size of a hex pattern in cpp - c++

I have a hex pattern stored in a variable, how to do I know what is the size of the hex pattern
E.g. --
#define MY_PATTERN 0xFFFF
now I want to know the size of MY_PATTERN, to use somewhere in my code.
sizeof (MY_PATTERN)
this is giving me warning -- "integer conversion resulted in truncation".
How can I fix this ? What is the way I should write it ?
The pattern can increase or decrease in size so I can't hard code it.

Don't do it.
There's no such thing in C++ as a "hex pattern". What you actually use is an integer literal. See paragraph "The type of the literal". Thus, sizeof (0xffff) is equal to sizeof(int). And the bad thing is: the exact size may vary.
From the design point of view, I can't really think of a situation where such a solution is acceptable. You're not even deriving a type from a literal value, which would be a suspicious as well, but at least, a typesafe solution. Sizes of values are mostly used in operations working with memory buffers directly, like memcpy() or fwrite(). Sizes defined in such indirect ways lead to a very brittle binary interface and maintenance difficulties. What if you compile a program on both x86 and Motorola 68000 machines and want them to interoperate via a network protocol, or want to write some files on the first machine, and read them on another? sizeof(int) is 4 for the first and 2 for the second. It will break.
Instead, explicitly use the exactly sized types, like int8_t, uint32_t, etc. They're defined in the <cstdint> header.

This will solve your problem:
#define MY_PATTERN 0xFFFF
struct TypeInfo
{
template<typename T>
static size_t SizeOfType(T) { return sizeof(T); }
};
void main()
{
size_t size_of_type = TypeInfo::SizeOfType(MY_PATTERN);
}
as pointed out by Nighthawk441 you can just do:
sizeof(MY_PATTERN);
Just make sure to use a size_t wherever you are getting a warning and that should solve your problem.

You could explicitly typedef various types to hold hex numbers with restricted sizes such that:
typedef unsigned char one_byte_hex;
typedef unsigned short two_byte_hex;
typedef unsigned int four_byte_hex;
one_byte_hex pattern = 0xFF;
two_byte_hex bigger_pattern = 0xFFFF;
four_byte_hex big_pattern = 0xFFFFFFFF;
//sizeof(pattern) == 1
//sizeof(bigger_pattern) == 2
//sizeof(biggest_pattern) == 4
four_byte_hex new_pattern = static_cast<four_byte_hex>(pattern);
//sizeof(new_pattern) == 4
It would be easier to just treat all hex numbers as unsigned ints regardless of pattern used though.
Alternatively, you could put together a function which checks how many times it can shift the bits of the pattern until it's 0.
size_t sizeof_pattern(unsigned int pattern)
{
size_t bits = 0;
size_t bytes = 0;
unsigned int tmp = pattern;
while(tmp >> 1 != 0){
bits++;
tmp = tmp >> 1;
}
bytes = (bits + 1) / 8; //add 1 to bits to shift range from 0-31 to 1-32 so we can divide properly. 8 bits per byte.
if((bits + 1) % 8 != 0){
bytes++; //requires one more byte to store value since we have remaining bits.
}
return bytes;
}

Related

Using Unions to convert an IP address into one 32-bit, two 16-bit, and four 8-bit values

I'm working on a school assignment. I am writing a program that utilizes unions to convert a given IP address in "192.168.1.10" format into its 32-bit single value, two 16-bit values, and four 8-bit values.
I'm having trouble with implementing my structs and unions appropriately, and am looking for insight on the subject. To my understanding, unions point to the same location as the referenced struct, but can look at specified pieces.
Any examples showing how a struct with four 8-bit values and a union can be used together would help. Also, any articles or books that might help me would also be appreciated.
Below is the assignment outline:
Create a program that manages an IP address. Allow the user to enter the IP address as four 8 bit unsigned integer values (just use 4 sequential CIN statements). The program should output the IP address upon the users request as any of the following. As a single 32 bit unsigned integer value, or as four 8 bit unsigned integer values, or as 32 individual bit values which can be requested as a single bit by the user (by entering an integer 0 to 31). Or as all 32 bits assigned into 2 variable sized groups (host group and network group) and outputted as 2 unsigned integer values from 1 bit to 31 bits each.
I was going to cin to int pt1,pt2,pt3,pt4 and assign them to the IP_Adress.pt1, .... etc.
struct IP_Adress {
unsigned int pt1 : 8;
unsigned int pt2 : 8;
unsigned int pt3 : 8;
unsigned int pt4 : 8;
};
I have not gotten anything to work appropriately yet. I think I am lacking a true understanding of the implementation of unions.
A union is not a good fit for this assignment. In fact, nothing in the text you quoted even says to use a union at all. And, a union will not help you with the parts of the assignment that deal with "32 individual bit values" or with "32 bits assigned into 2 variable sized groups". Those parts of the assignment will require bit shifting instead. Bit shifting is the better way to solve the other parts of the assignment, as well.
That being said, if you absolutely must use a union, you are probably looking for something more like this instead:
union IP_Adress {
uint8_t u8[4]; // four 8-bit values
uint16_t u16[2]; // two 16-bit values
uint32_t u32; // one 32-bit value
};
Except that C++ does not allow you to write into one union field and then read from another. C allows that kind of type puning, but it is undefined behavior in C++.
Why is type punning considered UB?
The asker already knows that doing this can blow up in their face a number of different ways, but here's a simple example for 4 byte, 4x1 byte, and 32x1 bit.
union bad_idea
{
uint32_t ints; // 32 bit unsigned integer
uint8_t bytes[sizeof(uint32_t)]; // 4 8 bit unsigned integers
};
and then
uint32_t get_int(const bad_idea & in)
{
return in.ints;
}
uint8_t get_byte(const bad_idea & in,
size_t offset)
{
if (offset >= sizeof(uint32_t)) // trap typos and idiots
{
throw std::runtime_error("invalid offset");
}
return in.bytes[offset];
}
bool get_bit(const bad_idea & in,
size_t offset)
{
if (offset >= sizeof(uint32_t)*8)
{
throw std::runtime_error("invalid offset");
}
return (in.ints >> offset) & 1; // shift the required bit to the end (in.ints >> offset)
// then mask off all of the other bits (& 1)
}
Things get a bit ugly getting input because you can't simply
std::cin >> bad.bytes[0];
because it reads a single character. Type in 127 for the first octet and you'll wind up filling bad.bytes[0] through bad.bytes[2] with '1', '2', and '7'.
You need to involve a temporary variable
int temp;
std::cin >> temp;
// tests for valid range in temp
bad.bytes[0] = temp
or risk some explosive batsmurf like
std::cin >> *(int*)&bad.bytes[0];
// tests for valid value in bad.bytes[0] impossible because aliasing has been broken
// and everything is already <expletive deleted>ed
pardon my C. The more respectable
std::cin >> *reinterpret_cast<int*>(&bad.bytes[0]);
isn't any better. As ugly as it is, use the temporary variable and bundle it up in a function to eliminate the duplication. Frankly this is a time when I'd probably fall back into C and pull out good ol' scanf.
The assignment doesn't say c++, you can just use typecasting instead of a union. I like to print the 32bit address out in hex also as it's easier to make sure you have right 32bit value;
#define word8 uint8_t
#define word16 uint16_t
#define word32 uint32_t
char *sIP = "192.168.0.11";
main(){
word32 ip, *pIP;
pIP = &ip;
inet_pton(AF_INET, sIP, pIP);
printf("32bit:%u %x\n", *pIP, *pIP);
printf("16bit:%u %u\n", *(word16*)pIP, *(((word16*)pIP)+1));
printf("8bit:%u %u %u %u\n", *(word8*)pIP, *(((word8*)pIP)+1),*(((word8*)pIP)+2),*(((word8*)pIP)+3));
}
Output:
32bit:184592576 b00a8c0
16bit:43200 2816
8bit:192 168 0 11
You could also store the IP as a 4 byte string and do math to get the 16 bit and 32 bit answers. Its a pretty dumb assignment IMO; I would never use a union to do it.

How to iterate over every bit of a type in C++

I wanted to write the Digital Search Tree in C++ using templates. To do that given a type T and data of type T I have to iterate over bits of this data. Doing this on integers is easy, one can just shift the number to the right an appropriate number of positions and "&" the number with 1, like it was described for example here How to get nth bit values . The problem starts when one tries to do get i'th bit from the templated data. I wrote something like this
#include <iostream>
template<typename T>
bool getIthBit (T data, unsigned int bit) {
return ((*(((char*)&data)+(bit>>3)))>>(bit&7))&1;
}
int main() {
uint32_t a = 16;
for (int i = 0; i < 32; i++) {
std::cout << getIthBit (a, i);
}
std::cout << std::endl;
}
Which works, but I am not exactly sure if it is not undefined behavior. The problem with this is that to iterate over all bits of the data, one has to know how many of them are, which is hard for struct data types because of padding. For example here
#include <iostream>
struct s {
uint32_t i;
char c;
};
int main() {
std::cout << sizeof (s) << std::endl;
}
The actual data has 5 bytes, but the output of the program says it has 8. I don't know how to get the actual size of the data, or if it is at all possible. A question about this was asked here How to check the size of struct w/o padding? , but the answers are just "don't".
It's easy to know know how many bits there are in a type. There's exactly CHAR_BIT * sizeof(T). sizeof(T) is the actual size of the type in bytes. But indeed, there isn't a general way within standard C++ to know which of those bits - that are part of the type - are padding.
I recommend not attempting to support types that have padding as keys of your DST.
Following trick might work for finding padding bits of trivially copyable classes:
Use std::memset to set all bits of the object to 0.
For each sub object with no sub objects of their own, set all bits to 1 using std::memset.
For each sub object with their own sub objects, perform the previous and this step recursively.
Check which bits stayed 0.
I'm not sure if there are any technical guarantees that the padding actually stays 0, so whether this works may be unspecified. Furthermore, there can be non-classes that have padding, and the described trick won't detect those. long double is typical example; I don't know if there are others. This probably won't detect unused bits of integers that underlie bitfields.
So, there are a lot of caveats, but it should work in your example case:
s sobj;
std::memset(&sobj, 0, sizeof sobj);
std::memset(&sobj.i, -1, sizeof sobj.i);
std::memset(&sobj.c, -1, sizeof sobj.c);
std::cout << "non-padding bits:\n";
unsigned long long ull;
std::memcpy(&ull, &sobj, sizeof sobj);
std::cout << std::bitset<sizeof sobj * CHAR_BIT>(ull) << std::endl;
There's a Standard way to know if a type has unique representation or not. It is std::has_unique_object_representations, available since C++17.
So if an object has unique representations, it is safe to assume that every bit is significant.
There's no standard way to know if non-unique representation caused by padding bytes/bits like in struct { long long a; char b; }, or by equivalent representations¹. And no standard way to know padding bits/bytes offsets.
Note that "actual size" concept may be misleading, as padding can be in the middle, like in struct { char a; long long b; }
Internally compiler has to distinguish padding bits from value bits to implement C++20 atomic<T>::compare_exchange_*. MSVC does this by zeroing padding bits with __builtin_zero_non_value_bits. Other compiler may use other name, another approach, or not expose atomic<T>::compare_exchange_* internals to this level.
¹ like multiple NaN floating point values

How can i use 6 bits to store value?

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 a 32-bit integer overflows, can we use a 40-bit structure instead of a 64-bit long one?

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.

C/C++ pointer type pointing to single bit?

I want to modify individual bits of data, (for e.g. ints or chars). I want to do this by making a pointer, say ptr. by assigning it to some int or char, and then after incrementing ptr n times, I want to access the nth bit of that data.
Something like
// If i want to change all the 8 bits in a char variable
char c="A";
T *ptr=&c; //T is the data type of pointer I want..
int index=0;
for(index;index<8;index++)
{
*ptr=1; //Something like assigning 1 to the bit pointed by ptr...
}
There no such thing as a bit pointer in C++. You need to use two things, a byte pointer and an offset to the bit. That seems to be what you are getting towards in your code. Here's how you do the individual bit operations.
// set a bit
*ptr |= 1 << index;
// clear a bit
*ptr &= ~(1 << index);
// test a bit
if (*ptr & (1 << index))
...
The smallest addressable memory unit in C and C++ is 1 byte. So You cannot have a pointer to anything less than a byte.If you want to perform bitwise operations C and C++ provide the bitwise operators for these operations.
It is impossible to have address of individual bit, but you can utilize structures with bit fields. Like in this example from Wikipedia so:
struct box_props
{
unsigned int opaque : 1;
unsigned int fill_color : 3;
unsigned int : 4; // fill to 8 bits
unsigned int show_border : 1;
unsigned int border_color : 3;
unsigned int border_style : 2;
unsigned int : 2; // fill to 16 bits
};
Then by manipulating individual fields you will change sets of bits inside unsigned int. Technically this is identical to bitwise operations, but in this case compiler will generate the code (and you have lower chances of bug).
Be advised that you have to be cautious using bit fields.
C and C++ doesn't have a "bit pointer", technically speaking, C and C++ as such, deosn't know about "bits". You could build your own type, to do this, you need two things: A pointer to some type (char, int - probably unsigned) and a bit number. You'd then use the pointer and the bit number, along with the bitwise operators, to actually access the values.
There is nothing like a pointer to a bit
If you want all bits set to 1 then c = 0xff; is what you want, if you want to set a bit under some condition:
for(index;index<8;index++)
{
if (condition) c |= 1 << index;
}
As you can see there is no need to use a pointer
You can not read a single bit from the memory, CPU always read a full cache line, which could have different sizes for different CPUs.
But from the language point of view you can use bit fields
http://publications.gbdirect.co.uk/c_book/chapter6/bitfields.html
http://en.wikipedia.org/wiki/Bit_field