Get part of specific length of allocated memory space - c

I have some billions of bits loaded into RAM by the use of malloc() - will call it big_set. I also have another amount of bits (will call it small_set) in RAM which are all set to 1 and I know its size (how many bits - I will call it ss_size), but can't predict it, as varies on each execution. ss_size can be sometimes as small as 100 or large as hundreds of millions.
I need to do some bitwise operations between small_set and some unpredictable parts of big_set of ss_size bits length. I can't just extend small_set with zeros on both most-significant and least-significant sides to make its size equal big_set's size, as that would be very RAM and CPU expensive (same operations will be done at same time with a lot of differently sized small_sets and also will do shift operations over small_set, expanding it would lead in much more bits to CPU work on).
Example:
big_set: 100111001111100011000111110001100 (would be billions of bits in reality)
small_set: 111111, so ss_size is 6. (may be an unpredictable number of bits).
I need to take 6 bits length parts of big_set, e.g.: 001100, 000111, etc. Obs.: not necessarily Nth 6 bits, it could be from 3rd to 9th bits, for instance. I don't know how can I get it.
I don't want to get a big_set copy with everything zeroed except the 6 bits I would be taking, like on 000000001111100000000000000000000, as that would be also very RAM expensive.
The question is: how can I get N bits from anywhere inside big_set, so I can do bitwise operations between they and small_set? Being N = ss_size.

I'm not sure that the example given below will give an answer to your question, also I am not sure that the realized XOR will work correctly.
But I have tried to show how confusing can be the implementation of the algorithm, if the task is to save memory.
This is my example for case of 40 bit in big_set and 6 bit in small_set:
#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>
void setBitsInMemory(uint8_t * memPtr, size_t from, size_t to)
// sets bits in the memory allocated from memPtr (pointer to the first byte)
// where from and to are numbers of bits to be set
{
for (size_t i = from; i <= to; i++)
{
size_t block = i / 8;
size_t offset = i % 8;
*(memPtr + block) |= 0x1 << offset;
}
}
uint8_t * allocAndBuildSmallSet(size_t bitNum)
// Allocate memory to store bitNum bits and set them to 1
{
uint8_t * ptr = NULL;
size_t byteNum = 1 + bitNum / 8; // determine number of bytes for
ptr = (uint8_t*) malloc(byteNum);
if (ptr != NULL)
{
for (size_t i = 0; i < byteNum; i++) ptr[i] = 0;
setBitsInMemory(ptr, 0, bitNum - 1);
}
return ptr;
}
void printBits(uint8_t * memPtr, size_t from, size_t to)
{
for (size_t i = from; i <= to; i++)
{
size_t block = i / 8;
size_t offset = i % 8;
if (*(memPtr + block) & (0x1 << offset) )
printf("1");
else
printf("0");
}
}
void applyXOR(uint8_t * mainMem, size_t start, size_t cnt, uint8_t * pattern, size_t ptrnSize)
// Applys bitwise XOR between cnt bits of mainMem and pattern
// starting from start bit in mainMem and 0 bit in pattern
// if pattern is smaller than cnt, it will be applyed cyclically
{
size_t ptrnBlk = 0;
size_t ptrnOff = 0;
for (size_t i = start; i < start + cnt; i++)
{
size_t block = i / 8;
size_t offset = i % 8;
*(mainMem + block) ^= ((*(pattern + ptrnBlk) & (0x1 << ptrnOff)) ? 1 : 0) << offset;
ptrnOff++;
if ((ptrnBlk * 8 + ptrnOff) >= ptrnSize)
{
ptrnBlk = 0;
ptrnOff = 0;
}
if (ptrnOff % 8 == 0)
{
ptrnBlk++;
ptrnOff = 0;
}
}
}
int main(void)
{
uint8_t * big_set;
size_t ss_size;
uint8_t * small_set;
big_set = (uint8_t*)malloc(5); // 5 bytes (40 bit) without initialization
ss_size = 6;
small_set = allocAndBuildSmallSet(ss_size);
printf("Initial big_set:\n");
printBits(big_set, 0, 39);
// some operation for ss_size bits starting from 12th
applyXOR(big_set, 12, ss_size, small_set, ss_size);
// output for visual analysis
printf("\nbig_set after XOR with small_set:\n");
printBits(big_set, 0, 39);
printf("\n");
// free memory
free(big_set);
free(small_set);
}
At my PC I can see the following:

Related

Implement BigInt bit shifting using intrinsics over an array

I'd like to implement bit shifting over a block of memory using SIMD. I have figured out this solution, that essentially follows these steps:
Byte shifts of shift / CHAR_BIT using memmove if the shift is greater than CHAR_BIT
Bit shifts iterating over each char of the memory block
This is essentially an O(n). I know I can just iterate over data types bigger than char, however I'm pretty sure this can be done even faster using intrinsics.
void rshift(void *self, int other, size_t size) {
if (other < size * CHAR_BIT) {
unsigned char *s = self,
*pass = calloc(2, sizeof(unsigned char));
if (pass != NULL) {
int chars = other / CHAR_BIT,
bits = other % CHAR_BIT;
if (chars > 0) {
memmove(s + chars, s, size - chars);
memset(s, 0, chars);
}
if (bits > 0) {
for (size_t i = 0; i < size;
i++, pass[0] = pass[1] << (CHAR_BIT - bits)) {
pass[1] = s[i] & ((1 << bits) - 1);
s[i] = (s[i] >> bits) | pass[0];
}
}
free(pass);
}
} else memset(self, 0, size);
}
I have tried taking a look at the different intrinsic operations, like _mm_srli_si128, that however byte shifts, doesn't bit shift.
Is there a way to implement the previous function using SIMD instructions?

memset and (&, >>) symbols in C

I found one project where I can't uderstand one piece of code.Virtual memory
#include <assert.h>
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#define TLB_SIZE 16
#define PAGE_SIZE 256
#define FRAME_SIZE 256
#define PHYSICAL_MEMORY_SIZE PAGE_SIZE *FRAME_SIZE
int logicalAddress = 0;
int offsetNumber = 0;
int pageNumber = 0;
int physicalAddress = 0;
int Frame = 0;
int Value = 0;
int Hit = 0;
int tlbIndex = 0;
int tlbSize = 0;
unsigned pageNumberMask = 65280; // 1111111100000000
unsigned offsetMask = 255; // 11111111
int tlbHitCount = 0;
float tlbHitRate = 0;
int addressCount = 0;
int pageFaultCount = 0;
float pageFaultRate = 0;
struct tlbTable {
unsigned int pageNum;
unsigned int frameNum;
};
int main(int argc, char *argv[]) {
// Check to see if user inputs addresses.txt
if (argc != 2) {
fprintf(stderr, "Usage ./VirtualMem_Manager <Filename.txt> \n");
exit(1);
}
// Open addresses.txt, BACKING_STORE.bin, and
// Create Output.txt to store program results
FILE *addresses = fopen(argv[1], "r");
FILE *BACKINGSTORE = fopen("BACKING_STORE.bin", "rb");
FILE *Output = fopen("addressOutput.txt", "w");
int physicalMemory[PHYSICAL_MEMORY_SIZE];
char Buffer[256];
int Index;
// Declare and initialize pageTable[] array to -1
int pageTable[PAGE_SIZE];
memset(pageTable, -1, 256 * sizeof(int));
// Declare and initialize tlb[] structure to -1
struct tlbTable tlb[TLB_SIZE];
memset(pageTable, -1, 16 * sizeof(char));
// Read each address from addresses.txt
while (fscanf(addresses, "%d", &logicalAddress) == 1) {
addressCount++;
// set the page number and offset for each logical address
pageNumber = logicalAddress & pageNumberMask;
pageNumber = pageNumber >> 8;
offsetNumber = logicalAddress & offsetMask;
Hit = -1;
// Check to see if the page number is already in the tlb
// If it is in tlb, then it is tlb hit
for (Index = 0; Index < tlbSize; Index++) {
if (tlb[Index].pageNum == pageNumber) {
Hit = tlb[Index].frameNum;
physicalAddress = Hit * 256 + offsetNumber;
}
}
if (!(Hit == -1)) {
tlbHitCount++;
}
// This "else if" loop is the tlb miss
// Gets the physical page number from page table
else if (pageTable[pageNumber] == -1) {
fseek(BACKINGSTORE, pageNumber * 256, SEEK_SET);
fread(Buffer, sizeof(char), 256, BACKINGSTORE);
pageTable[pageNumber] = Frame;
for (Index = 0; Index < 256; Index++) {
physicalMemory[Frame * 256 + Index] = Buffer[Index];
}
pageFaultCount++;
Frame++;
// FIFO algorithm for the tlb
if (tlbSize == 16) tlbSize--;
for (tlbIndex = tlbSize; tlbIndex > 0; tlbIndex--) {
tlb[tlbIndex].pageNum = tlb[tlbIndex - 1].pageNum;
tlb[tlbIndex].frameNum = tlb[tlbIndex - 1].frameNum;
}
if (tlbSize <= 15) tlbSize++;
tlb[0].pageNum = pageNumber;
tlb[0].frameNum = pageTable[pageNumber];
physicalAddress = pageTable[pageNumber] * 256 + offsetNumber;
} else {
physicalAddress = pageTable[pageNumber] * 256 + offsetNumber;
}
// Gets the value from the bin file provided
Value = physicalMemory[physicalAddress];
// print the addresses and value to Output.txt
fprintf(Output, "Virtual Address: %d Physical Address: %d Value: %d \n",
logicalAddress, physicalAddress, Value);
}
// The statistics of the program
pageFaultRate = pageFaultCount * 1.0f / addressCount;
tlbHitRate = tlbHitCount * 1.0f / addressCount;
// Close files provided for the project
fclose(addresses);
fclose(BACKINGSTORE);
// Print the statistics of the program to Output.txt
fprintf(Output, "Number of Addresses: %d\n", addressCount);
fprintf(Output, "Number of Page Faults: %d\n", pageFaultCount);
fprintf(Output, "Page Fault Rate: %f\n", pageFaultRate);
fprintf(Output, "TLB Hits: %d\n", tlbHitCount);
fprintf(Output, "TLB Hit Rate %f\n", tlbHitRate);
// Close Output.txt
fclose(Output);
return 0;
}
I can't understand these three lines.
pageNumber = logicalAddress & pageNumberMask;
pageNumber = pageNumber >> 8;
offsetNumber = logicalAddress & offsetMask;
As I understand papeNumber is value where logicalAddress and pageNuberMask a stored together. And this char & join them. Then we do offset on 8(>> 8).
And why is using memset dangerous.
(For purposes of illustration, this answer assumes the addresses are 32 bits, which appears to be consistent with the code in the question.)
pageNumber = logicalAddress & pageNumberMask;
pageNumber = pageNumber >> 8;
offsetNumber = logicalAddress & offsetMask;
The logical address is partitioned into three sets of bits: 16 upper bits that are ignored in this code, 8 bits used for a page number, and 8 lower bits that are used for an offset within a page.
pageNumberMask is a value that has been prepared to have 1s in the page number bits and 0s elsewhere. Then the operation logicalAddress & pageNumberMask performs a bitwise AND, which produces a value in which the only bits set are those that were set in the page number bits of logicalAddress. In other words, it produces a value in which the upper bits and the offset bits are cleared, effectively isolating the page number bits.
Then pageNumber >> 8 shifts those page number bits down to the low bits, so the result is a page number. That page number can then be used to index arrays.
Similarly, offsetMask is a prepared value with 1s in the offset bits and 0s elsewhere, and logicalAddress & offsetMask extracts just the offset bits.
Note that logicalAddress & pageNumberMask serves a purpose only if bits above the page number bits might be set. This is because pageNumber >> 8 removes the lower bits anyway, so there is no point in clearing them with an & operation. So this code separates the page number bits from the upper bits but makes no use of those upper bits. This might be because it is a simplified example for teaching, and actual code for deployment would make use of the upper bits.
And why is using memset dangerous.
memset is a powerful routine that is easy to use incorrectly. Given incorrect arguments, it will modify any accessible memory, and the compiler often cannot detect that it has been used incorrectly, so it cannot provide a warning or error message.
Defects in the Code
The code in the question has a number of defects, so it is not a great example for teaching:
struct tlbTable tlb[TLB_SIZE]; followed by memset(pageTable, -1, 16*sizeof(char)); looks like a bug. pageTable was initialized previously, and this was probably intended to initialize tlb. And the size is wrong; tlb has 16 elements (TLB_SIZE is 16), and each is several bytes. That line should probably be memset(tlb, -1, sizeof tlb);, and the previous memset should be memset(pageTable, -1, sizeof pageTable);.
Preprocessor macros are defined for various parameters, such as 256 for PAGE_SIZE, but then parts of the code ignore these and use hard-coded constants, as in Frame*256 + Index. This is a recipe for bugs.
For performance and memory reasons pageNumber and offset in the page are stored in single unsigned value.
I will use c++14 bit representation with 0b prefix in comments to explain.
unsigned logicalAddress = someValue; // <- 0bPPPPPPPPFFFFFFFF
where P denotes bits for pageNumber and F bits for offsetNumber;
pageNumber = logicalAddress & pageNumberMask; // <- 0bPPPPPPPP00000000
pageNumber = pageNumber >> 8; // <- 0bPPPPPPPP
offsetNumber = logicalAddress & offsetMask; // <- 0bFFFFFFFF
About memset using it is not dangerous if used correctly.

Efficient algorithm for finding a byte in a bit array

Given a bytearray uint8_t data[N] what is an efficient method to find a byte uint8_t search within it even if search is not octet aligned? i.e. the first three bits of search could be in data[i] and the next 5 bits in data[i+1].
My current method involves creating a bool get_bit(const uint8_t* src, struct internal_state* state) function (struct internal_state contains a mask that is bitshifted right, &ed with src and returned, maintaining size_t src_index < size_t src_len) , leftshifting the returned bits into a uint8_t my_register and comparing it with search every time, and using state->src_index and state->src_mask to get the position of the matched byte.
Is there a better method for this?
If you're searching an eight bit pattern within a large array you can implement a sliding window over 16 bit values to check if the searched pattern is part of the two bytes forming that 16 bit value.
To be portable you have to take care of endianness issues which is done by my implementation by building the 16 bit value to search for the pattern manually. The high byte is always the currently iterated byte and the low byte is the following byte. If you do a simple conversion like value = *((unsigned short *)pData) you will run into trouble on x86 processors...
Once value, cmp and mask are setup cmp and mask are shifted. If the pattern was not found within hi high byte the loop continues by checking the next byte as start byte.
Here is my implementation including some debug printouts (the function returns the bit position or -1 if pattern was not found):
int findPattern(unsigned char *data, int size, unsigned char pattern)
{
int result = -1;
unsigned char *pData;
unsigned char *pEnd;
unsigned short value;
unsigned short mask;
unsigned short cmp;
int tmpResult;
if ((data != NULL) && (size > 0))
{
pData = data;
pEnd = data + size;
while ((pData < pEnd) && (result == -1))
{
printf("\n\npData = {%02x, %02x, ...};\n", pData[0], pData[1]);
if ((pData + 1) < pEnd) /* still at least two bytes to check? */
{
tmpResult = (int)(pData - data) * 8; /* calculate bit offset according to current byte */
/* avoid endianness troubles by "manually" building value! */
value = *pData << 8;
pData++;
value += *pData;
/* create a sliding window to check if search patter is within value */
cmp = pattern << 8;
mask = 0xFF00;
while (mask > 0x00FF) /* the low byte is checked within next iteration! */
{
printf("cmp = %04x, mask = %04x, tmpResult = %d\n", cmp, mask, tmpResult);
if ((value & mask) == cmp)
{
result = tmpResult;
break;
}
tmpResult++; /* count bits! */
mask >>= 1;
cmp >>= 1;
}
}
else
{
/* only one chance left if there is only one byte left to check! */
if (*pData == pattern)
{
result = (int)(pData - data) * 8;
}
pData++;
}
}
}
return (result);
}
I don't think you can do much better than this in C:
/*
* Searches for the 8-bit pattern represented by 'needle' in the bit array
* represented by 'haystack'.
*
* Returns the index *in bits* of the first appearance of 'needle', or
* -1 if 'needle' is not found.
*/
int search(uint8_t needle, int num_bytes, uint8_t haystack[num_bytes]) {
if (num_bytes > 0) {
uint16_t window = haystack[0];
if (window == needle) return 0;
for (int i = 1; i < num_bytes; i += 1) {
window = window << 8 + haystack[i];
/* Candidate for unrolling: */
for (int j = 7; j >= 0; j -= 1) {
if ((window >> j) & 0xff == needle) {
return 8 * i - j;
}
}
}
}
return -1;
}
The main idea is to handle the 87.5% of cases that cross the boundary between consecutive bytes by pairing bytes in a wider data type (uint16_t in this case). You could adjust it to use an even wider data type, but I'm not sure that would gain anything.
What you cannot safely or easily do is anything involving casting part or all of your array to a wider integer type via a pointer (i.e. (uint16_t *)&haystack[i]). You cannot be ensured of proper alignment for such a cast, nor of the byte order with which the result might be interpreted.
I don't know if it would be better, but i would use sliding window.
uint counter = 0, feeder = 8;
uint window = data[0];
while (search ^ (window & 0xff)){
window >>= 1;
feeder--;
if (feeder < 8){
counter++;
if (counter >= data.length) {
feeder = 0;
break;
}
window |= data[counter] << feeder;
feeder += 8;
}
}
//Returns index of first bit of first sequence occurrence or -1 if sequence is not found
return (feeder > 0) ? (counter+1)*8-feeder : -1;
Also with some alterations you can use this method to search for arbitrary length (1 to 64-array_element_size_in_bits) bits sequence.
If AVX2 is acceptable (with earlier versions it didn't work out so well, but you can still do something there), you can search in a lot of places at the same time. I couldn't test this on my machine (only compile) so the following is more to give to you an idea of how it could be approached than copy&paste code, so I'll try to explain it rather than just code-dump.
The main idea is to read an uint64_t, shift it right by all values that make sense (0 through 7), then for each of those 8 new uint64_t's, test whether the byte is in there. Small complication: for the uint64_t's shifted by more than 0, the highest position should not be counted since it has zeroes shifted into it that might not be in the actual data. Once this is done, the next uint64_t should be read at an offset of 7 from the current one, otherwise there is a boundary that is not checked across. That's fine though, unaligned loads aren't so bad anymore, especially if they're not wide.
So now for some (untested, and incomplete, see below) code,
__m256i needle = _mm256_set1_epi8(find);
size_t i;
for (i = 0; i < n - 6; i += 7) {
// unaligned load here, but that's OK
uint64_t d = *(uint64_t*)(data + i);
__m256i x = _mm256_set1_epi64x(d);
__m256i low = _mm256_srlv_epi64(x, _mm256_set_epi64x(3, 2, 1, 0));
__m256i high = _mm256_srlv_epi64(x, _mm256_set_epi64x(7, 6, 5, 4));
low = _mm256_cmpeq_epi8(low, needle);
high = _mm256_cmpeq_epi8(high, needle);
// in the qword right-shifted by 0, all positions are valid
// otherwise, the top position corresponds to an incomplete byte
uint32_t lowmask = 0x7f7f7fffu & _mm256_movemask_epi8(low);
uint32_t highmask = 0x7f7f7f7fu & _mm256_movemask_epi8(high);
uint64_t mask = lowmask | ((uint64_t)highmask << 32);
if (mask) {
int bitindex = __builtin_ffsl(mask);
// the bit-index and byte-index are swapped
return 8 * (i + (bitindex & 7)) + (bitindex >> 3);
}
}
The funny "bit-index and byte-index are swapped" thing is because searching within a qword is done byte by byte and the results of those comparisons end up in 8 adjacent bits, while the search for "shifted by 1" ends up in the next 8 bits and so on. So in the resulting masks, the index of the byte that contains the 1 is a bit-offset, but the bit-index within that byte is actually the byte-offset, for example 0x8000 would correspond to finding the byte at the 7th byte of the qword that was right-shifted by 1, so the actual index is 8*7+1.
There is also the issue of the "tail", the part of the data left over when all blocks of 7 bytes have been processed. It can be done much the same way, but now more positions contain bogus bytes. Now n - i bytes are left over, so the mask has to have n - i bits set in the lowest byte, and one fewer for all other bytes (for the same reason as earlier, the other positions have zeroes shifted in). Also, if there is exactly 1 byte "left", it isn't really left because it would have been tested already, but that doesn't really matter. I'll assume the data is sufficiently padded that accessing out of bounds doesn't matter. Here it is, untested:
if (i < n - 1) {
// make n-i-1 bits, then copy them to every byte
uint32_t validh = ((1u << (n - i - 1)) - 1) * 0x01010101;
// the lowest position has an extra valid bit, set lowest zero
uint32_t validl = (validh + 1) | validh;
uint64_t d = *(uint64_t*)(data + i);
__m256i x = _mm256_set1_epi64x(d);
__m256i low = _mm256_srlv_epi64(x, _mm256_set_epi64x(3, 2, 1, 0));
__m256i high = _mm256_srlv_epi64(x, _mm256_set_epi64x(7, 6, 5, 4));
low = _mm256_cmpeq_epi8(low, needle);
high = _mm256_cmpeq_epi8(high, needle);
uint32_t lowmask = validl & _mm256_movemask_epi8(low);
uint32_t highmask = validh & _mm256_movemask_epi8(high);
uint64_t mask = lowmask | ((uint64_t)highmask << 32);
if (mask) {
int bitindex = __builtin_ffsl(mask);
return 8 * (i + (bitindex & 7)) + (bitindex >> 3);
}
}
If you are searching a large amount of memory and can afford an expensive setup, another approach is to use a 64K lookup table. For each possible 16-bit value, the table stores a byte containing the bit shift offset at which the matching octet occurs (+1, so 0 can indicate no match). You can initialize it like this:
uint8_t* g_pLookupTable = malloc(65536);
void initLUT(uint8_t octet)
{
memset(g_pLookupTable, 0, 65536); // zero out
for(int i = 0; i < 65536; i++)
{
for(int j = 7; j >= 0; j--)
{
if(((i >> j) & 255) == octet)
{
g_pLookupTable[i] = j + 1;
break;
}
}
}
}
Note that the case where the value is shifted 8 bits is not included (the reason will be obvious in a minute).
Then you can scan through your array of bytes like this:
int findByteMatch(uint8_t* pArray, uint8_t octet, int length)
{
if(length >= 0)
{
uint16_t index = (uint16_t)pArray[0];
if(index == octet)
return 0;
for(int bit, i = 1; i < length; i++)
{
index = (index << 8) | pArray[i];
if(bit = g_pLookupTable[index])
return (i * 8) - (bit - 1);
}
}
return -1;
}
Further optimization:
Read 32 or however many bits at a time from pArray into a uint32_t and then shift and AND each to get byte one at a time, OR with index and test, before reading another 4.
Pack the LUT into 32K by storing a nybble for each index. This might help it squeeze into the cache on some systems.
It will depend on your memory architecture whether this is faster than an unrolled loop that doesn't use a lookup table.

Strip parity bits in C from 8 bits of data followed by 1 parity bit

I have a buffer of bits with 8 bits of data followed by 1 parity bit. This pattern repeats itself. The buffer is currently stored as an array of octets.
Example (p are parity bits):
0001 0001 p000 0100 0p00 0001 00p01 1100 ...
should become
0001 0001 0000 1000 0000 0100 0111 00 ...
Basically, I need to strip of every ninth bit to just obtain the data bits. How can I achieve this?
This is related to another question asked here sometime back.
This is on a 32 bit machine so the solution to the related question may not be applicable. The maximum possible number of bits is 45 i.e. 5 data octets
This is what I have tried so far. I have created a "boolean" array and added the bits into the array based on the the bitset of the octet. I then look at every ninth index of the array and through it away. Then move the remaining array down one index. Then I've got only the data bits left. I was thinking there may be better ways of doing this.
Your idea of having an array of bits is good. Just implement the array of bits by a 32-bit number (buffer).
To remove a bit from the middle of the buffer:
void remove_bit(uint32_t* buffer, int* occupancy, int pos)
{
assert(*occupancy > 0);
uint32_t high_half = *buffer >> pos >> 1;
uint32_t low_half = *buffer << (32 - pos) >> (32 - pos);
*buffer = high_half | low_half;
--*occupancy;
}
To add a byte to the buffer:
void add_byte(uint32_t* buffer, int* occupancy, uint8_t byte)
{
assert(*occupancy <= 24);
*buffer = (*buffer << 8) | byte;
*occupancy += 8;
}
To remove a byte from the buffer:
uint8_t remove_byte(uint32_t* buffer, int* occupancy)
{
uint8_t result = *buffer >> (*occupancy - 8);
assert(*occupancy >= 8);
*occupancy -= 8;
return result;
}
You will have to arrange the calls so that the buffer never overflows. For example:
buffer = 0;
occupancy = 0;
add_byte(buffer, occupancy, *input++);
add_byte(buffer, occupancy, *input++);
remove_bit(buffer, occupancy, 7);
*output++ = remove_byte(buffer, occupancy);
add_byte(buffer, occupancy, *input++);
remove_bit(buffer, occupancy, 6);
*output++ = remove_byte(buffer, occupancy);
... (there are only 6 input bytes, so this should be easy)
In pseudo-code (since you're not providing any proof you've tried something), I would probably do it like this, for simplicity:
View the data (with parity bits included) as a stream of bits
While there are bits left to read:
Read the next 8 bits
Write to the output
Read one more bit, and discard it
This "lifts you up" from worrying about reading bytes, which no longer is a useful operation since your bytes are interleaved with bits you want to discard.
I have written helper functions to read unaligned bit buffers (this was for AVC streams, see original source here). The code itself is GPL, I'm pasting interesting (modified) bits here.
typedef struct bit_buffer_ {
uint8_t * start;
size_t size;
uint8_t * current;
uint8_t read_bits;
} bit_buffer;
/* reads one bit and returns its value as a 8-bit integer */
uint8_t get_bit(bit_buffer * bb) {
uint8_t ret;
ret = (*(bb->current) >> (7 - bb->read_bits)) & 0x1;
if (bb->read_bits == 7) {
bb->read_bits = 0;
bb->current++;
}
else {
bb->read_bits++;
}
return ret;
}
/* reads up to 32 bits and returns the value as a 32-bit integer */
uint32_t get_bits(bit_buffer * bb, size_t nbits) {
uint32_t i, ret;
ret = 0;
for (i = 0; i < nbits; i++) {
ret = (ret << 1) + get_bit(bb);
}
return ret;
}
You can use the structure like this:
uint_8 * buffer;
size_t buffer_size;
/* assumes buffer points to your data */
bit_buffer bb;
bb.start = buffer;
bb.size = buffer_size;
bb.current = buffer;
bb.read_bits = 0;
uint32_t value = get_bits(&bb, 8);
uint8_t parity = get_bit(&bb);
uint32_t value2 = get_bits(&bb, 8);
uint8_t parity2 = get_bit(&bb);
/* etc */
I must stress that this code is quite perfectible, proper bound checking must be implemented, but it works fine in my use-case.
I leave it as an exercise to you to implement a proper bit buffer reader using this for inspiration.
This also works
void RemoveParity(unsigned char buffer[], int size)
{
int offset = 0;
int j = 0;
for(int i = 1; i + j < size; i++)
{
if (offset == 0)
{
printf("%u\n", buffer[i + j - 1]);
}
else
{
unsigned char left = buffer[i + j - 1] << offset;
unsigned char right = buffer[i + j] >> (8 - offset);
printf("%u\n", (unsigned char)(left | right));
}
offset++;
if (offset == 8)
{
offset = 0;
j++; // advance buffer (8 parity bit consumed)
}
}
}

Efficient bitshifting an array of int?

To be on the same page, let's assume sizeof(int)=4 and sizeof(long)=8.
Given an array of integers, what would be an efficient method to logically bitshift the array to either the left or right?
I am contemplating an auxiliary variable such as a long, that will compute the bitshift for the first pair of elements (index 0 and 1) and set the first element (0). Continuing in this fashion the bitshift for elements (index 1 and 2) will be computer, and then index 1 will be set.
I think this is actually a fairly efficient method, but there are drawbacks. I cannot bitshift greater than 32 bits. I think using multiple auxiliary variables would work, but I'm envisioning recursion somewhere along the line.
There's no need to use a long as an intermediary. If you're shifting left, start with the highest order int, shifting right start at the lowest. Add in the carry from the adjacent element before you modify it.
void ShiftLeftByOne(int * arr, int len)
{
int i;
for (i = 0; i < len - 1; ++i)
{
arr[i] = (arr[i] << 1) | ((arr[i+1] >> 31) & 1);
}
arr[len-1] = arr[len-1] << 1;
}
This technique can be extended to do a shift of more than 1 bit. If you're doing more than 32 bits, you take the bit count mod 32 and shift by that, while moving the result further along in the array. For example, to shift left by 33 bits, the code will look nearly the same:
void ShiftLeftBy33(int * arr, int len)
{
int i;
for (i = 0; i < len - 2; ++i)
{
arr[i] = (arr[i+1] << 1) | ((arr[i+2] >> 31) & 1);
}
arr[len-2] = arr[len-1] << 1;
arr[len-1] = 0;
}
For anyone else, this is a more generic version of Mark Ransom's answer above for any number of bits and any type of array:
/* This function shifts an array of byte of size len by shft number of
bits to the left. Assumes array is big endian. */
#define ARR_TYPE uint8_t
void ShiftLeft(ARR_TYPE * arr_out, ARR_TYPE * arr_in, int arr_len, int shft)
{
const int int_n_bits = sizeof(ARR_TYPE) * 8;
int msb_shifts = shft % int_n_bits;
int lsb_shifts = int_n_bits - msb_shifts;
int byte_shft = shft / int_n_bits;
int last_byt = arr_len - byte_shft - 1;
for (int i = 0; i < arr_len; i++){
if (i <= last_byt){
int msb_idx = i + byte_shft;
arr_out[i] = arr_in[msb_idx] << msb_shifts;
if (i != last_byt)
arr_out[i] |= arr_in[msb_idx + 1] >> lsb_shifts;
}
else arr_out[i] = 0;
}
}
Take a look at BigInteger implementation in Java, which internally stores data as an array of bytes. Specifically you can check out the funcion leftShift(). Syntax is the same as in C, so it wouldn't be too difficult to write a pair of funciontions like those. Take into account too, that when it comes to bit shifting you can take advange of unsinged types in C. This means that in Java to safely shift data without messing around with sign you usually need bigger types to hold data (i.e. an int to shift a short, a long to shift an int, ...)

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