I should be able to store a value in a data structure that could go from 0 to 3.. so I need 2 bits. This data structure I will be great 2 ^ 16 locations. So, i want to have 2 ^ 16 * 2 (bits). In C + + do you use to have exactly 2 bits in memory?
You need two bits per unit (not three), so you can pack four units into one byte, or 16 units into one 32-bit integer.
So you will need a std::array<uint32_t, 4096> to accomodate 216 units of 2-bit values.
You access the nth value as follows:
unsigned int get(std::size_t n, std::array<uint32_t, 4096> const & arr)
{
const uint32_t u = arr[n / 16];
return (u >> (2 * (n % 16))) & 0x3;
}
Alternatively, you could go with a bitfield:
struct BF32 {
uint32_t u0 : 2;
uint32_t u1 : 2;
//...
uint32_t uF : 2;
}
And then make an std::array<BF32, 4096>.
You cannot allocate a single object that is less than 1 byte (because 1 byte is the smallest addressable unit in the system).
You can, however, have portions of a structure that are smaller than a byte using bitfields. You could create one of these to hold 8 of your values, the size of this is exactly 3 bytes:
#pragma pack(1) // MSVC requires this
struct three_by_eight {
unsigned value1 : 3;
unsigned value2 : 3;
unsigned value3 : 3;
unsigned value4 : 3;
unsigned value5 : 3;
unsigned value6 : 3;
unsigned value7 : 3;
unsigned value8 : 3;
}
__attribute__ ((packed)) // GCC requires this
;
These can be clumsy to work with since they can't be accessed using [].... Your best be would be to create your own class that works similar to a bitset but works on 3 bits instead of 1.
If you are not working on an embedded system and resources are sufficient, you can have a look at std::bitset<> which will make your job as a programmer easier.
But if you are working on an embedded system, the bitset is probably not good for you (your compiler probably doesn't even support templates). There are a number of techniques for manipulating bits, each with its own quirks; here's an article that might help you:
> http://www.atmel.com/dyn/resources/prod_documents/avr_3_04.pdf
0 to 3 has 4 possible values. Since log2(4) == 2, or because 2^2 == 4, you need two bits, no three.
You might want to use bit fields
There was a discussion on the size allocated to bit-field structs last night. A struct cannot be smaller than a byte, and with most machines and compilers will be either 2 or 4, depending on the compiler and word size. So, no, you can't get a 3-bit struct (2-bit as you actually need). You can, however, pack bits yourself into an array of, say, uint64_ts. Or you could make a struct with 16 2-bit members and see if gcc makes that a 4-byte struct, then use an array of those.
There a very old trick to sneak a couple of bits around if you already have some data structures. This is quite nasty and unless you have extremely good reasons, it is most likely not at all a good idea. I'm just pointing this out in case you really really need to save a couple of bits.
Due to alignment, pointers on x86 or x64 are often multiples of 4, hence the two least significant bits of such pointers (e.g. pointers to int) are always 0. You can exploit this and sneak your two bits in there, but you have to make sure to remove them, when accessing those pointers (depending on the architecture, I'm not sure here).
Again, this is nasty, dangerous and pretty UB but perhaps it is worth it in your case.
3^5 = 243
and can fit 5 entries in 8bits. You spend like 20% less space storing lot of data this way. All you need is lookup table for 2 directional lookups and manipulations.
Related
My current project involves working with arrays of 5+ dimensions, but the individual elements of the array do not need to have 256 possible values. I was wondering if I could save on memory space by using a custom data type with, for example, only 4 or 6 bits to represent the value of an element, and if these memory savings would come at some significant performance cost.
Multidimensional arrays in C are really basically arrays of arrays.
(It can't be any other way as RAM is inherently linear).
You can emulate them on linear arrays in terms of pointer arithmetic:
#undef NDEBUG
#include <assert.h>
#include <stdint.h>
int main()
{
typedef uint32_t TYPE;
enum{A=3,B=4,C=5};
TYPE a[A][B][C];
assert((char*)&a[1][2][3] == ((char*)&a) + \
3*sizeof(TYPE) + 2 *C*sizeof(TYPE) + 1 *B*C*sizeof(TYPE));
}
Computers don't let you address sub-char types but it's not difficult to imagine a sub-char type.
The above char offset calculation for addressing a[1][2][3] could be rewriten like
char_ix = (3*sizeof(TYPE)*CHAR_BIT + 2 *C*sizeof(TYPE)*CHAR_BIT + 1 *B*C*sizeof(TYPE)*CHAR_BIT)/CHAR_BIT;
and if instead of chars (8-bits) you wanted to address e.g., 4-bits, you'd change it to
char_ix_of_4_bit =
(3*sizeof(TYPE)*CHAR_BIT/2 +
2 *C*sizeof(TYPE)*CHAR_BIT/2 +
1 *B*C*sizeof(TYPE)*CHAR_BIT/2) \
/ CHAR_BIT; //2 4-bits per octet
char_ix_of_4_bit_remainder =
(3*sizeof(TYPE)*CHAR_BIT/2 +
2 *C*sizeof(TYPE)*CHAR_BIT/2 +
1 *B*C*sizeof(TYPE)*CHAR_BIT/2) \
% CHAR_BIT; //2 4-bits per octet
The 4 bit value at the destination would then be
((unsigned char*)&a)[char_ix_of_4_bit] >> (4*char_ix_of_4_bit_remainder)
Similar for other bit groups.
In short, you can think of multidimensional bit arrays, reimagine them as linear bit arrays and then use regular indexing and bit shifting
to address the appropriate bit group or individual bits (IIRC, C++'s std::bitset/std::vector<bool> hide the last part under bit indexing
with an overloaded [] operator, but it's not hard to do it manually (which is what you'll need to do in pure C anyway, as pure C doesn't have operator overloading).
Bit ops are said to be slower and generate larger code than operations with whole types, but this might be well be offset by better cache locality, which using sub-char bit arrays might buy you depending on your data (you'd better have lots of data if you're attempting to do this).
My data unit (a network packet header) i am currently working on has 2 flags in its definition, stored in a byte field and accessed via bitwise operators. Unfortunately, i need only 2 bits and thinking what i can do with other 6 bits? Can i use them to store number?
Can i use them to store some internal state code (value range smaller than char?) and do not just waste them.
Is there any data types smaller than byte and how can i use them in C++? If no, should i waste those bits and left them without meaning?
You could use a bit field, as described here.
Adapted from that page:
#include <iostream>
struct S {
// 6-bit unsigned field,
// allowed values are 0...63
unsigned int b : 6;
};
int main()
{
S s = {7};
++s.b;
std::cout << s.b << '\n'; // output: 8
}
In C++, there is no datatype smaller than a char, which is - by definition - one byte. However, you do not need a dedicated datatype to access the bits of a value. Bitwise logic and Bitwise Shift operators are sufficient.
If you cannot live with sacrificing 6 bits (this is assuming 8-bit bytes) you might want to consider the std::vector<bool> specialization. Note, though, that there are a number of restrictions and differences to a regular std::vector.
Another option to make individual (consecutive) bits of a datatype accessible by name is to use bit fields:
struct S {
unsigned int flags : 2;
unsigned int state : 6;
};
static_assert( sizeof( S ) == 1, "Packing is implementation-defined." );
This declares a structure that can hold two pieces of information: flags and state, which occupy 2 and 6 bits, respectively. Adjacent bit fields are usually packed together (although this behavior is implementation-defined).
If, say, a 32-bit integer is overflowing, instead of upgrading int to long, can we make use of some 40-bit type if we need a range only within 240, so that we save 24 (64-40) bits for every integer?
If so, how?
I have to deal with billions and space is a bigger constraint.
Yes, but...
It is certainly possible, but it is usually nonsensical (for any program that doesn't use billions of these numbers):
#include <stdint.h> // don't want to rely on something like long long
struct bad_idea
{
uint64_t var : 40;
};
Here, var will indeed have a width of 40 bits at the expense of much less efficient code generated (it turns out that "much" is very much wrong -- the measured overhead is a mere 1-2%, see timings below), and usually to no avail. Unless you have need for another 24-bit value (or an 8 and 16 bit value) which you wish to pack into the same structure, alignment will forfeit anything that you may gain.
In any case, unless you have billions of these, the effective difference in memory consumption will not be noticeable (but the extra code needed to manage the bit field will be noticeable!).
Note:
The question has in the mean time been updated to reflect that indeed billions of numbers are needed, so this may be a viable thing to do, presumed that you take measures not to lose the gains due to structure alignment and padding, i.e. either by storing something else in the remaining 24 bits or by storing your 40-bit values in structures of 8 each or multiples thereof).
Saving three bytes a billion times is worthwhile as it will require noticeably fewer memory pages and thus cause fewer cache and TLB misses, and above all page faults (a single page fault weighting tens of millions instructions).
While the above snippet does not make use of the remaining 24 bits (it merely demonstrates the "use 40 bits" part), something akin to the following will be necessary to really make the approach useful in a sense of preserving memory -- presumed that you indeed have other "useful" data to put in the holes:
struct using_gaps
{
uint64_t var : 40;
uint64_t useful_uint16 : 16;
uint64_t char_or_bool : 8;
};
Structure size and alignment will be equal to a 64 bit integer, so nothing is wasted if you make e.g. an array of a billion such structures (even without using compiler-specific extensions). If you don't have use for an 8-bit value, you could also use an 48-bit and a 16-bit value (giving a bigger overflow margin).
Alternatively you could, at the expense of usability, put 8 40-bit values into a structure (least common multiple of 40 and 64 being 320 = 8*40). Of course then your code which accesses elements in the array of structures will become much more complicated (though one could probably implement an operator[] that restores the linear array functionality and hides the structure complexity).
Update:
Wrote a quick test suite, just to see what overhead the bitfields (and operator overloading with bitfield refs) would have. Posted code (due to length) at gcc.godbolt.org, test output from my Win7-64 machine is:
Running test for array size = 1048576
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 2 1 35 35 1
uint64_t 0 3 3 35 35 1
bad40_t 0 5 3 35 35 1
packed40_t 0 7 4 48 49 1
Running test for array size = 16777216
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 38 14 560 555 8
uint64_t 0 81 22 565 554 17
bad40_t 0 85 25 565 561 16
packed40_t 0 151 75 765 774 16
Running test for array size = 134217728
what alloc seq(w) seq(r) rand(w) rand(r) free
-----------------------------------------------------------
uint32_t 0 312 100 4480 4441 65
uint64_t 0 648 172 4482 4490 130
bad40_t 0 682 193 4573 4492 130
packed40_t 0 1164 552 6181 6176 130
What one can see is that the extra overhead of bitfields is neglegible, but the operator overloading with bitfield reference as a convenience thing is rather drastic (about 3x increase) when accessing data linearly in a cache-friendly manner. On the other hand, on random access it barely even matters.
These timings suggest that simply using 64-bit integers would be better since they are still faster overall than bitfields (despite touching more memory), but of course they do not take into account the cost of page faults with much bigger datasets. It might look very different once you run out of physical RAM (I didn't test that).
You can quite effectively pack 4*40bits integers into a 160-bit struct like this:
struct Val4 {
char hi[4];
unsigned int low[4];
}
long getLong( const Val4 &pack, int ix ) {
int hi= pack.hi[ix]; // preserve sign into 32 bit
return long( (((unsigned long)hi) << 32) + (unsigned long)pack.low[i]);
}
void setLong( Val4 &pack, int ix, long val ) {
pack.low[ix]= (unsigned)val;
pack.hi[ix]= (char)(val>>32);
}
These again can be used like this:
Val4[SIZE] vals;
long getLong( int ix ) {
return getLong( vals[ix>>2], ix&0x3 )
}
void setLong( int ix, long val ) {
setLong( vals[ix>>2], ix&0x3, val )
}
You might want to consider Variable-Lenght Encoding (VLE)
Presumably, you have store a lot of those numbers somewhere (in RAM, on disk, send them over the network, etc), and then take them one by one and do some processing.
One approach would be to encode them using VLE.
From Google's protobuf documentation (CreativeCommons licence)
Varints are a method of serializing integers using
one or more bytes. Smaller numbers take a smaller number of bytes.
Each byte in a varint, except the last byte, has the most significant
bit (msb) set – this indicates that there are further bytes to come.
The lower 7 bits of each byte are used to store the two's complement
representation of the number in groups of 7 bits, least significant
group first.
So, for example, here is the number 1 – it's a single byte, so the msb
is not set:
0000 0001
And here is 300 – this is a bit more complicated:
1010 1100 0000 0010
How do you figure out that this is 300? First you drop the msb from
each byte, as this is just there to tell us whether we've reached the
end of the number (as you can see, it's set in the first byte as there
is more than one byte in the varint)
Pros
If you have lots of small numbers, you'll probably use less than 40 bytes per integer, in average. Possibly much less.
You are able to store bigger numbers (with more than 40 bits) in the future, without having to pay a penalty for the small ones
Cons
You pay an extra bit for each 7 significant bits of your numbers. That means a number with 40 significant bits will need 6 bytes. If most of your numbers have 40 significant bits, you are better of with a bit field approach.
You will lose the ability to easily jump to a number given its index (you have to at least partially parse all previous elements in an array in order to access the current one.
You will need some form of decoding before doing anything useful with the numbers (although that is true for other approaches as well, like bit fields)
(Edit: First of all - what you want is possible, and makes sense in some cases; I have had to do similar things when I tried to do something for the Netflix challenge and only had 1GB of memory; Second - it is probably best to use a char array for the 40-bit storage to avoid any alignment issues and the need to mess with struct packing pragmas; Third - this design assumes that you're OK with 64-bit arithmetic for intermediate results, it is only for large array storage that you would use Int40; Fourth: I don't get all the suggestions that this is a bad idea, just read up on what people go through to pack mesh data structures and this looks like child's play by comparison).
What you want is a struct that is only used for storing data as 40-bit ints but implicitly converts to int64_t for arithmetic. The only trick is doing the sign extension from 40 to 64 bits right. If you're fine with unsigned ints, the code can be even simpler. This should be able to get you started.
#include <cstdint>
#include <iostream>
// Only intended for storage, automatically promotes to 64-bit for evaluation
struct Int40
{
Int40(int64_t x) { set(static_cast<uint64_t>(x)); } // implicit constructor
operator int64_t() const { return get(); } // implicit conversion to 64-bit
private:
void set(uint64_t x)
{
setb<0>(x); setb<1>(x); setb<2>(x); setb<3>(x); setb<4>(x);
};
int64_t get() const
{
return static_cast<int64_t>(getb<0>() | getb<1>() | getb<2>() | getb<3>() | getb<4>() | signx());
};
uint64_t signx() const
{
return (data[4] >> 7) * (uint64_t(((1 << 25) - 1)) << 39);
};
template <int idx> uint64_t getb() const
{
return static_cast<uint64_t>(data[idx]) << (8 * idx);
}
template <int idx> void setb(uint64_t x)
{
data[idx] = (x >> (8 * idx)) & 0xFF;
}
unsigned char data[5];
};
int main()
{
Int40 a = -1;
Int40 b = -2;
Int40 c = 1 << 16;
std::cout << "sizeof(Int40) = " << sizeof(Int40) << std::endl;
std::cout << a << "+" << b << "=" << (a+b) << std::endl;
std::cout << c << "*" << c << "=" << (c*c) << std::endl;
}
Here is the link to try it live: http://rextester.com/QWKQU25252
You can use a bit-field structure, but it's not going to save you any memory:
struct my_struct
{
unsigned long long a : 40;
unsigned long long b : 24;
};
You can squeeze any multiple of 8 such 40-bit variables into one structure:
struct bits_16_16_8
{
unsigned short x : 16;
unsigned short y : 16;
unsigned short z : 8;
};
struct bits_8_16_16
{
unsigned short x : 8;
unsigned short y : 16;
unsigned short z : 16;
};
struct my_struct
{
struct bits_16_16_8 a1;
struct bits_8_16_16 a2;
struct bits_16_16_8 a3;
struct bits_8_16_16 a4;
struct bits_16_16_8 a5;
struct bits_8_16_16 a6;
struct bits_16_16_8 a7;
struct bits_8_16_16 a8;
};
This will save you some memory (in comparison with using 8 "standard" 64-bit variables), but you will have to split every operation (and in particular arithmetic ones) on each of these variables into several operations.
So the memory-optimization will be "traded" for runtime-performance.
As the comments suggest, this is quite a task.
Probably an unnecessary hassle unless you want to save alot of RAM - then it makes much more sense. (RAM saving would be the sum total of bits saved across millions of long values stored in RAM)
I would consider using an array of 5 bytes/char (5 * 8 bits = 40 bits). Then you will need to shift bits from your (overflowed int - hence a long) value into the array of bytes to store them.
To use the values, then shift the bits back out into a long and you can use the value.
Then your RAM and file storage of the value will be 40 bits (5 bytes), BUT you must consider data alignment if you plan to use a struct to hold the 5 bytes. Let me know if you need elaboration on this bit shifting and data alignment implications.
Similarly, you could use the 64 bit long, and hide other values (3 chars perhaps) in the residual 24 bits that you do not want to use. Again - using bit shifting to add and remove the 24 bit values.
Another variation that may be helpful would be to use a structure:
typedef struct TRIPLE_40 {
uint32_t low[3];
uint8_t hi[3];
uint8_t padding;
};
Such a structure would take 16 bytes and, if 16-byte aligned, would fit entirely within a single cache line. While identifying which of the parts of the structure to use may be more expensive than it would be if the structure held four elements instead of three, accessing one cache line may be much cheaper than accessing two. If performance is important, one should use some benchmarks since some machines may perform a divmod-3 operation cheaply and have a high cost per cache-line fetch, while others might have have cheaper memory access and more expensive divmod-3.
If you have to deal with billions of integers, I'd try to encapuslate arrays of 40-bit numbers instead of single 40-bit numbers. That way, you can test different array implementations (e.g. an implementation that compresses data on the fly, or maybe one that stores less-used data to disk.) without changing the rest of your code.
Here's a sample implementation (http://rextester.com/SVITH57679):
class Int64Array
{
char* buffer;
public:
static const int BYTE_PER_ITEM = 5;
Int64Array(size_t s)
{
buffer=(char*)malloc(s*BYTE_PER_ITEM);
}
~Int64Array()
{
free(buffer);
}
class Item
{
char* dataPtr;
public:
Item(char* dataPtr) : dataPtr(dataPtr){}
inline operator int64_t()
{
int64_t value=0;
memcpy(&value, dataPtr, BYTE_PER_ITEM); // Assumes little endian byte order!
return value;
}
inline Item& operator = (int64_t value)
{
memcpy(dataPtr, &value, BYTE_PER_ITEM); // Assumes little endian byte order!
return *this;
}
};
inline Item operator[](size_t index)
{
return Item(buffer+index*BYTE_PER_ITEM);
}
};
Note: The memcpy-conversion from 40-bit to 64-bit is basically undefined behavior, as it assumes litte-endianness. It should work on x86-platforms, though.
Note 2: Obviously, this is proof-of-concept code, not production-ready code. To use it in real projects, you'd have to add (among other things):
error handling (malloc can fail!)
copy constructor (e.g. by copying data, add reference counting or by making the copy constructor private)
move constructor
const overloads
STL-compatible iterators
bounds checks for indices (in debug build)
range checks for values (in debug build)
asserts for the implicit assumptions (little-endianness)
As it is, Item has reference semantics, not value semantics, which is unusual for operator[]; You could probably work around that with some clever C++ type conversion tricks
All of those should be straightforward for a C++ programmer, but they would make the sample code much longer without making it clearer, so I've decided to omit them.
I'll assume that
this is C, and
you need a single, large array of 40 bit numbers, and
you are on a machine that is little-endian, and
your machine is smart enough to handle alignment
you have defined size to be the number of 40-bit numbers you need
unsigned char hugearray[5*size+3]; // +3 avoids overfetch of last element
__int64 get_huge(unsigned index)
{
__int64 t;
t = *(__int64 *)(&hugearray[index*5]);
if (t & 0x0000008000000000LL)
t |= 0xffffff0000000000LL;
else
t &= 0x000000ffffffffffLL;
return t;
}
void set_huge(unsigned index, __int64 value)
{
unsigned char *p = &hugearray[index*5];
*(long *)p = value;
p[4] = (value >> 32);
}
It may be faster to handle the get with two shifts.
__int64 get_huge(unsigned index)
{
return (((*(__int64 *)(&hugearray[index*5])) << 24) >> 24);
}
For the case of storing some billions of 40-bit signed integers, and assuming 8-bit bytes, you can pack 8 40-bit signed integers in a struct (in the code below using an array of bytes to do that), and, since this struct is ordinarily aligned, you can then create a logical array of such packed groups, and provide ordinary sequential indexing of that:
#include <limits.h> // CHAR_BIT
#include <stdint.h> // int64_t
#include <stdlib.h> // div, div_t, ptrdiff_t
#include <vector> // std::vector
#define STATIC_ASSERT( e ) static_assert( e, #e )
namespace cppx {
using Byte = unsigned char;
using Index = ptrdiff_t;
using Size = Index;
// For non-negative values:
auto roundup_div( const int64_t a, const int64_t b )
-> int64_t
{ return (a + b - 1)/b; }
} // namespace cppx
namespace int40 {
using cppx::Byte;
using cppx::Index;
using cppx::Size;
using cppx::roundup_div;
using std::vector;
STATIC_ASSERT( CHAR_BIT == 8 );
STATIC_ASSERT( sizeof( int64_t ) == 8 );
const int bits_per_value = 40;
const int bytes_per_value = bits_per_value/8;
struct Packed_values
{
enum{ n = sizeof( int64_t ) };
Byte bytes[n*bytes_per_value];
auto value( const int i ) const
-> int64_t
{
int64_t result = 0;
for( int j = bytes_per_value - 1; j >= 0; --j )
{
result = (result << 8) | bytes[i*bytes_per_value + j];
}
const int64_t first_negative = int64_t( 1 ) << (bits_per_value - 1);
if( result >= first_negative )
{
result = (int64_t( -1 ) << bits_per_value) | result;
}
return result;
}
void set_value( const int i, int64_t value )
{
for( int j = 0; j < bytes_per_value; ++j )
{
bytes[i*bytes_per_value + j] = value & 0xFF;
value >>= 8;
}
}
};
STATIC_ASSERT( sizeof( Packed_values ) == bytes_per_value*Packed_values::n );
class Packed_vector
{
private:
Size size_;
vector<Packed_values> data_;
public:
auto size() const -> Size { return size_; }
auto value( const Index i ) const
-> int64_t
{
const auto where = div( i, Packed_values::n );
return data_[where.quot].value( where.rem );
}
void set_value( const Index i, const int64_t value )
{
const auto where = div( i, Packed_values::n );
data_[where.quot].set_value( where.rem, value );
}
Packed_vector( const Size size )
: size_( size )
, data_( roundup_div( size, Packed_values::n ) )
{}
};
} // namespace int40
#include <iostream>
auto main() -> int
{
using namespace std;
cout << "Size of struct is " << sizeof( int40::Packed_values ) << endl;
int40::Packed_vector values( 25 );
for( int i = 0; i < values.size(); ++i )
{
values.set_value( i, i - 10 );
}
for( int i = 0; i < values.size(); ++i )
{
cout << values.value( i ) << " ";
}
cout << endl;
}
Yes, you can do that, and it will save some space for large quantities of numbers
You need a class that contains a std::vector of an unsigned integer type.
You will need member functions to store and to retrieve an integer. For example, if you want do store 64 integers of 40 bit each, use a vector of 40 integers of 64 bits each. Then you need a method that stores an integer with index in [0,64] and a method to retrieve such an integer.
These methods will execute some shift operations, and also some binary | and & .
I am not adding any more details here yet because your question is not very specific. Do you know how many integers you want to store? Do you know it during compile time? Do you know it when the program starts? How should the integers be organized? Like an array? Like a map? You should know all this before trying to squeeze the integers into less storage.
There are quite a few answers here covering implementation, so I'd like to talk about architecture.
We usually expand 32-bit values to 64-bit values to avoid overflowing because our architectures are designed to handle 64-bit values.
Most architectures are designed to work with integers whose size is a power of 2 because this makes the hardware vastly simpler. Tasks such as caching are much simpler this way: there are a large number of divisions and modulus operations which can be replaced with bit masking and shifts if you stick to powers of 2.
As an example of just how much this matters, The C++11 specification defines multithreading race-cases based on "memory locations." A memory location is defined in 1.7.3:
A memory location is either an object of scalar type or a maximal
sequence of adjacent bit-fields all having non-zero width.
In other words, if you use C++'s bitfields, you have to do all of your multithreading carefully. Two adjacent bitfields must be treated as the same memory location, even if you wish computations across them could be spread across multiple threads. This is very unusual for C++, so likely to cause developer frustration if you have to worry about it.
Most processors have a memory architecture which fetches 32-bit or 64-bit blocks of memory at a time. Thus use of 40-bit values will have a surprising number of extra memory accesses, dramatically affecting run-time. Consider the alignment issues:
40-bit word to access: 32-bit accesses 64bit-accesses
word 0: [0,40) 2 1
word 1: [40,80) 2 2
word 2: [80,120) 2 2
word 3: [120,160) 2 2
word 4: [160,200) 2 2
word 5: [200,240) 2 2
word 6: [240,280) 2 2
word 7: [280,320) 2 1
On a 64 bit architecture, one out of every 4 words will be "normal speed." The rest will require fetching twice as much data. If you get a lot of cache misses, this could destroy performance. Even if you get cache hits, you are going to have to unpack the data and repack it into a 64-bit register to use it (which might even involve a difficult to predict branch).
It is entirely possible this is worth the cost
There are situations where these penalties are acceptable. If you have a large amount of memory-resident data which is well indexed, you may find the memory savings worth the performance penalty. If you do a large amount of computation on each value, you may find the costs are minimal. If so, feel free to implement one of the above solutions. However, here are a few recommendations.
Do not use bitfields unless you are ready to pay their cost. For example, if you have an array of bitfields, and wish to divide it up for processing across multiple threads, you're stuck. By the rules of C++11, the bitfields all form one memory location, so may only be accessed by one thread at a time (this is because the method of packing the bitfields is implementation defined, so C++11 can't help you distribute them in a non-implementation defined manner)
Do not use a structure containing a 32-bit integer and a char to make 40 bytes. Most processors will enforce alignment and you wont save a single byte.
Do use homogenous data structures, such as an array of chars or array of 64-bit integers. It is far easier to get the alignment correct. (And you also retain control of the packing, which means you can divide an array up amongst several threads for computation if you are careful)
Do design separate solutions for 32-bit and 64-bit processors, if you have to support both platforms. Because you are doing something very low level and very ill-supported, you'll need to custom tailor each algorithm to its memory architecture.
Do remember that multiplication of 40-bit numbers is different from multiplication of 64-bit expansions of 40-bit numbers reduced back to 40-bits. Just like when dealing with the x87 FPU, you have to remember that marshalling your data between bit-sizes changes your result.
This begs for streaming in-memory lossless compression. If this is for a Big Data application, dense packing tricks are tactical solutions at best for what seems to require fairly decent middleware or system-level support. They'd need thorough testing to make sure one is able to recover all the bits unharmed. And the performance implications are highly non-trivial and very hardware-dependent because of interference with the CPU caching architecture (e.g. cache lines vs packing structure). Someone mentioned complex meshing structures : these are often fine-tuned to cooperate with particular caching architectures.
It's not clear from the requirements whether the OP needs random access. Given the size of the data it's more likely one would only need local random access on relatively small chunks, organised hierarchically for retrieval. Even the hardware does this at large memory sizes (NUMA). Like lossless movie formats show, it should be possible to get random access in chunks ('frames') without having to load the whole dataset into hot memory (from the compressed in-memory backing store).
I know of one fast database system (kdb from KX Systems to name one but I know there are others) that can handle extremely large datasets by seemlessly memory-mapping large datasets from backing store. It has the option to transparently compress and expand the data on-the-fly.
If what you really want is an array of 40 bit integers (which obviously you can't have), I'd just combine one array of 32 bit and one array of 8 bit integers.
To read a value x at index i:
uint64_t x = (((uint64_t) array8 [i]) << 32) + array32 [i];
To write a value x to index i:
array8 [i] = x >> 32; array32 [i] = x;
Obviously nicely encapsulated into a class using inline functions for maximum speed.
There is one situation where this is suboptimal, and that is when you do truly random access to many items, so that each access to an int array would be a cache miss - here you would get two cache misses every time. To avoid this, define a 32 byte struct containing an array of six uint32_t, an array of six uint8_t, and two unused bytes (41 2/3rd bits per number); the code to access an item is slightly more complicated, but both components of the item are in the same cache line.
Say, i have binary protocol, where first 4 bits represent a numeric value which can be less than or equal to 10 (ten in decimal).
In C++, the smallest data type available to me is char, which is 8 bits long. So, within my application, i can hold the value represented by 4 bits in a char variable. My question is, if i have to pack the char value back into 4 bits for network transmission, how do i pack my char's value back into 4 bits?
You do bitwise operation on the char;
so
unsigned char packedvalue = 0;
packedvalue |= 0xF0 & (7 <<4);
packedvalue |= 0x0F & (10);
Set the 4 upper most bit to 7 and the lower 4 bits to 10
Unpacking these again as
int upper, lower;
upper = (packedvalue & 0xF0) >>4;
lower = packedvalue & 0x0F;
As an extra answer to the question -- you may also want to look at protocol buffers for a way of encoding and decoding data for binary transfers.
Sure, just use one char for your value:
std::ofstream outfile("thefile.bin", std::ios::binary);
unsigned int n; // at most 10!
char c = n << 4; // fits
outfile.write(&c, 1); // we wrote the value "10"
The lower 4 bits will be left at zero. If they're also used for something, you'll have to populate c fully before writing it. To read:
infile.read(&c, 1);
unsigned int n = c >> 4;
Well, there's the popular but non-portable "Bit Fields". They're standard-compliant, but may create a different packing order on different platforms. So don't use them.
Then, there are the highly portable bit shifting and bitwise AND and OR operators, which you should prefer. Essentially, you work on a larger field (usually 32 bits, for TCP/IP protocols) and extract or replace subsequences of bits. See Martin's link and Soren's answer for those.
Are you familiar with C's bitfields? You simply write
struct my_bits {
unsigned v1 : 4;
...
};
Be warned, various operations are slower on bitfields because the compiler must unpack them for things like addition. I'd imagine unpacking a bitfield will still be faster than the addition operation itself, even though it requires multiple instructions, but it's still overhead. Bitwise operations should remain quite fast. Equality too.
You must also take care with endianness and threads (see the wikipedia article I linked for details, but the issues are kinda obvious). You should leearn about endianness anyways since you said "binary protocol" (see this previous questions)
I want to define my own datatype that can hold a single one of six possible values in order to learn more about memory management in c++. In numbers, I want to be able to hold 0 through 5. Binary, It would suffice with three bits (101=5), although some (6 and 7) wont be used. The datatype should also consume as little memory as possible.
Im not sure on how to accomplish this. First, I tried an enum with defined values for all the fields. As far as I know, the values are in hex there, so one "hexbit" should allow me to store 0 through 15. But comparing it to a char (with sizeof) it stated that its 4 times the size of a char, and a char holds 0 through 255 if Im not misstaken.
#include <iostream>
enum Foo
{
a = 0x0,
b = 0x1,
c = 0x2,
d = 0x3,
e = 0x4,
f = 0x5,
};
int main()
{
Foo myfoo = a;
char mychar = 'a';
std::cout << sizeof(myfoo); // prints 4
std::cout << sizeof(mychar); // prints 1
return 1;
}
Ive clearly misunderstood something, but fail to see what, so I turn to SO. :)
Also, when writing this post I realised that I clearly lack some parts of the vocabulary. Ive made this post a community wiki, please edit it so I can learn the correct words for everything.
A char is the smallest possible type.
If you happen to know that you need several such 3 bit values in a single place you get use a structure with bitfield syntax:
struct foo {
unsigned int val1:3;
unsigned int val2:3;
};
and hence get 2 of them within one byte. In theory you could pack 10 such fields into a 32-bit "int" value.
C++ 0x will contain Strongly typed enumerations where you can specify the underlying datatype (in your example char), but current C++ does not support this. The standard is not clear about the use of a char here (the examples are with int, short and long), but they mention the underlying integral type and that would include char as well.
As of today Neil Butterworth's answer to create a class for your problem seems the most elegant, as you can even extend it to contain a nested enumeration if you want symbolical names for the values.
C++ does not express units of memory smaller than bytes. If you're producing them one at a time, That's the best you can do. Your own example works well. If you need to get just a few, You can use bit-fields as Alnitak suggests. If you're planning on allocating them one at a time, then you're even worse off. Most archetectures allocate page-size units, 16 bytes being common.
Another choice might be to wrap std::bitset to do your bidding. This will waste very little space, if you need many such values, only about 1 bit for every 8.
If you think about your problem as a number, expressed in base-6, and convert that number to base two, possibly using an Unlimited precision integer (for example GMP), you won't waste any bits at all.
This assumes, of course, that you're values have a uniform, random distribution. If they follow a different distribution, You're best bet will be general compression of the first example, with something like gzip.
You can store values smaller than 8 or 32 bits. You just need to pack them into a struct (or class) and use bit fields.
For example:
struct example
{
unsigned int a : 3; //<Three bits, can be 0 through 7.
bool b : 1; //<One bit, the stores 0 or 1.
unsigned int c : 10; //<Ten bits, can be 0 through 1023.
unsigned int d : 19; //<19 bits, can be 0 through 524287.
}
In most cases, your compiler will round up the total size of your structure to 32 bits on a 32 bit platform. The other problem is, like you pointed out, that your values may not have a power of two range. This will make for wasted space. If you read the entire struct as one number, you will find values that will be impossible to set, if your input ranges aren't all powers of 2.
Another feature you may find interesting is a union. They work like a struct, but share memory. So if you write to one field it overwrites the others.
Now, if you are really tight for space, and you want to push each bit to the maximum, there is a simple encoding method. Let's say you want to store 3 numbers, each can be from 0 to 5. Bit fields are wasteful, because if you use 3 bits each, you'll waste some values (i.e. you could never set 6 or 7, even though you have room to store them). So, lets do an example:
//Here are three example values, each can be from 0 to 5:
const int one = 3, two = 4, three = 5;
To pack them together most efficiently, we should think in base 6 (since each value is from 0-5). So packed into the smallest possible space is:
//This packs all the values into one int, from 0 - 215.
//pack could be any value from 0 - 215. There are no 'wasted' numbers.
int pack = one + (6 * two) + (6 * 6 * three);
See how it looks like we're encoding in base six? Each number is multiplied by it's place like 6^n, where n is the place (starting at 0).
Then to decode:
const int one = pack % 6;
pack /= 6;
const int two = pack % 6;
pack /= 6;
const int three = pack;
Theses schemes are extremely handy when you have to encode some fields in a bar code or in an alpha numeric sequence for human typing. Just saying those few partial bits can make a huge difference. Also, the fields don't all have to have the same range. If one field is from 0 through 7, you'd use 8 instead of 6 in the proper place. There is no requirement that all fields have the same range.
Minimal size what you can use - 1 byte.
But if you use group of enum values ( writing in file or storing in container, ..), you can pack this group - 3 bits per value.
You don't have to enumerate the values of the enum:
enum Foo
{
a,
b,
c,
d,
e,
f,
};
Foo myfoo = a;
Here Foo is an alias of int, which on your machine takes 4 bytes.
The smallest type is char, which is defined as the smallest addressable data on the target machine. The CHAR_BIT macro yields the number of bits in a char and is defined in limits.h.
[Edit]
Note that generally speaking you shouldn't ask yourself such questions. Always use [unsigned] int if it's sufficient, except when you allocate quite a lot of memory (e.g. int[100*1024] vs char[100*1024], but consider using std::vector instead).
The size of an enumeration is defined to be the same of an int. But depending on your compiler, you may have the option of creating a smaller enum. For example, in GCC, you may declare:
enum Foo {
a, b, c, d, e, f
}
__attribute__((__packed__));
Now, sizeof(Foo) == 1.
The best solution is to create your own type implemented using a char. This should have sizeof(MyType) == 1, though this is not guaranteed.
#include <iostream>
using namespace std;
class MyType {
public:
MyType( int a ) : val( a ) {
if ( val < 0 || val > 6 ) {
throw( "bad value" );
}
}
int Value() const {
return val;
}
private:
char val;
};
int main() {
MyType v( 2 );
cout << sizeof(v) << endl;
cout << v.Value() << endl;
}
It is likely that packing oddly sized values into bitfields will incur a sizable performance penalty due to the architecture not supporting bit-level operations (thus requiring several processor instructions per operation). Before you implement such a type, ask yourself if it is really necessary to use as little space as possible, or if you are committing the cardinal sin of programming that is premature optimization. At most, I would encapsulate the value in a class whose backing store can be changed transparently if you really do need to squeeze every last byte for some reason.
You can use an unsigned char. Probably typedef it into an BYTE. It will occupy only one byte.