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  <title>Bitmap Allocator</title>
  <meta content="Dhruv Matani" name="author">
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<h1 style="text-align: center;">Bitmap Allocator</h1>
<em><br>
<small><small>The latest version of this document is always available
at <a
 href="http://gcc.gnu.org/onlinedocs/libstdc++/ext/ballocator_doc.html">
http://gcc.gnu.org/onlinedocs/libstdc++/ext/ballocator_doc.html</a>.</small></small></em><br>
<br>
<em> To the <a href="http://gcc.gnu.org/libstdc++/">libstdc++-v3
homepage</a>.</em><br>
<br>
<hr style="width: 100%; height: 2px;"><br>
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.<br>
<br>
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.<br>
<br>
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.<br>
<br>
<hr style="width: 100%; height: 2px;"><br>
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.<br>
<br>
<hr style="width: 100%; height: 2px;">
<h3 style="text-align: center;">What is the Free List Store?</h3>
<br>
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.<br>
<br>
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.<br>
<br>
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.<br>
<br>
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.<br>
<br>
<ol>
  <li>Equal size check. Return true only when the 2 blocks are of equal
size.</li>
  <li>Difference Threshold: Return true only when the _block_size is
greater than or equal to the _required_size, and if the _BS is &gt; _RS
by a difference of less than some THRESHOLD value, then return true,
else return false. </li>
  <li>Percentage Threshold. Return true only when the _block_size is
greater than or equal to the _required_size, and if the _BS is &gt; _RS
by a percentage of less than some THRESHOLD value, then return true,
else return false.</li>
</ol>
<br>
Currently, (3) is being used with a value of 36% Maximum wastage per
Super Block.<br>
<br>
<hr style="width: 100%; height: 2px;"><span style="font-weight: bold;">1)
What is a super block? Why is it needed?</span><br>
<br>
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. <br>
<br>
<span style="font-weight: bold;">2) How does it interact with the free
list store?</span><br>
<br>
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.<br>
<br>
<hr style="width: 100%; height: 2px;">
<h3 style="text-align: center;">How does the allocate function Work?</h3>
<br>
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.<br>
<br>
Suppose n == 1, then the allocator does the following:<br>
<br>
<ol>
  <li>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.</li>
  <li>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.</li>
  <li>Is there any block in whatever region of memory that we own free?
This is done by checking <br>
    <div style="margin-left: 40px;">
    <ul>
      <li>The use count for each super block, and if that fails then </li>
      <li>The individual bit-maps for each super block. </li>
    </ul>
    </div>
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).</li>
  <li>This process involves Refilling the internal exponentially
growing memory pool. The said effect is achieved by calling
_S_refill_pool which does the following: <br>
    <div style="margin-left: 40px;">
    <ul>
      <li>Gets more memory from the Global Free List of the Required
size. </li>
      <li>Adjusts the size for the next call to itself. </li>
      <li>Writes the appropriate headers in the bit-maps.</li>
      <li>Sets the use count for that super-block just allocated to 0
(zero). </li>
      <li>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.</li>
    </ul>
    </div>
  </li>
</ol>
<br>
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.<br>
<br>
<br>
<hr style="width: 100%; height: 2px;">
<h3 style="text-align: center;">How does the deallocate function work?</h3>
<br>
The deallocate function again is specialized for single objects ONLY.
For all n belonging to &gt; 1, the operator delete is called without
further ado, and the deallocate function returns.<br>
<br>
However for n == 1, a series of steps are performed:<br>
<br>
<ol>
  <li>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.</li>
  <li>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.</li>
</ol>
<br>
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.<br>
<br>
<hr style="width: 100%; height: 2px;"><br>
<h3 style="text-align: center;">Data Layout for a Super Block:</h3>
<br>
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 &nbsp;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.<br>
<br>
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.<br>
<br>
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).<br>
<br>
<table cellpadding="0" cellspacing="0" border="1"
 style="text-align: left; width: 763px; height: 21px;">
  <tbody>
    <tr>
      <td style="vertical-align: top; text-align: center;">268<br>
      </td>
      <td style="vertical-align: top; text-align: center;">0<br>
      </td>
      <td style="vertical-align: top; text-align: center;">4294967295<br>
      </td>
      <td style="vertical-align: top; text-align: center;">4294967295<br>
      </td>
      <td style="vertical-align: top; text-align: center;">Data -&gt;
Space for 64 ints<br>
      </td>
    </tr>
  </tbody>
</table>
<br>
<br>
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.<br>
<br>
<span style="font-weight: bold;">What do the remaining values represent?</span><br>
<br>
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.<br>
<br>
The value 4294967295 represents the integer corresponding to the bit
representation of all bits set: 11111111111111111111111111111111.<br>
<br>
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).<br>
<br>
<hr style="width: 100%; height: 2px;"><br>
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. <br>
<br>
<ol>
  <li>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.</li>
  <li>And also this would preserve the cache as far as possible.</li>
</ol>
<br>
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.<br>
<br>
<hr style="width: 100%; height: 2px;"><br>
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.<br>
<br>
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.<br>
<br>
<hr style="width: 100%; height: 2px;">After reading all this, you may
still have a few questions about the internal working of this
allocator, like my friend had!<br>
<br>
Well here are the exact questions that he posed:<br>
<br>
<span style="font-weight: bold;">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?</span><br>
<br>
<div style="margin-left: 40px;"> 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.<br>
</div>
<br>
<span style="font-weight: bold;">Q2) 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?</span><br style="font-weight: bold;">
<br>
<div style="margin-left: 40px;"> Good question! 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.<br>
</div>
<br>
<span style="font-weight: bold;">Q3) How do the allocate and deallocate
functions work in regard to bitmaps?</span><br>
<br>
<div style="margin-left: 40px;"> 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.<br>
<br>
e.g.: Consider a bit-map of 64-bits as represented below:<br>
1111111111111111111111111111111111111111111111111111111111111111<br>
<br>
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.<br>
<br>
The bit-map now looks like this:<br>
1111111111111111111111111111111111111111111111111111111111111110<br>
</div>
<br>
<br>
<hr style="width: 100%; height: 2px;"><br>
(Tech-Stuff, Please stay out if you are not interested in the selection
of certain constants. 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).<br>
<br>
The formula for calculating the maximum wastage as a percentage:<br>
<br>
(32 x k + 1) / (2 x (32 x k + 1 + 32 x c)) x 100.<br>
<br>
Where,<br>
k =&gt; The constant overhead per node. eg. for list, it is 8 bytes,
and for map it is 12 bytes.<br>
c =&gt; 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.<br>
<br>
Plugging-in: For List: k = 8 and c = 4 (int and double), we get:<br>
33.376%<br>
<br>
For map/multimap: k = 12, and c = 4 (int and double), we get:<br>
37.524%<br>
<br>
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.<br>
<br>
<hr style="width: 100%; height: 2px;"><small><small><em> See <a
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for copying conditions. Comments and suggestions are welcome, and may
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