using operator delete so that at any given time the
76 Free List's size does not exceed 64 entries. This is done because
77 a Binary Search is used to locate an entry in a free list when a
78 request for memory comes along. Thus, the run-time complexity of
79 the search would go up given an increasing size, for 64 entries
80 however, lg(64) == 6 comparisons are enough to locate the correct
81 free list if it exists.
84 Suppose the free list size has reached its threshold, then the
85 largest block from among those in the list and the new block will
86 be selected and given back to the OS. This is done because it
87 reduces external fragmentation, and allows the OS to use the
88 larger blocks later in an orderly fashion, possibly merging them
89 later. Also, on some systems, large blocks are obtained via calls
90 to mmap, so giving them back to free system resources becomes most
94 The function _S_should_i_give decides the policy that determines
95 whether the current block of memory should be given to the
96 allocator for the request that it has made. That's because we may
97 not always have exact fits for the memory size that the allocator
98 requests. We do this mainly to prevent external fragmentation at
99 the cost of a little internal fragmentation. Now, the value of
100 this internal fragmentation has to be decided by this function. I
101 can see 3 possibilities right now. Please add more as and when you
102 find better strategies.
106 <listitem><para>Equal size check. Return true only when the 2 blocks are of equal
107 size.</para></listitem>
108 <listitem><para>Difference Threshold: Return true only when the _block_size is
109 greater than or equal to the _required_size, and if the _BS is > _RS
110 by a difference of less than some THRESHOLD value, then return true,
111 else return false. </para></listitem>
112 <listitem><para>Percentage Threshold. Return true only when the _block_size is
113 greater than or equal to the _required_size, and if the _BS is > _RS
114 by a percentage of less than some THRESHOLD value, then return true,
115 else return false.</para></listitem>
119 Currently, (3) is being used with a value of 36% Maximum wastage per
124 <section xml:id="bitmap.impl.super_block" xreflabel="Super Block"><info><title>Super Block</title></info>
128 A super block is the block of memory acquired from the FLS from
129 which the bitmap allocator carves out memory for single objects
130 and satisfies the user's requests. These super blocks come in
131 sizes that are powers of 2 and multiples of 32
132 (_Bits_Per_Block). Yes both at the same time! That's because the
133 next super block acquired will be 2 times the previous one, and
134 also all super blocks have to be multiples of the _Bits_Per_Block
138 How does it interact with the free list store?
141 The super block is contained in the FLS, and the FLS is responsible for
142 getting / returning Super Bocks to and from the OS using operator new
143 as defined by the C++ standard.
147 <section xml:id="bitmap.impl.super_block_data" xreflabel="Super Block Data"><info><title>Super Block Data Layout</title></info>
150 Each Super Block will be of some size that is a multiple of the
151 number of Bits Per Block. Typically, this value is chosen as
152 Bits_Per_Byte x sizeof(size_t). On an x86 system, this gives the
153 figure 8 x 4 = 32. Thus, each Super Block will be of size 32
154 x Some_Value. This Some_Value is sizeof(value_type). For now, let
155 it be called 'K'. Thus, finally, Super Block size is 32 x K bytes.
158 This value of 32 has been chosen because each size_t has 32-bits
159 and Maximum use of these can be made with such a figure.
162 Consider a block of size 64 ints. In memory, it would look like this:
163 (assume a 32-bit system where, size_t is a 32-bit entity).
166 <table frame="all" xml:id="table.bitmap_alloc">
167 <title>Bitmap Allocator Memory Map</title>
169 <tgroup cols="5" align="left" colsep="1" rowsep="1">
170 <colspec colname="c1"/>
171 <colspec colname="c2"/>
172 <colspec colname="c3"/>
173 <colspec colname="c4"/>
174 <colspec colname="c5"/>
180 <entry>4294967295</entry>
181 <entry>4294967295</entry>
182 <entry>Data -> Space for 64 ints</entry>
189 The first Column(268) represents the size of the Block in bytes as
190 seen by the Bitmap Allocator. Internally, a global free list is
191 used to keep track of the free blocks used and given back by the
192 bitmap allocator. It is this Free List Store that is responsible
193 for writing and managing this information. Actually the number of
194 bytes allocated in this case would be: 4 + 4 + (4x2) + (64x4) =
195 272 bytes, but the first 4 bytes are an addition by the Free List
196 Store, so the Bitmap Allocator sees only 268 bytes. These first 4
197 bytes about which the bitmapped allocator is not aware hold the
202 What do the remaining values represent?</para>
204 The 2nd 4 in the expression is the sizeof(size_t) because the
205 Bitmapped Allocator maintains a used count for each Super Block,
206 which is initially set to 0 (as indicated in the diagram). This is
207 incremented every time a block is removed from this super block
208 (allocated), and decremented whenever it is given back. So, when
209 the used count falls to 0, the whole super block will be given
210 back to the Free List Store.
213 The value 4294967295 represents the integer corresponding to the bit
214 representation of all bits set: 11111111111111111111111111111111.
217 The 3rd 4x2 is size of the bitmap itself, which is the size of 32-bits
219 which is 8-bytes, or 2 x sizeof(size_t).
223 <section xml:id="bitmap.impl.max_wasted" xreflabel="Max Wasted Percentage"><info><title>Maximum Wasted Percentage</title></info>
227 This has nothing to do with the algorithm per-se,
228 only with some vales that must be chosen correctly to ensure that the
229 allocator performs well in a real word scenario, and maintains a good
230 balance between the memory consumption and the allocation/deallocation
234 The formula for calculating the maximum wastage as a percentage:
238 (32 x k + 1) / (2 x (32 x k + 1 + 32 x c)) x 100.
242 where k is the constant overhead per node (e.g., for list, it is
243 8 bytes, and for map it is 12 bytes) and c is the size of the
244 base type on which the map/list is instantiated. Thus, suppose the
245 type1 is int and type2 is double, they are related by the relation
246 sizeof(double) == 2*sizeof(int). Thus, all types must have this
247 double size relation for this formula to work properly.
250 Plugging-in: For List: k = 8 and c = 4 (int and double), we get:
255 For map/multimap: k = 12, and c = 4 (int and double), we get: 37.524%
258 Thus, knowing these values, and based on the sizeof(value_type), we may
259 create a function that returns the Max_Wastage_Percentage for us to use.
264 <section xml:id="bitmap.impl.allocate" xreflabel="Allocate"><info><title><function>allocate</function></title></info>
268 The allocate function is specialized for single object allocation
269 ONLY. Thus, ONLY if n == 1, will the bitmap_allocator's
270 specialized algorithm be used. Otherwise, the request is satisfied
271 directly by calling operator new.
274 Suppose n == 1, then the allocator does the following:
279 Checks to see whether a free block exists somewhere in a region
280 of memory close to the last satisfied request. If so, then that
281 block is marked as allocated in the bit map and given to the
282 user. If not, then (2) is executed.
287 Is there a free block anywhere after the current block right
288 up to the end of the memory that we have? If so, that block is
289 found, and the same procedure is applied as above, and
290 returned to the user. If not, then (3) is executed.
295 Is there any block in whatever region of memory that we own
296 free? This is done by checking
301 The use count for each super block, and if that fails then
306 The individual bit-maps for each super block.
312 Note: Here we are never touching any of the memory that the
313 user will be given, and we are confining all memory accesses
314 to a small region of memory! This helps reduce cache
315 misses. If this succeeds then we apply the same procedure on
316 that bit-map as (1), and return that block of memory to the
317 user. However, if this process fails, then we resort to (4).
322 This process involves Refilling the internal exponentially
323 growing memory pool. The said effect is achieved by calling
324 _S_refill_pool which does the following:
329 Gets more memory from the Global Free List of the Required
335 Adjusts the size for the next call to itself.
340 Writes the appropriate headers in the bit-maps.
345 Sets the use count for that super-block just allocated to 0
351 All of the above accounts to maintaining the basic invariant
352 for the allocator. If the invariant is maintained, we are
353 sure that all is well. Now, the same process is applied on
354 the newly acquired free blocks, which are dispatched
363 Thus, you can clearly see that the allocate function is nothing but a
364 combination of the next-fit and first-fit algorithm optimized ONLY for
365 single object allocations.
370 <section xml:id="bitmap.impl.deallocate" xreflabel="Deallocate"><info><title><function>deallocate</function></title></info>
373 The deallocate function again is specialized for single objects ONLY.
374 For all n belonging to > 1, the operator delete is called without
375 further ado, and the deallocate function returns.
378 However for n == 1, a series of steps are performed:
383 We first need to locate that super-block which holds the memory
384 location given to us by the user. For that purpose, we maintain
385 a static variable _S_last_dealloc_index, which holds the index
386 into the vector of block pairs which indicates the index of the
387 last super-block from which memory was freed. We use this
388 strategy in the hope that the user will deallocate memory in a
389 region close to what he/she deallocated the last time around. If
390 the check for belongs_to succeeds, then we determine the bit-map
391 for the given pointer, and locate the index into that bit-map,
392 and mark that bit as free by setting it.
395 If the _S_last_dealloc_index does not point to the memory block
396 that we're looking for, then we do a linear search on the block
397 stored in the vector of Block Pairs. This vector in code is
398 called _S_mem_blocks. When the corresponding super-block is
399 found, we apply the same procedure as we did for (1) to mark the
400 block as free in the bit-map.
405 Now, whenever a block is freed, the use count of that particular
406 super block goes down by 1. When this use count hits 0, we remove
407 that super block from the list of all valid super blocks stored in
408 the vector. While doing this, we also make sure that the basic
409 invariant is maintained by making sure that _S_last_request and
410 _S_last_dealloc_index point to valid locations within the vector.
414 <section xml:id="bitmap.impl.questions" xreflabel="Questions"><info><title>Questions</title></info>
417 <section xml:id="bitmap.impl.question.1" xreflabel="Question 1"><info><title>1</title></info>
420 Q1) The "Data Layout" section is
421 cryptic. I have no idea of what you are trying to say. Layout of what?
422 The free-list? Each bitmap? The Super Block?
425 The layout of a Super Block of a given
426 size. In the example, a super block of size 32 x 1 is taken. The
427 general formula for calculating the size of a super block is
428 32 x sizeof(value_type) x 2^n, where n ranges from 0 to 32 for 32-bit
433 <section xml:id="bitmap.impl.question.2" xreflabel="Question 2"><info><title>2</title></info>
436 And since I just mentioned the
437 term `each bitmap', what in the world is meant by it? What does each
438 bitmap manage? How does it relate to the super block? Is the Super
439 Block a bitmap as well?
442 Each bitmap is part of a Super Block which is made up of 3 parts
443 as I have mentioned earlier. Re-iterating, 1. The use count,
444 2. The bit-map for that Super Block. 3. The actual memory that
445 will be eventually given to the user. Each bitmap is a multiple
446 of 32 in size. If there are 32 x (2^3) blocks of single objects
447 to be given, there will be '32 x (2^3)' bits present. Each 32
448 bits managing the allocated / free status for 32 blocks. Since
449 each size_t contains 32-bits, one size_t can manage up to 32
450 blocks' status. Each bit-map is made up of a number of size_t,
451 whose exact number for a super-block of a given size I have just
456 <section xml:id="bitmap.impl.question.3" xreflabel="Question 3"><info><title>3</title></info>
459 How do the allocate and deallocate functions work in regard to
463 The allocate and deallocate functions manipulate the bitmaps and
464 have nothing to do with the memory that is given to the user. As
465 I have earlier mentioned, a 1 in the bitmap's bit field
466 indicates free, while a 0 indicates allocated. This lets us
467 check 32 bits at a time to check whether there is at lease one
468 free block in those 32 blocks by testing for equality with
469 (0). Now, the allocate function will given a memory block find
470 the corresponding bit in the bitmap, and will reset it (i.e.,
471 make it re-set (0)). And when the deallocate function is called,
472 it will again set that bit after locating it to indicate that
473 that particular block corresponding to this bit in the bit-map
474 is not being used by anyone, and may be used to satisfy future
478 e.g.: Consider a bit-map of 64-bits as represented below:
479 1111111111111111111111111111111111111111111111111111111111111111
483 Now, when the first request for allocation of a single object
484 comes along, the first block in address order is returned. And
485 since the bit-maps in the reverse order to that of the address
486 order, the last bit (LSB if the bit-map is considered as a
487 binary word of 64-bits) is re-set to 0.
491 The bit-map now looks like this:
492 1111111111111111111111111111111111111111111111111111111111111110
497 <section xml:id="bitmap.impl.locality" xreflabel="Locality"><info><title>Locality</title></info>
500 Another issue would be whether to keep the all bitmaps in a
501 separate area in memory, or to keep them near the actual blocks
502 that will be given out or allocated for the client. After some
503 testing, I've decided to keep these bitmaps close to the actual
504 blocks. This will help in 2 ways.
508 <listitem><para>Constant time access for the bitmap themselves, since no kind of
509 look up will be needed to find the correct bitmap list or its
510 equivalent.</para></listitem>
511 <listitem><para>And also this would preserve the cache as far as possible.</para></listitem>
515 So in effect, this kind of an allocator might prove beneficial from a
516 purely cache point of view. But this allocator has been made to try and
517 roll out the defects of the node_allocator, wherein the nodes get
518 skewed about in memory, if they are not returned in the exact reverse
519 order or in the same order in which they were allocated. Also, the
520 new_allocator's book keeping overhead is too much for small objects and
521 single object allocations, though it preserves the locality of blocks
522 very well when they are returned back to the allocator.
526 <section xml:id="bitmap.impl.grow_policy" xreflabel="Grow Policy"><info><title>Overhead and Grow Policy</title></info>
529 Expected overhead per block would be 1 bit in memory. Also, once
530 the address of the free list has been found, the cost for
531 allocation/deallocation would be negligible, and is supposed to be
532 constant time. For these very reasons, it is very important to
533 minimize the linear time costs, which include finding a free list
534 with a free block while allocating, and finding the corresponding
535 free list for a block while deallocating. Therefore, I have
536 decided that the growth of the internal pool for this allocator
537 will be exponential as compared to linear for
538 node_allocator. There, linear time works well, because we are
539 mainly concerned with speed of allocation/deallocation and memory
540 consumption, whereas here, the allocation/deallocation part does
541 have some linear/logarithmic complexity components in it. Thus, to
542 try and minimize them would be a good thing to do at the cost of a
543 little bit of memory.
547 Another thing to be noted is the pool size will double every time
548 the internal pool gets exhausted, and all the free blocks have
549 been given away. The initial size of the pool would be
550 sizeof(size_t) x 8 which is the number of bits in an integer,
551 which can fit exactly in a CPU register. Hence, the term given is
552 exponential growth of the internal pool.