Bitwise shifting in C++ - c++

Trying to hide data within a PPM Image using C++:
void PPMObject::hideData(string phrase)
{
phrase += '\0';
size_t size = phrase.size() * 8;
bitset<8> binary_phrase (phrase.c_str()[0]);
//We need 8 channels for each letter
for (size_t index = 0; index < size; index += 3)
{
//convert red channel to bits
bitset<8> r (this->m_Ptr[index]);
if (r.at(7) != binary_phrase.at(index))
{
r.flip(7);
}
this->m_Ptr[index] = (char) r.to_ulong();
//convert blue channel to bits and find LSB
bitset<8> g (this->m_Ptr[index+1]);
if (g.at(7) != binary_phrase.at(index+1))
{
g.flip(7);
}
this->m_Ptr[index+1] = (char) g.to_ulong();
//convert green channel to bits and find LSB
bitset<8> b (this->m_Ptr[index+2]);
if (b.at(7) != binary_phrase.at(index+2))
{
b.flip(7);
}
this->m_Ptr[index+2] = (char) b.to_ulong();
}
//this->m_Ptr[index+1] = (r.to_ulong() & 0xFF);
}
Then extracting the data by reversing the above process:
string PPMObject::recoverData()
{
size_t size = this->width * this->height * 3;
string message("");
//We need 8 channels for each letter
for (size_t index = 0; index < size; index += 3)
{
//retreive our hidden data from the LSB in red channel
bitset<8> r (this->m_Ptr[index]);
message += r.to_string()[7];
//retreive our hidden data from the LSB in green channel
bitset<8> g (this->m_Ptr[index+1]);
message += g.to_string()[7];
//retreive our hidden data from the LSB in blue channel
bitset<8> b (this->m_Ptr[index+2]);
message += b.to_string()[7];
}
return message;
}
The above hide data function converts each channel (RGB) to binary. It then attempts to find the least significant bit and flips it if it does not match the nth bit of the phrase (starting at zero). It then assigns that new converted binary string back into the pointer as a casted char.
Is using the bitset library a "best practice" technique? I am all ears to a more straightforward, efficient technique. Perhaps, using bitwise maniuplations?
There are no logic errors or problems whatsoever with reading and writing the PPM Image. The pixel data is assigned to a char pointer: this->m_Ptr (above).

Here's some more compact code that does bit manipulation. It doesn't bounds check m_Ptr, but neither does your code.
#include <iostream>
#include <string>
using namespace std;
struct PPMObject
{
void hideData(const string &phrase);
string recoverData(size_t size);
char m_Ptr[256];
};
void PPMObject::hideData(const string &phrase)
{
size_t size = phrase.size();
for (size_t p_index = 0, i_index = 0; p_index < size; ++p_index)
for (int i = 0, bits = phrase[p_index]; i < 8; ++i, bits >>= 1, ++i_index)
{
m_Ptr[i_index] &= 0xFE; // set lsb to 0
m_Ptr[i_index] |= (bits & 0x1); // set lsb to lsb of bits
}
}
string PPMObject::recoverData(size_t size)
{
string ret(size, ' ');
for (size_t p_index = 0, i_index = 0; p_index < size; ++p_index)
{
int i, bits;
for (i = 0, bits = 0; i < 8; ++i, ++i_index)
bits |= ((m_Ptr[i_index] & 0x1) << i);
ret[p_index] = (char) bits;
}
return ret;
}
int main()
{
PPMObject p;
p.hideData("Hello World!");
cout << p.recoverData(12) << endl;
return 0;
}
Note that this code encodes from lsb to msb of each byte of the phrase.

Related

Remove nth bit from buffer, and shift the rest

Giving a uint8_t buffer of x length, I am trying to come up with a function or a macro that can remove nth bit (or n to n+i), then left-shift the remaining bits.
example #1:
for input 0b76543210 0b76543210 ... then output should be 0b76543217 0b654321 ...
example #2: if the input is:
uint8_t input[8] = {
0b00110011,
0b00110011,
...
};
the output without the first bit, should be
uint8_t output[8] = {
0b00110010,
0b01100100,
...
};
I have tried the following to remove the first bit, but it did not work for the second group of bits.
/* A macro to extract (a-b) range of bits without shifting */
#define BIT_RANGE(N,x,y) ((N) & ((0xff >> (7 - (y) + (x))) << ((x))))
void removeBit0(uint8_t *n) {
for (int i=0; i < 7; i++) {
n[i] = (BIT_RANGE(n[i], i + 1, 7)) << (i + 1) |
(BIT_RANGE(n[i + 1], 1, i + 1)) << (7 - i); /* This does not extract the next element bits */
}
n[7] = 0;
}
Update #1
In my case, the input will be uint64_t number, then I will use memmov to shift it one place to the left.
Update #2
The solution can be in C/C++, assembly(x86-64) or inline assembly.
This is really 2 subproblems: remove bits from each byte and pack the results. This is the flow of the code below. I wouldn't use a macro for this. Too much going on. Just inline the function if you're worried about performance at that level.
#include <stdio.h>
#include <stdint.h>
// Remove bits n to n+k-1 from x.
unsigned scrunch_1(unsigned x, int n, int k) {
unsigned hi_bits = ~0u << n;
return (x & ~hi_bits) | ((x >> k) & hi_bits);
}
// Remove bits n to n+k-1 from each byte in the buffer,
// then pack left. Return number of packed bytes.
size_t scrunch(uint8_t *buf, size_t size, int n, int k) {
size_t i_src = 0, i_dst = 0;
unsigned src_bits = 0; // Scrunched source bit buffer.
int n_src_bits = 0; // Initially it's empty.
for (;;) {
// Get scrunched bits until the buffer has at least 8.
while (n_src_bits < 8) {
if (i_src >= size) { // Done when source bytes exhausted.
// If there are left-over bits, add one more byte to output.
if (n_src_bits > 0) buf[i_dst++] = src_bits << (8 - n_src_bits);
return i_dst;
}
// Pack 'em in.
src_bits = (src_bits << (8 - k)) | scrunch_1(buf[i_src++], n, k);
n_src_bits += 8 - k;
}
// Write the highest 8 bits of the buffer to the destination byte.
n_src_bits -= 8;
buf[i_dst++] = src_bits >> n_src_bits;
}
}
int main(void) {
uint8_t x[] = { 0xaa, 0xaa, 0xaa, 0xaa };
size_t n = scrunch(x, 4, 2, 3);
for (size_t i = 0; i < n; i++) {
printf("%x ", x[i]);
}
printf("\n");
return 0;
}
This writes b5 ad 60, which by my reckoning is correct. A few other test cases work as well.
Oops I coded it the first time shifting the wrong way, but include that here in case it's useful to someone.
#include <stdio.h>
#include <stdint.h>
// Remove bits n to n+k-1 from x.
unsigned scrunch_1(unsigned x, int n, int k) {
unsigned hi_bits = 0xffu << n;
return (x & ~hi_bits) | ((x >> k) & hi_bits);
}
// Remove bits n to n+k-1 from each byte in the buffer,
// then pack right. Return number of packed bytes.
size_t scrunch(uint8_t *buf, size_t size, int n, int k) {
size_t i_src = 0, i_dst = 0;
unsigned src_bits = 0; // Scrunched source bit buffer.
int n_src_bits = 0; // Initially it's empty.
for (;;) {
// Get scrunched bits until the buffer has at least 8.
while (n_src_bits < 8) {
if (i_src >= size) { // Done when source bytes exhausted.
// If there are left-over bits, add one more byte to output.
if (n_src_bits > 0) buf[i_dst++] = src_bits;
return i_dst;
}
// Pack 'em in.
src_bits |= scrunch_1(buf[i_src++], n, k) << n_src_bits;
n_src_bits += 8 - k;
}
// Write the lower 8 bits of the buffer to the destination byte.
buf[i_dst++] = src_bits;
src_bits >>= 8;
n_src_bits -= 8;
}
}
int main(void) {
uint8_t x[] = { 0xaa, 0xaa, 0xaa, 0xaa };
size_t n = scrunch(x, 4, 2, 3);
for (size_t i = 0; i < n; i++) {
printf("%x ", x[i]);
}
printf("\n");
return 0;
}
This writes d6 5a b. A few other test cases work as well.
Something similar to this should work:
template<typename S> void removeBit(S* buffer, size_t length, size_t index)
{
const size_t BITS_PER_UNIT = sizeof(S)*8;
// first we find which data unit contains the desired bit
const size_t unit = index / BITS_PER_UNIT;
// and which index has the bit inside the specified unit, starting counting from most significant bit
const size_t relativeIndex = (BITS_PER_UNIT - 1) - index % BITS_PER_UNIT;
// then we unset that bit
buffer[unit] &= ~(1 << relativeIndex);
// now we have to shift what's on the right by 1 position
// we create a mask such that if 0b00100000 is the bit removed we use 0b00011111 as mask to shift the rest
const S partialShiftMask = (1 << relativeIndex) - 1;
// now we keep all bits left to the removed one and we shift left all the others
buffer[unit] = (buffer[unit] & ~partialShiftMask) | ((buffer[unit] & partialShiftMask) << 1);
for (int i = unit+1; i < length; ++i)
{
//we set rightmost bit of previous unit according to last bit of current unit
buffer[i-1] |= buffer[i] >> (BITS_PER_UNIT-1);
// then we shift current unit by one
buffer[i] <<= 1;
}
}
I just tested it on some basic cases so maybe something is not exactly correct but this should move you onto the right track.

Shifting arrays of bytes and skipping bits

I'm trying to make a function that would return N number of bits of a given memory chunk, and optionally skipping M bits.
Example:
unsigned char *data = malloc(3);
data[0] = 'A'; data[1] = 'B'; data[2] = 'C';
read(data, 8, 4);
would skip 12 bits and then read 8 bits from the data chunk "ABC".
"Skipping" bits means it would actually bitshift the entire array, carrying bits from the right to the left.
In this example ABC is
01000001 01000010 01000011
and the function would need to return
0001 0100
This question is a follow up of my previous question
Minimal compilable code
#include <ios>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
using namespace std;
typedef unsigned char byte;
typedef struct bit_data {
byte *data;
size_t length;
} bit_data;
/*
Asume skip_n_bits will be 0 >= skip_n_bits <= 8
*/
bit_data *read(size_t n_bits, size_t skip_n_bits) {
bit_data *bits = (bit_data *) malloc(sizeof(struct bit_data));
size_t bytes_to_read = ceil(n_bits / 8.0);
size_t bytes_to_read_with_skip = ceil(n_bits / 8.0) + ceil(skip_n_bits / 8.0);
bits->data = (byte *) calloc(1, bytes_to_read);
bits->length = n_bits;
/* Hardcoded for the sake of this example*/
byte *tmp = (byte *) malloc(3);
tmp[0] = 'A'; tmp[1] = 'B'; tmp[2] = 'C';
/*not working*/
if(skip_n_bits > 0){
unsigned char *tmp2 = (unsigned char *) calloc(1, bytes_to_read_with_skip);
size_t i;
for(i = bytes_to_read_with_skip - 1; i > 0; i--) {
tmp2[i] = tmp[i] << skip_n_bits;
tmp2[i - 1] = (tmp[i - 1] << skip_n_bits) | (tmp[i] >> (8 - skip_n_bits));
}
memcpy(bits->data, tmp2, bytes_to_read);
free(tmp2);
}else{
memcpy(bits->data, tmp, bytes_to_read);
}
free(tmp);
return bits;
}
int main(void) {
//Reading "ABC"
//01000001 01000010 01000011
bit_data *res = read(8, 4);
cout << bitset<8>(*res->data);
cout << " -> Should be '00010100'";
return 0;
}
The current code returns 00000000 instead of 00010100.
I feel like the error is something small, but I'm missing it. Where is the problem?
Your code is tagged as C++, and indeed you're already using C++ constructs like bitset, however it's very C-like. The first thing to do I think would be to use more C++.
Turns out bitset is pretty flexible already. My approach would be to create one to store all the bits in our input data, and then grab a subset of that based on the number you wish to skip, and return the subset:
template<size_t N, size_t M, typename T = unsigned char>
std::bitset<N> read(size_t skip_n_bits, const std::array<T, M>& data)
{
const size_t numBits = sizeof(T) * 8;
std::bitset<N> toReturn; // initially all zeros
// if we want to skip all bits, return all zeros
if (M*numBits <= skip_n_bits)
return toReturn;
// create a bitset to store all the bits represented in our data array
std::bitset<M*numBits> tmp;
// set bits in tmp based on data
// convert T into bit representations
size_t pos = M*numBits-1;
for (const T& element : data)
{
for (size_t i=0; i < numBits; ++i)
{
tmp.set(pos-i, (1 << (numBits - i-1)) & element);
}
pos -= numBits;
}
// grab just the bits we need
size_t startBit = tmp.size()-skip_n_bits-1;
for (size_t i = 0; i < N; ++i)
{
toReturn[N-i-1] = tmp[startBit];
tmp <<= 1;
}
return toReturn;
}
Full working demo
And now we can call it like so:
// return 8-bit bitset, skip 12 bits
std::array<unsigned char, 3> data{{'A', 'B', 'C'}};
auto&& returned = read<8>(12, data);
std::cout << returned << std::endl;
Prints
00100100
which is precisely our input 01000001 01000010 01000011 skipping the first twelve bits (from the left towards the right), and only grabbing the next 8 available.
I'd argue this is a bit easier to read than what you've got, esp. from a C++ programmer's point of view.

Convert integer to binary and store it in an integer array of specified size:c++

I want to convert an integer to binary string and then store each bit of the integer string to an element of a integer array of a given size. I am sure that the input integer's binary expression won't exceed the size of the array specified. How to do this in c++?
Pseudo code:
int value = ???? // assuming a 32 bit int
int i;
for (i = 0; i < 32; ++i) {
array[i] = (value >> i) & 1;
}
template<class output_iterator>
void convert_number_to_array_of_digits(const unsigned number,
output_iterator first, output_iterator last)
{
const unsigned number_bits = CHAR_BIT*sizeof(int);
//extract bits one at a time
for(unsigned i=0; i<number_bits && first!=last; ++i) {
const unsigned shift_amount = number_bits-i-1;
const unsigned this_bit = (number>>shift_amount)&1;
*first = this_bit;
++first;
}
//pad the rest with zeros
while(first != last) {
*first = 0;
++first;
}
}
int main() {
int number = 413523152;
int array[32];
convert_number_to_array_of_digits(number, std::begin(array), std::end(array));
for(int i=0; i<32; ++i)
std::cout << array[i] << ' ';
}
Proof of compilation here
You could use C++'s bitset library, as follows.
#include<iostream>
#include<bitset>
int main()
{
int N;//input number in base 10
cin>>N;
int O[32];//The output array
bitset<32> A=N;//A will hold the binary representation of N
for(int i=0,j=31;i<32;i++,j--)
{
//Assigning the bits one by one.
O[i]=A[j];
}
return 0;
}
A couple of points to note here:
First, 32 in the bitset declaration statement tells the compiler that you want 32 bits to represent your number, so even if your number takes fewer bits to represent, the bitset variable will have 32 bits, possibly with many leading zeroes.
Second, bitset is a really flexible way of handling binary, you can give a string as its input or a number, and again you can use the bitset as an array or as a string.It's a really handy library.
You can print out the bitset variable A as
cout<<A;
and see how it works.
You can do like this:
while (input != 0) {
if (input & 1)
result[index] = 1;
else
result[index] =0;
input >>= 1;// dividing by two
index++;
}
As Mat mentioned above, an int is already a bit-vector (using bitwise operations, you can check each bit). So, you can simply try something like this:
// Note: This depends on the endianess of your machine
int x = 0xdeadbeef; // Your integer?
int arr[sizeof(int)*CHAR_BIT];
for(int i = 0 ; i < sizeof(int)*CHAR_BIT ; ++i) {
arr[i] = (x & (0x01 << i)) ? 1 : 0; // Take the i-th bit
}
Decimal to Binary: Size independent
Two ways: both stores binary represent into a dynamic allocated array bits (in msh to lsh).
First Method:
#include<limits.h> // include for CHAR_BIT
int* binary(int dec){
int* bits = calloc(sizeof(int) * CHAR_BIT, sizeof(int));
if(bits == NULL) return NULL;
int i = 0;
// conversion
int left = sizeof(int) * CHAR_BIT - 1;
for(i = 0; left >= 0; left--, i++){
bits[i] = !!(dec & ( 1u << left ));
}
return bits;
}
Second Method:
#include<limits.h> // include for CHAR_BIT
int* binary(unsigned int num)
{
unsigned int mask = 1u << ((sizeof(int) * CHAR_BIT) - 1);
//mask = 1000 0000 0000 0000
int* bits = calloc(sizeof(int) * CHAR_BIT, sizeof(int));
if(bits == NULL) return NULL;
int i = 0;
//conversion
while(mask > 0){
if((num & mask) == 0 )
bits[i] = 0;
else
bits[i] = 1;
mask = mask >> 1 ; // Right Shift
i++;
}
return bits;
}
I know it doesn't add as many Zero's as you wish for positive numbers. But for negative binary numbers, it works pretty well.. I just wanted to post a solution for once :)
int BinToDec(int Value, int Padding = 8)
{
int Bin = 0;
for (int I = 1, Pos = 1; I < (Padding + 1); ++I, Pos *= 10)
{
Bin += ((Value >> I - 1) & 1) * Pos;
}
return Bin;
}
This is what I use, it also lets you give the number of bits that will be in the final vector, fills any unused bits with leading 0s.
std::vector<int> to_binary(int num_to_convert_to_binary, int num_bits_in_out_vec)
{
std::vector<int> r;
// make binary vec of minimum size backwards (LSB at .end() and MSB at .begin())
while (num_to_convert_to_binary > 0)
{
//cout << " top of loop" << endl;
if (num_to_convert_to_binary % 2 == 0)
r.push_back(0);
else
r.push_back(1);
num_to_convert_to_binary = num_to_convert_to_binary / 2;
}
while(r.size() < num_bits_in_out_vec)
r.push_back(0);
return r;
}

C++ Bytes To Bits Conversion And Then Print

Code Taken From: Bytes to Binary in C Credit: BSchlinker
The following code I modified to take more than 1 Byte at a time. I modified it, and got it half working and then got really confused on my loops. :( Ive spent the last day and a half trying to figure it out... but my C++ skills are not really that good (still learning!)
#include <iostream>
using namespace std;
char show_binary(unsigned char u, unsigned char *result,int len);
int main()
{
unsigned char p40[3] = {0x40, 0x00, 0x0a};
unsigned char bits[8*(sizeof(p40))];
int c;
c=sizeof(p40);
show_binary(*p40, bits, 3);
cout << "\n\n";
cout << "BIN = ";
do{
for (int i = 0; i < 8; i++)
printf("%d",bits[i+(8*c)]);
c++;
}while(c < 3);
cout << "\n";
int a;
cin >> a;
return 0;
}
char show_binary(unsigned char u, unsigned char *result, int len)
{
unsigned char mask = 1;
unsigned char bits[8*sizeof(result)];
int a,b,c;
a=0;
b=0;
c=len;
do{
for (int i = 0; i < 8; i++)
bits[i+(8*a)] = (u[&a] & (mask << i)) != 0;
a++;
}while(a < len);
//Need to reverse it?
do{
for (int i = 8; i != -1; i--)
result[i+(8*c)] = bits[i+(8*c)];
b++;
c--;
}while(b < len);
return *result;
}
After I spit out:
cout << "BIN = ";
do{
for (int i = 0; i < 8; i++)
printf("%d",bits[i+(8*c)]);
c++;
}while(c < 3);
Id like to take bit[11] ~ bit[the end] and compute a BYTE every 8 bits. If that makes sense. But first the function should work. Any pro tips on how this should be done? And of course, rip my code apart. I like to learn.
Man, there is a lot going on in this code, so it's hard to know where to start. Suffice to say, you're trying a bit too hard. It sounds like you are trying to 1) pass in a byte array; 2) turn those bytes into a string representation of the binary; and 3) turn that string representation back into a value?
It just so happens I recently did something similar to this in C, which should still work using a C++ compiler.
#include <stdio.h>
#include <string.h>
/* A macro to get a substring */
#define substr(dest, src, dest_size, startPos, strLen) snprintf(dest, dest_size, "%.*s", strLen, src+startPos)
/* Pass in char* array of bytes, get binary representation as string in bitStr */
void str2bs(const char *bytes, size_t len, char *bitStr) {
size_t i;
char buffer[9] = "";
for(i = 0; i < len; i++) {
sprintf(buffer,
"%c%c%c%c%c%c%c%c",
(bytes[i] & 0x80) ? '1':'0',
(bytes[i] & 0x40) ? '1':'0',
(bytes[i] & 0x20) ? '1':'0',
(bytes[i] & 0x10) ? '1':'0',
(bytes[i] & 0x08) ? '1':'0',
(bytes[i] & 0x04) ? '1':'0',
(bytes[i] & 0x02) ? '1':'0',
(bytes[i] & 0x01) ? '1':'0');
strncat(bitStr, buffer, 8);
buffer[0] = '\0';
}
}
To get the string of binary back into a value it can by done with bit shifting:
unsigned char bs2uc(char *bitStr) {
unsigned char val = 0;
int toShift = 0;
int i;
for(i = strlen(bitStr)-1; i >= 0; i--) {
if(bitStr[i] == '1') {
val = (1 << toShift) | val;
}
toShift++;
}
return val;
}
Once you had a binary string you could then take substrings of any arbitrary 8 bits (or less, I guess) and turn them back into bytes.
char *bitStr; /* Let's pretend this is populated with a valid string */
char byte[9] = "";
substr(byte, bitStr, 9, 4, 8);
/* This would create a substring of length 8 starting from index 4 of bitStr */
unsigned char b = bs2uc(byte);
I've actually created a whole suite of value -> binary string -> value functions if you'd like to take a look at them. GitHub - binstr

Does anyone have an easy solution to parsing Exp-Golomb codes using C++?

Trying to decode the SDP sprop-parameter-sets values for an H.264 video stream and have found to access some of the values will involve parsing of Exp-Golomb encoded data and my method contains the base64 decoded sprop-parameter-sets data in a byte array which I now bit walking but have come up to the first part of Exp-Golomb encoded data and looking for a suitable code extract to parse these values.
Exp.-Golomb codes of what order ??
If it you need to parse H.264 bit stream (I mean transport layer) you can write a simple functions to make an access to scecified bits in the endless bit stream. Bits indexing from left to right.
inline u_dword get_bit(const u_byte * const base, u_dword offset)
{
return ((*(base + (offset >> 0x3))) >> (0x7 - (offset & 0x7))) & 0x1;
}
This function implement decoding of exp-Golomb codes of zero range (used in H.264).
u_dword DecodeUGolomb(const u_byte * const base, u_dword * const offset)
{
u_dword zeros = 0;
// calculate zero bits. Will be optimized.
while (0 == get_bit(base, (*offset)++)) zeros++;
// insert first 1 bit
u_dword info = 1 << zeros;
for (s_dword i = zeros - 1; i >= 0; i--)
{
info |= get_bit(base, (*offset)++) << i;
}
return (info - 1);
}
u_dword means unsigned 4 bytes integer.
u_byte means unsigned 1 byte integer.
Note that first byte of each NAL Unit is a specified structure with forbidden bit, NAL reference, and NAL type.
Accepted answer is not a correct implementation. It is giving wrong output. Correct implementation as per pseudo code from
"Sec 9.1 Parsing process for Exp-Golomb codes" spec T-REC-H.264-201304
int32_t getBitByPos(unsigned char *buffer, int32_t pos) {
return (buffer[pos/8] >> (8 - pos%8) & 0x01);
}
uint32_t decodeGolomb(unsigned char *byteStream, uint32_t *index) {
uint32_t leadingZeroBits = -1;
uint32_t codeNum = 0;
uint32_t pos = *index;
if (byteStream == NULL || pos == 0 ) {
printf("Invalid input\n");
return 0;
}
for (int32_t b = 0; !b; leadingZeroBits++)
b = getBitByPos(byteStream, pos++);
for (int32_t b = leadingZeroBits; b > 0; b--)
codeNum = codeNum | (getBitByPos(byteStream, pos++) << (b - 1));
*index = pos;
return ((1 << leadingZeroBits) - 1 + codeNum);
}
I wrote a c++ jpeg-ls compression library that uses golomb codes. I don't know if Exp-Golomb codes is exactly the same. The library is open source can be found at http://charls.codeplex.com. I use a lookup table to decode golomb codes <= 8 bits in length. Let me know if you have problems finding your way around.
Revised with a function to get N bits from the stream; works parsing H.264 NALs
inline uint32_t get_bit(const uint8_t * const base, uint32_t offset)
{
return ((*(base + (offset >> 0x3))) >> (0x7 - (offset & 0x7))) & 0x1;
}
inline uint32_t get_bits(const uint8_t * const base, uint32_t * const offset, uint8_t bits)
{
uint32_t value = 0;
for (int i = 0; i < bits; i++)
{
value = (value << 1) | (get_bit(base, (*offset)++) ? 1 : 0);
}
return value;
}
// This function implement decoding of exp-Golomb codes of zero range (used in H.264).
uint32_t DecodeUGolomb(const uint8_t * const base, uint32_t * const offset)
{
uint32_t zeros = 0;
// calculate zero bits. Will be optimized.
while (0 == get_bit(base, (*offset)++)) zeros++;
// insert first 1 bit
uint32_t info = 1 << zeros;
for (int32_t i = zeros - 1; i >= 0; i--)
{
info |= get_bit(base, (*offset)++) << i;
}
return (info - 1);
}