I have just started to study the CRC and how to implement it in software. My information source is mainly following document. Here is mentioned some simple algorithm for calculating CRC for any generator polynomial. I have attempted to write this algorithm in C++ language. I have tested it for generator polynomial x^5 + x^4 + x^2 + 1 (CRC-5) (generator polynomial used by chip) with usage of this online calculator and it works.
#include <iostream>
using namespace std;
int main() {
uint8_t data_byte = 0x31;
// polynom x^5 + x^4 + x^2 + 1
uint16_t polynom = 0x35;
// register contains 0 at the beginning
uint32_t crc = 0;
uint32_t message = 0;
// shift the message byte to left by so many bits which is needed for generator polynomial
message = (data_byte << 5);
// now the message byte is 13 bits long
uint8_t processed_bit = 13;
while(processed_bit > 0) {
// prepare free space for new bit from the message byte
crc = crc << 1;
// find out value of current msb in the message byte
message = message << 1;
if(message & 0x2000) {
// msb in message byte is "1"
// lsb in register is set to "1"
crc |= 1U;
} else {
// msb in message byte is "0"
// lsb in register is set to "0"
crc &= ~1U;
}
// remove msb from message byte
message = message & ~0x2000;
if(crc & 0x20) {
// subtract current multiple of the generator polynomial
crc = crc ^ polynom;
}
// remove msb from the register
crc = crc & ~0x20;
processed_bit--;
}
cout << "CRC: " << (int)crc << endl;
return 0;
}
It is obvious that this program is uneffective as far as execution time. So I have been thinking about a possibility how to improve it in this perspective. I know that there is a variant to use the look-up table containing the precalculated reminders but I would like to avoid this method. Does anybody know how to improve the above mentioned algorithm from the execution time perspective? Thanks in advance for any suggestions.
Just a quick glance shows several unnecessary statements. You don't need crc &= ~1U;, since the crc = crc << 1; already put a zero there. You don't need message = message & ~0x2000;, since you are only ever looking at one bit in there. Just let the other bits shift up and away. You don't need the crc = crc & ~0x20;, since the exclusive-or with the polynomial already did that.
If you read the document you linked, you will find that you do not need to process five more bits (13 total). You only need to process the eight message bits. Also reading that document, you do not need to feed in the message bits one at a time. You can exclusive-or the message byte directly onto the CRC register, and then process the eight bits all in the register.
Finally, you can speed up the calculation significantly with a table look up, processing eight bits at a time instead of one bit at a time. This is also described beautifully in the document you linked. You can find code here to automatically generate the table and C code for the calculation.
In the end though, none of this matters if you're not calculating the right thing to begin with. You need to verify the calculation with data from the chip first. I found this document with details on the CRC calculation for that chip. You need to spend some time with it and understand it in detail.
To answer your question directly, here is some code that does what your code does, but is much simpler. Also it is extended to work on n bits, not just eight. It does n loops instead of n+5 loops:
// Return a CRC-5 of the low n bits of data. The remaining bits of data must be
// zero. n must be in [5..32].
uint8_t crc5(uint32_t data, int n) {
int shift = n - 5;
uint32_t poly = (uint32_t)0x15 << shift;
uint32_t top = (uint32_t)1 << (n - 1);
do {
data = data & top ? (data << 1) ^ poly : data << 1;
} while (--n);
return (data >> shift) & 0x1f;
}
Simpler and faster still is the equivalent of yours restricted to eight bits, unrolled:
uint8_t crc5_8(uint8_t data) {
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
data = data & 0x80 ? (data << 1) ^ 0xa8 : data << 1;
return data >> 3;
}
However neither can calculate what you need for your chip.
Related
I am analyzing the MODBUS protocol (rs232 com port) used in the DPS5020 power supply module and I cannot understand the CRC calculation method in RTU mode (page 3) https://cloud.kyme32.ro/ftp_backup/DPS5020%20PC%20Software(2017.11.04)/DPS5020%20CNC%20Communication%20%20Protocol%20V1.2.pdf.
In the first example on page 4 for sending bytes 1, 3,0,2,0,2 the value CRC = 65CB (Hex) is indicated (2 byte swapped).
I've also tried several CRC calculators online but can't find the right value.
I also did a step-by-step diagram of the calculation and the right rotation of the bits, but the values do not return to me.
Is it necessary to use all the bytes of the frame (6) for the calculation or only the data values (4)? I have tried both without success...
Could you kindly put a little diagram of how the calculation is done and the return values step by step (16 bit xor with A001 value, rotate right yes / no ... etc)?
I know that in the end you have to swap the 2 bytes between them but the single values do not come back to me anyway.
Or is it simply a misprint of the manual?
All bytes in the frame are used in the CRC calculation.
Here is a C implementation of the CRC, which should answer your question about exactly what to shift and exclusive-or when:
#include <stddef.h>
#include <stdint.h>
uint16_t crc16modbus_bit(uint16_t crc, void const *mem, size_t len) {
unsigned char const *data = mem;
if (data == NULL)
return 0xffff;
for (size_t i = 0; i < len; i++) {
crc ^= data[i];
for (unsigned k = 0; k < 8; k++) {
crc = crc & 1 ? (crc >> 1) ^ 0xa001 : crc >> 1;
}
}
return crc;
}
(The initial CRC value is returned when called with mem equal to NULL.)
I'm currently attempting to build my own Operating System and have run into an issue when trying to test out my kernel code using VirtualBox.
The real issue arises when I call the assembly instruction sti as I'm currently attempting to implement an interrupt descriptor table and communicate with the PICs.
Here is the code that calls it. It's a function called kernel_main that is called from another assembly file. That file simply sets up the stack before executing any code from the OS, but there hasn't been any issues there, and everything works fine until I add the instruction asm("sti"); to the following code:
/* main function of our kernal
* accepts the pointer to multiboot and the magic code (no particular reason to take the magic number)
*
* use extern C to prevent gcc from changing the name
*/
extern "C" void kernel_main(void *multiboot_structure, uint32_t magic_number)
{
// can't use the standard printf as we're outside an OS currently
// we don't have access to glibc so we write our own printf
printf_boot_message("kernel.....\n");
// create the global descriptor table
GlobalDescriptorTable gdt;
// create the interrupt descriptor table
InterruptHandler interrupt_handler(&gdt);
// enable interrupts (test)
asm("sti"); // <- causes crash
// random debug printf
printf_boot_message("sti called\n");
// kernal never really stops, inf loop
while (1)
;
}
Below is the virtual box debug output, I've googled around for VINF_EM_TRIPLE_FAULT but mostly found RAM related issues that I don't think apply to me. The printf calls in the above code execute as expected followed by the VM immediately crashing stating the following:
Link to output as it's too large to post here: https://pastebin.com/jfPfhJUQ
Here is my interrupt handling code:
* Implementations of the interrupt handling routines in sys_interrupts.h
*/
#include "sys_interrupts.h"
#include "misc.h"
//handle() is used to take the interrupt number,
//i_number, and the address to the current CPU stack frame.
uint32_t InterruptHandler::handle(uint8_t i_number, uint32_t crnt_stkptr)
{
// debug
printf(" INTERRUPT");
// after the interrupt code has been executed,
// return the stack pointer so that the CPU can resume
// where it left off.
// this works for now as we do not have multiple
// concurrent processes running, so there is no issue
// of handling the threat number.
return crnt_stkptr;
}
// define the global descriptor table
InterruptHandler::_gate_descriptor InterruptHandler::interrupt_desc_table[N_ENTRIES];
// define the constructor. Takes a pointer to the global
// descriptor table
InterruptHandler::InterruptHandler(GlobalDescriptorTable* global_desc_table)
{
// grab the offset of the usable memory within our global segment
uint16_t seg = global_desc_table->CodeSegmentSelector();
// set all the entries in the IDT to block request initially
for (uint16_t i = 0; i < N_ENTRIES; i++)
{
// create an a gate for a system level interrupt, calling the block function (does nothing) using seg as its memory.
create_entry(i, seg, &block_request, PRIV_LVL_KERNEL, GATE_INTERRUPT);
}
// create a couple interrupts for 0x00 and 0x01, really 0x20 and 0x21 in memory
//create_entry(BASE_I_NUM + 0x00, seg, &isr0x00, PRIV_LVL_KERNEL, GATE_INTERRUPT);
//create_entry(BASE_I_NUM + 0x01, seg, &isr0x01, PRIV_LVL_KERNEL, GATE_INTERRUPT);
// init the PICs
pic_controller.send_master_cmd(PIC_INIT);
pic_controller.send_slave_cmd(PIC_INIT);
// tell master pic to add 0x20 to any interrupt number it sends to CPU, while slave pic sends 0x28 + i_number
pic_controller.send_master_data(PIC_OFFSET_MASTER);
pic_controller.send_slave_data(PIC_OFFSET_SLAVE);
// set the interrupt vectoring to cascade and tell master that there is a slave PIC at IRQ2
pic_controller.send_master_data(ICW1_INTERVAL4);
pic_controller.send_slave_data(ICW1_SINGLE);
// set the PICs to work in 8086 mode
pic_controller.send_master_data(ICW1_8086);
pic_controller.send_slave_data(ICW1_8086);
// send 0s
pic_controller.send_master_data(DEFAULT_MASK);
pic_controller.send_slave_data(DEFAULT_MASK);
// tell the cpu to use the table
interrupt_desc_table_pointerdata idt_ptr;
//set the size
idt_ptr.table_size = N_ENTRIES * sizeof(_gate_descriptor) - 1;
// set the base address
idt_ptr.base_addr = (uint32_t)interrupt_desc_table;
// use lidt instruction to load the table
// the cpu will map interrupts to the table
asm volatile("lidt %0" : : "m" (idt_ptr));
// issue debug print
printf_boot_message(" 2: Created Interrupt Desc Table...\n");
}
// define the destructor of the class
InterruptHandler::~InterruptHandler()
{
}
// function to make entries in the IDT
// takes the interrupt number as an index, the segment offset it used to specify which memory segment to use
// a pointer to the function to call, the flags and access level.
void InterruptHandler::create_entry(uint8_t i_number, uint16_t segment_desc_offset, void (*isr)(), uint8_t priv_lvl, uint8_t desc_type)
{
// set the i_number'th entry to the given params
// take the lower bits of the pointer
interrupt_desc_table[i_number].handler_lower_bits = ((uint32_t)isr) & 0xFFFF;
// take the upper bits
interrupt_desc_table[i_number].handler_upper_bits = (((uint32_t)isr) >> 16) & 0xFFFF;
// calculate the privilage byte, setting the correct bits
interrupt_desc_table[i_number].priv_lvl = 0x80 | ((priv_lvl & 3) << 5) | desc_type;
interrupt_desc_table[i_number].segment_desc_offset = segment_desc_offset;
// reserved byte is always 0
interrupt_desc_table[i_number].reserved_byte = 0;
}
// need a function to block or ignore any requests
// that we dont want to service. Requests could be caused
// by devices we haven't yet configured when testing the os.
void InterruptHandler::block_request()
{
// do nothing
}
// function to tell the CPU to send interrupts
// to this table
void InterruptHandler::set_active()
{
// call sti assembly to start interrup poling at the CPU level
asm volatile("sti"); // <- calling this crashes the kernel
// issue debug print
printf_boot_message(" 4: Activated sys interrupts...\n");
}
And here is the code for my GDT, I followed the os dev wiki guide for this:
#include "global_desc_table.h"
/**
* A code segment is identified by flag 0x9A, cannot write to a code segment
* while a data segment is identified by flag 0x92
*
* Based on the C code present on OSDEV Wiki
*/
GlobalDescriptorTable::GlobalDescriptorTable() : nullSegmentSelector(0, 0, 0),
unusedSegmentSelector(0, 0, 0),
codeSegmentSelector(0, 64*1024*1024, 0x9A),
dataSegmentSelector(0, 64*1024*1024, 0x92)
{
//8 bytes defined, but processor expects 6 bytes only
uint32_t i[2];
//first 4 bytes is address of table
i[0] = (uint32_t)this;
//second 4 bytes, the high bytes, are size of global desc table
i[1] = sizeof(GlobalDescriptorTable) << 16;
// tell processor to use this table using its ldgt function
asm volatile("lgdt (%0)" : : "p" (((uint8_t *) i) + 2));
// issue debug print
printf_boot_message(" 1: Created Global Desc Table...\n");
}
// function to get the offset of the datasegment selector
uint16_t GlobalDescriptorTable::DataSegmentSelector()
{
// calculate the offset by subtracting the table's address from the datasegment's address
return (uint8_t *) &dataSegmentSelector - (uint8_t*)this;
}
// function to get the offset of the code segment
uint16_t GlobalDescriptorTable::CodeSegmentSelector()
{
// calculate the offset by subtracting the table's address from the code segment's address
return (uint8_t *) &codeSegmentSelector - (uint8_t*)this;
}
// default destructor
GlobalDescriptorTable::~GlobalDescriptorTable()
{
}
/**
* The constructor to create a new entry segment, set the flags, determine the formatting for the limit, and set the base
*/
GlobalDescriptorTable::SegmentDescriptor::SegmentDescriptor(uint32_t base, uint32_t limit, uint8_t flags)
{
uint8_t* target = (uint8_t*)this;
//if 16 bit limit
if (limit <= 65536)
{
// tell processor that this is a 16bit entry
target[6] = 0x40;
} else {
// if the last 12 bits of limit are not 1s
if ((limit & 0xFFF) != 0xFFF)
{
limit = (limit >> 12) - 1;
} else {
limit >>= 12;
}
// indicate that there was a shift of 12 done
target[6] = 0xC0;
}
// set the lower and upper 2 lowest bytes of limit
target[0] = limit & 0xFF;
target[1] = (limit >> 8) & 0xFF;
//the rest of limit must go in lower 4 bit of byte 6, and byte 5
target[6] |= (limit >> 16) & 0xF;
//encode the pointer
target[2] = base & 0xFF;
target[3] = (base >> 8) & 0xFF;
target[4] = (base >> 16) & 0xFF;
target[7] = (base >> 24) & 0xFF;
// set the flags
target[5] = flags;
}
/**
* Define the methods to get the base pointer from an segment and
* the limit for a segment, taken from os wiki
*/
uint32_t GlobalDescriptorTable::SegmentDescriptor::Base()
{
// simply do the reverse of wht was done to place the pointer in
uint8_t* target = (uint8_t*) this;
uint32_t result = target[7];
result = (result << 8) + target[4];
result = (result << 8) + target[3];
result = (result << 8) + target[2];
return result;
}
uint32_t GlobalDescriptorTable::SegmentDescriptor::Limit()
{
uint8_t* target = (uint8_t *)this;
uint32_t result = target[6] & 0xF;
result = (result << 8) + target[1];
result = (result << 8) + target[0];
//check if there was a shift of 12
if (target[6] & 0xC0 == 0xC0)
{
result = (result << 12) & 0xFFF;
}
return result;
}
i[0] = (uint32_t)this;
//second 4 bytes, the high bytes, are size of global desc table
i[1] = sizeof(GlobalDescriptorTable) << 16;
I've had the same problem, just swap the 0 and 1 in between:
i[1] = (uint32_t)this;
//second 4 bytes, the high bytes, are size of global desc table
i[0] = sizeof(GlobalDescriptorTable) << 16;
That's the problem if you are following the same tutorial and I think you do if you came here.
Sometimes due to wrong idtr value also(invalid pointer causing crash)
check the idtr reg value in vbox log
if u load idt in protected mode address of idt shows some wierd changes(shifted left 16bits or some value in lower 16 bit)
try changing pointer according to that(thats how i did) or use lidt in before entering protected mode(this is also tested)
There was a bug in my GDT that forced the kernel to read an invalid pointer from the segment. This caused a seg fault.
I got a quite simple problem today. I have a matrix float gradient[COLS][ROWS]. As you probably know the float type includes 32 bits.
In my code I do 4 different checks on another table. For each of them I want to write in gradient[][] the results.
What I would like to do is write these results on 8 bits in gradient[][].
So the LSB would contain the result of the first Check, the 8 following bits the results of the second Check, and so on.
As for the reason I want to do this, it is because I'm trying to synthetize this code using HLS and make it run on a Xilinx ZedBoard. There is however not much memory available on the FPGA, so instead of storing the results of my 4 functions into 4 different matrix I woul like to store them in the same matrix using bit operations.
I know I can use masks with an AND operator like gradient[][]&0xFF. What I'm not sure however is when and how do I apply this mask ?
As an example here is the code for one of the Checks (sry for the spanish names i didn't write this) :
void FullCheck(float brightness_tab[COLS][ROWS]){
for(int i=0;i<ROWS;i++){
int previous_point = (int)(brightness_tab[0][i]);
for(int j=1;j<COLS-1;j++){
float brightness=brightness_tab[i][j];
int brightnessi=(int)(brightness);
gradient[i][j]=brightnessi- previous_point;
if(!(gradient[i][j]>VALOR_PENDIENTE || gradient[i][j]<-VALOR_PENDIENTE)){
if(!(gradient[i][j] + gradient[i][j-1] >VALOR_PENDIENTE_TRUNCAR || gradient[i][j] + gradient[i][j+1]<-VALOR_PENDIENTE_TRUNCAR)){
gradient[i][j]=0;
}
}
if(j<2 || i<2 || COLS-1 ==i){gradient[i][j]=0;}
previous_point=brightnessi;
}
}
}
Thank you in advance for your answers !
Deducing from your comments, I'll assume that gradient will be declared as an int array.
In your sample code, there are 2 cases for writing something to the matrix. In the first case, you want to write some value, such as this line:
gradient[i][j] = brightnessi - previous_point;
If you want to write some data to a specific byte, the data you want to write should be a 1-byte data itself.
gradient[i][j] = 0; // initialize to all zero bits
int data1 = 0x12; // 1-byte value
gradient[i][j] |= data1; // writing to the 1st byte (LSB)
int data2 = 0x34;
gradient[i][j] |= data2 << 8; // writing to the 2nd byte
int data3 = 0x56;
gradient[i][j] |= data3 << 16; // writing to the 3rd byte
int data4 = 0x78;
gradient[i][j] |= data4 << 24; // writing to the 4th byte
After executing above code, the value of gradient[i][j] will be 0x78563412.
The second case is clearing what you have written before by writing 0, such as this line:
gradient[i][j] = 0;
In this case you can do
gradient[i][j] &= 0xffffff00; // clearing the 1st byte (LSB)
gradient[i][j] &= 0xffff00ff; // clearing the 2nd byte
gradient[i][j] &= 0xff00ffff; // clearing the 3rd byte
gradient[i][j] &= 0x00ffffff; // clearing the 4th byte
You could also do a struct that has the same memory layout
struct Bytes
{
uint8_t a;
uint8_t b;
uint8_t c;
uint8_t d;
} ;
Bytes* g = reinterpret_cast<Bytes*>(&gradient[i][j]);
That way you can access the individual bytes easily like g->a
The dataFile.bin is a binary file with 6-byte records. The first 3
bytes of each record contain the latitude and the last 3 bytes contain
the longitude. Each 24 bit value represents radians multiplied by
0X1FFFFF
This is a task I've been working on. I havent done C++ in years so its taking me way longer than I thought it would -_-. After googling around I saw this algorthim which made sense to me.
int interpret24bitAsInt32(byte[] byteArray) {
int newInt = (
((0xFF & byteArray[0]) << 16) |
((0xFF & byteArray[1]) << 8) |
(0xFF & byteArray[2])
);
if ((newInt & 0x00800000) > 0) {
newInt |= 0xFF000000;
} else {
newInt &= 0x00FFFFFF;
}
return newInt;
}
The problem is a syntax issue I am restricting to working by the way the other guy had programmed this. I am not understanding how I can store the CHAR "data" into an INT. Wouldn't it make more sense if "data" was an Array? Since its receiving 24 integers of information stored into a BYTE.
double BinaryFile::from24bitToDouble(char *data) {
int32_t iValue;
// ****************************
// Start code implementation
// Task: Fill iValue with the 24bit integer located at data.
// The first byte is the LSB.
// ****************************
//iValue +=
// ****************************
// End code implementation
// ****************************
return static_cast<double>(iValue) / FACTOR;
}
bool BinaryFile::readNext(DataRecord &record)
{
const size_t RECORD_SIZE = 6;
char buffer[RECORD_SIZE];
m_ifs.read(buffer,RECORD_SIZE);
if (m_ifs) {
record.latitude = toDegrees(from24bitToDouble(&buffer[0]));
record.longitude = toDegrees(from24bitToDouble(&buffer[3]));
return true;
}
return false;
}
double BinaryFile::toDegrees(double radians) const
{
static const double PI = 3.1415926535897932384626433832795;
return radians * 180.0 / PI;
}
I appreciate any help or hints even if you dont understand a clue or hint will help me alot. I just need to talk to someone.
I am not understanding how I can store the CHAR "data" into an INT.
Since char is a numeric type, there is no problem combining them into a single int.
Since its receiving 24 integers of information stored into a BYTE
It's 24 bits, not bytes, so there are only three integer values that need to be combined.
An easier way of producing the same result without using conditionals is as follows:
int interpret24bitAsInt32(byte[] byteArray) {
return (
(byteArray[0] << 24)
| (byteArray[1] << 16)
| (byteArray[2] << 8)
) >> 8;
}
The idea is to store the three bytes supplied as an input into the upper three bytes of the four-byte int, and then shift it down by one byte. This way the program would sign-extend your number automatically, avoiding conditional execution.
Note on portability: This code is not portable, because it assumes 32-bit integer size. To make it portable use <cstdint> types:
int32_t interpret24bitAsInt32(const std::array<uint8_t,3> byteArray) {
return (
(const_cast<int32_t>(byteArray[0]) << 24)
| (const_cast<int32_t>(byteArray[1]) << 16)
| (const_cast<int32_t>(byteArray[2]) << 8)
) >> 8;
}
It also assumes that the most significant byte of the 24-bit number is stored in the initial element of byteArray, then comes the middle element, and finally the least significant byte.
Note on sign extension: This code automatically takes care of sign extension by constructing the value in the upper three bytes and then shifting it to the right, as opposed to constructing the value in the lower three bytes right away. This additional shift operation ensures that C++ takes care of sign-extending the result for us.
When an unsigned char is casted to an int the higher order bits are filled with 0's
When a signed char is casted to a casted int, the sign bit is extended.
ie:
int x;
char y;
unsigned char z;
y=0xFF
z=0xFF
x=y;
/*x will be 0xFFFFFFFF*/
x=z;
/*x will be 0x000000FF*/
So, your algorithm, uses 0xFF as a mask to remove C' sign extension, ie
0xFF == 0x000000FF
0xABCDEF10 & 0x000000FF == 0x00000010
Then uses bit shifts and logical ands to put the bits in their proper place.
Lastly checks the most significant bit (newInt & 0x00800000) > 0 to decide if completing with 0's or ones the highest byte.
int32_t upperByte = ((int32_t) dataRx[0] << 24);
int32_t middleByte = ((int32_t) dataRx[1] << 16);
int32_t lowerByte = ((int32_t) dataRx[2] << 8);
int32_t ADCdata32 = (((int32_t) (upperByte | middleByte | lowerByte)) >> 8); // Right-shift of signed data maintains signed bit
I come across a very tricky problem with bit manipulation.
As far as I know, the smallest variable size to hold a value is one byte of 8 bits. The bit operations available in C/C++ apply to an entire unit of bytes.
Imagine that I have a map to replace a binary pattern 100100 (6 bits) with a signal 10000 (5 bits). If the 1st byte of input data from a file is 10010001 (8 bits) being stored in a char variable, part of it matches the 6 bit pattern and therefore be replaced by the 5 bit signal to give a result of 1000001 (7 bits).
I can use a mask to manipulate the bits within a byte to get a result of the left most bits to 10000 (5 bit) but the right most 3 bits become very tricky to manipulate. I cannot shift the right most 3 bits of the original data to get the correct result 1000001 (7 bit) followed by 1 padding bit in that char variable that should be filled by the 1st bit of next followed byte of input.
I wonder if C/C++ can actually do this sort of replacement of bit patterns of length that do not fit into a Char (1 byte) variable or even Int (4 bytes). Can C/C++ do the trick or we have to go for other assembly languages that deal with single bits manipulations?
I heard that Power Basic may be able to do the bit-by-bit manipulation better than C/C++.
If time and space are not important then you can convert the bits to a string representation and perform replaces on the string, then convert back when needed. Not an elegant solution but one that works.
<< shiftleft
^ XOR
>> shift right
~ one's complement
Using these operations, you could easily isolate the pieces that you are interested in and compare them as integers.
say the byte 001000100 and you want to check if it contains 1000:
char k = (char)68;
char c = (char)8;
int i = 0;
while(i<5){
if((k<<i)>>(8-3-i) == c){
//do stuff
break;
}
}
This is very sketchy code, just meant to be a demonstration.
I wonder if C/C++ can actually do this
sort of replacement of bit patterns of
length that do not fit into a Char (1
byte) variable or even Int (4 bytes).
What about std::bitset?
Here's a small bit reader class which may suit your needs. Of course, you may want to create a bit writer for your use case.
#include <iostream>
#include <sstream>
#include <cassert>
class BitReader {
public:
typedef unsigned char BitBuffer;
BitReader(std::istream &input) :
input(input), bufferedBits(8) {
}
BitBuffer peekBits(int numBits) {
assert(numBits <= 8);
assert(numBits > 0);
skipBits(0); // Make sure we have a non-empty buffer
return (((input.peek() << 8) | buffer) >> bufferedBits) & ((1 << numBits) - 1);
}
void skipBits(int numBits) {
assert(numBits >= 0);
numBits += bufferedBits;
while (numBits > 8) {
buffer = input.get();
numBits -= 8;
}
bufferedBits = numBits;
}
BitBuffer readBits(int numBits) {
assert(numBits <= 8);
assert(numBits > 0);
BitBuffer ret = peekBits(numBits);
skipBits(numBits);
return ret;
}
bool eof() const {
return input.eof();
}
private:
std::istream &input;
BitBuffer buffer;
int bufferedBits; // How many bits are buffered into 'buffer' (0 = empty)
};
Use a vector<bool> if you can read your data into the vector mostly at once. It may be more difficult to find-and-replace sequences of bits, though.
If I understood your questions correctly, you have an input stream and and output stream and you want to replace the 6bits of the input with 5 in the output - and your output still should be a bit stream?
So, the most important programmer's rule can be applied: Divide et impera!
You should split your component in three parts:
Input Stream converter: Convert every pattern in the input stream to a char array (ring) buffer. If I understood you correctly your input "commands" are 8bit long, so there is nothing special about this.
Do the replacement on the ring buffer in a way that you replace every matching 6-bit pattern with the 5bit one, but "pad" the 5 bit with a leading zero, so the total length is still 8bit.
Write an output handler that reads from the ring buffer and let this output handler write only the 7 LSB to the output stream from each input byte. Of course some bit manipulation is necessary again for this.
If your ring buffer size can be divided by 8 and 7 (= is a multiple of 56) you will have a clean buffer at the end and can start again with 1.
The most simplest way to implement this is to iterate over this 3 steps as long as input data is available.
If a performance really matters and you are running on a multi-core CPU you even could split the steps and 3 threads, but then you must carefully synchronize the access to the ring buffer.
I think the following does what you want.
PATTERN_LEN = 6
PATTERNMASK = 0x3F //6 bits
PATTERN = 0x24 //b100100
REPLACE_LEN = 5
REPLACEMENT = 0x10 //b10000
void compress(uint8* inbits, uint8* outbits, int len)
{
uint16 accumulator=0;
int nbits=0;
uint8 candidate;
while (len--) //for all input bytes
{
//for each bit (msb first)
for (i=7;i<=0;i--)
{
//add 1 bit to accumulator
accumulator<<=1;
accumulator|=(*inbits&(1<<i));
nbits++;
//check for pattern
candidate = accumulator&PATTERNMASK;
if (candidate==PATTERN)
{
//remove pattern
accumulator>>=PATTERN_LEN;
//add replacement
accumulator<<=REPLACE_LEN;
accumulator|=REPLACMENT;
nbits+= (REPLACE_LEN - PATTERN_LEN);
}
}
inbits++;
//move accumulator to output to prevent overflow
while (nbits>8)
{
//copy the highest 8 bits
nbits-=8;
*outbits++ = (accumulator>>nbits)&0xFF;
//clear them from accumulator
accumulator&= ~(0xFF<<nbits);
}
}
//copy remainder of accumulator to output
while (nbits>0)
{
nbits-=8;
*outbits++ = (accumulator>>nbits)&0xFF;
accumulator&= ~(0xFF<<nbits);
}
}
You could use a switch or a loop in the middle to check the candidate against multiple patterns. There might have to be some special handling after doing a replacment to ensure the replacement pattern is not re-checked for matches.
#include <iostream>
#include <cstring>
size_t matchCount(const char* str, size_t size, char pat, size_t bsize) noexcept
{
if (bsize > 8) {
return 0;
}
size_t bcount = 0; // curr bit number
size_t pcount = 0; // curr bit in pattern char
size_t totalm = 0; // total number of patterns matched
const size_t limit = size*8;
while (bcount < limit)
{
auto offset = bcount%8;
char c = str[bcount/8];
c >>= offset;
char tpat = pat >> pcount;
if ((c & 1) == (tpat & 1))
{
++pcount;
if (pcount == bsize)
{
++totalm;
pcount = 0;
}
}
else // mismatch
{
bcount -= pcount; // backtrack
//reset
pcount = 0;
}
++bcount;
}
return totalm;
}
int main(int argc, char** argv)
{
const char* str = "abcdefghiibcdiixyz";
char pat = 'i';
std::cout << "Num matches = " << matchCount(str, 18, pat, 7) << std::endl;
return 0;
}