I have array from serial read, named sensor_buffer. It contains 21 bytes.
gyro_out_X=((sensor_buffer[1]<<8)+sensor_buffer[2]);
gyro_out_Y=((sensor_buffer[3]<<8)+sensor_buffer[4]);
gyro_out_Z=((sensor_buffer[5]<<8)+sensor_buffer[6]);
acc_out_X=((sensor_buffer[7]<<8)+sensor_buffer[8]);
acc_out_Y=((sensor_buffer[9]<<8)+sensor_buffer[10]);
acc_out_Z=((sensor_buffer[11]<<8)+sensor_buffer[12]);
HMC_xo=((sensor_buffer[13]<<8)+sensor_buffer[14]);
HMC_yo=((sensor_buffer[15]<<8)+sensor_buffer[16]);
HMC_zo=((sensor_buffer[17]<<8)+sensor_buffer[18]);
adc_pressure=(((long)sensor_buffer[19]<<16)+(sensor_buffer[20]<<8)+sensor_buffer[21]);
What does this line do:
variable = (array_var<<8) + next_array_var
What effect does it have on the 8 bits?
<<8 ?
UPDATE:
Any example in another language (java, processing)?
Example for processing: (why use H like header?).
/*
* ReceiveBinaryData_P
*
* portIndex must be set to the port connected to the Arduino
*/
import processing.serial.*;
Serial myPort; // Create object from Serial class
short portIndex = 1; // select the com port, 0 is the first port
char HEADER = 'H';
int value1, value2; // Data received from the serial port
void setup()
{
size(600, 600);
// Open whatever serial port is connected to Arduino.
String portName = Serial.list()[portIndex];
println(Serial.list());
println(" Connecting to -> " + Serial.list()[portIndex]);
myPort = new Serial(this, portName, 9600);
}
void draw()
{
// read the header and two binary *(16 bit) integers:
if ( myPort.available() >= 5) // If at least 5 bytes are available,
{
if( myPort.read() == HEADER) // is this the header
{
value1 = myPort.read(); // read the least significant byte
value1 = myPort.read() * 256 + value1; // add the most significant byte
value2 = myPort.read(); // read the least significant byte
value2 = myPort.read() * 256 + value2; // add the most significant byte
println("Message received: " + value1 + "," + value2);
}
}
background(255); // Set background to white
fill(0); // set fill to black
// draw rectangle with coordinates based on the integers received from Arduino
rect(0, 0, value1,value2);
}
Your code has the same pattern:
value = (partial_value << 8) | (other_partial_value)
Your array has data stored in 8 bit bytes, but the values are in 16 bit bytes. Each of your data points are two bytes, with the most significant byte stored first in your array. This pattern simply builds the full 16 bit value by shifting the most significant byte 8 bits to the left, then OR'ing the least significant byte into the lower 8 bits.
Its a shift operator. It shifts the bits in you variable to the left by 8. Shift by 1 bit to the left is equivalent to multiplying by two (shifting to the right divides by 2). So essentially <<8 is equivalent to multiplying by 2^8.
See here for a list of C++ operators and what they do:
http://en.wikipedia.org/wiki/C%2B%2B_operators
<< is the left bit-shift operator, the result is the bits from the first operand moved to the left, with 0 bits filling in from the right.
A simple example in pseudocode:
x = 10000101;
x = x << 3;
now x is "00101000"
Study the Bitwise operation article on wikipedia for an introduction.
This is just a bit shift operator. If is basically taking the value and shitfing the bits a places to the left. This is equivalent to multiplying the value by 2^8. The code looks like its reading in 2 bytes of the array and creating a 16 bit integer from each pair.
It seems that sensor_buffer is a matrix of chars.
In order to get your value, e.g. gyro_out_X you have to combine sensor_buffer[1] and sensor_buffer[2],
where
sensor_buffer[1] holds the most significant byte and
sensor_buffer[2] holds the least significant byte
in that case
int gyro_out_X=((sensor_buffer[1]<<8)+sensor_buffer[2]);
combines the two bytes:
if sensor_buffer[1] is 0xFF
and sensor_buffer[2] is 0x10
then gyro_out_X is 0xFF10
It shifts the bits 8 places to the left, eg:
0000000001000100 << 8 = 0100010000000000
0000000001000100 << 1 =
0000000010001000 << 1 =
0000000100010000 << 1 =
0000001000100000 << 1 =
0000010001000000 << 1 =
0000100010000000 << 1 =
0001000100000000 << 1 =
0010001000000000 << 1 =
0100010000000000
Related
this is a test program to try to get my bitwise shift operations to work. I am hoping to add this to my cache simulator program, but I can't even get this part to work. My plan is to use bit shift (<<) and (>>) to isolate parts of a given memory address (tag, set, word, etc.) but it seems that in shifting the bits back right, it is filling with the values which were previously there, rather than with 0's. Here is the program first.
#include<iostream>
#include <cmath>
struct Address{
unsigned int tag;
unsigned int r;
unsigned int word;
};
int main(){
unsigned int tempAddress = 27; //0011011
int ramSize = 128;
int cacheSize = 64;
int blockSize = 8;
int cacheLines = cacheSize / blockSize;
int addressLength = log(ramSize)/log(2);
int wordBits = log(blockSize)/log(2);
int rBits = log(cacheLines)/log(2);
int tagBits = addressLength - (wordBits + rBits);
struct Address address;
address.tag = tempAddress >> (rBits + wordBits);
address.r = tempAddress << (tagBits) >> (tagBits + wordBits);
address.word = tempAddress << (rBits + tagBits) >> (rBits + tagBits);
std::cout << "tag is: " << address.tag << "\n";
std::cout << "r is: " << address.r << "\n";
std::cout << "word is: " << address.word << "\n";
}
I've found that when my tempAddress is [0-7] it works fine because binary 7 only affects the first 3 bits.
Similarly, when it is [8-63], tag and r are correct because 63 affects the first 6 bits.
Upon testing many addresses, I've found that when shifting right after shifting left, the bits are being replaced with what they were before, rather than with 0s as I think they should be.
(r is the part that is in the middle. I am calling it r because in direct mapping it is called line, and in set-associative mapping it is called set)
EDIT:
As someone pointed out, expected and produced outcome would be helpful. I'd first like to keep the cache size, ram size, and block size constant, and only change the address.
So, given tempAddress = 27(0011011 in binary), word should be 011 (first 3 bits), r should be 011 (next 3), and set should be 0 (remaining bits).
Output is this:
tag is: 0
r is: 3
word is: 27
I've found this to be the trend if every address between 0 and 63(inclusive) that tag and r are correct, but word is equal to address.
Now, for address = 65(1000001) Expected:
tag is: 1
r is: 0
word is: 1
Output:
tag is: 1
r is: 8
word is: 65
With these ram, cache, and block sizes, to find r, I am left shifting 1 time, and right shifting 4 times. To find word, I am left shifting 4 times, then right shifting 4 times. As I understand it, when left shifting the bits on the right are filled with 0, and when right shifting an unsigned integer, the bits on the left are filled with 0. My thought was that if I left shift until I only have the bits I need, then right shift them back to the first bits, I will have the correct values. However, consistent throughout numerous addresses, after left shifting then right shifting, the places that had 1s still do. That is why word is always equal to address, because I am shifting 4 bits both left then right. And r is always equal to the 7 bits shifted right 3 times (because I go left 1 then right 4). Am I misunderstanding how bitwise shifting works?
If I have a char array A, I use it to store hex
A = "0A F5 6D 02" size=11
The binary representation of this char array is:
00001010 11110101 01101101 00000010
I want to ask is there any function can random flip the bit?
That is:
if the parameter is 5
00001010 11110101 01101101 00000010
-->
10001110 11110001 01101001 00100010
it will random choose 5 bit to flip.
I am trying make this hex data to binary data and use bitmask method to achieve my requirement. Then turn it back to hex. I am curious is there any method to do this job more quickly?
Sorry, my question description is not clear enough. In simply, I have some hex data, and I want to simulate bit error in these data. For example, if I have 5 byte hex data:
"FF00FF00FF"
binary representation is
"1111111100000000111111110000000011111111"
If the bit error rate is 10%. Then I want to make these 40 bits have 4 bits error. One extreme random result: error happened in the first 4 bit:
"0000111100000000111111110000000011111111"
First of all, find out which char the bit represents:
param is your bit to flip...
char *byteToWrite = &A[sizeof(A) - (param / 8) - 1];
So that will give you a pointer to the char at that array offset (-1 for 0 array offset vs size)
Then get modulus (or more bit shifting if you're feeling adventurous) to find out which bit in here to flip:
*byteToWrite ^= (1u << param % 8);
So that should result for a param of 5 for the byte at A[10] to have its 5th bit toggled.
store the values of 2^n in an array
generate a random number seed
loop through x times (in this case 5) and go data ^= stored_values[random_num]
Alternatively to storing the 2^n values in an array, you could do some bit shifting to a random power of 2 like:
data ^= (1<<random%7)
Reflecting the first comment, you really could just write out that line 5 times in your function and avoid the overhead of a for loop entirely.
You have 32 bit number. You can treate the bits as parts of hte number and just xor this number with some random 5-bits-on number.
int count_1s(int )
{
int m = 0x55555555;
int r = (foo&m) + ((foo>>>1)&m);
m = 0x33333333;
r = (r&m) + ((r>>>2)&m);
m = 0x0F0F0F0F;
r = (r&m) + ((r>>>4)&m);
m = 0x00FF00FF;
r = (r&m) + ((r>>>8)&m);
m = 0x0000FFFF;
return r = (r&m) + ((r>>>16)&m);
}
void main()
{
char input[] = "0A F5 6D 02";
char data[4] = {};
scanf("%2x %2x %2x %2x", &data[0], &data[1], &data[2], &data[3]);
int *x = reinterpret_cast<int*>(data);
int y = rand();
while(count_1s(y) != 5)
{
y = rand(); // let's have this more random
}
*x ^= y;
printf("%2x %2x %2x %2x" data[0], data[1], data[2], data[3]);
return 0;
}
I see no reason to convert the entire string back and forth from and to hex notation. Just pick a random character out of the hex string, convert this to a digit, change it a bit, convert back to hex character.
In plain C:
#include <stdio.h>
#include <stdlib.h>
#include <sys/time.h>
int main (void)
{
char *hexToDec_lookup = "0123456789ABCDEF";
char hexstr[] = "0A F5 6D 02";
/* 0. make sure we're fairly random */
srand(time(0));
/* 1. loop 5 times .. */
int i;
for (i=0; i<5; i++)
{
/* 2. pick a random hex digit
we know it's one out of 8, grouped per 2 */
int hexdigit = rand() & 7;
hexdigit += (hexdigit>>1);
/* 3. convert the digit to binary */
int hexvalue = hexstr[hexdigit] > '9' ? hexstr[hexdigit] - 'A'+10 : hexstr[hexdigit]-'0';
/* 4. flip a random bit */
hexvalue ^= 1 << (rand() & 3);
/* 5. write it back into position */
hexstr[hexdigit] = hexToDec_lookup[hexvalue];
printf ("[%s]\n", hexstr);
}
return 0;
}
It might even be possible to omit the convert-to-and-from-ASCII steps -- flip a bit in the character string, check if it's still a valid hex digit and if necessary, adjust.
First randomly chose x positions (each position consist of array index and the bit position).
Now if you want to flip ith bit from right for a number n. Find the remainder of n by 2n as :
code:
int divisor = (2,i);
int remainder = n % divisor;
int quotient = n / divisor;
remainder = (remainder == 0) ? 1 : 0; // flip the remainder or the i th bit from right.
n = divisor * quotient + remainder;
Take mod 8 of input(5%8)
Shift 0x80 to right by input value (e.g 5)
XOR this value with (input/8)th element of your character array.
code:
void flip_bit(int bit)
{
Array[bit/8] ^= (0x80>>(bit%8));
}
I have a vector<char> and I want to be able to get an unsigned integer from a range of bits within the vector. E.g.
And I can't seem to be able to write the correct operations to get the desired output. My intended algorithm goes like this:
& the first byte with (0xff >> unused bits in byte on the left)
<< the result left the number of output bytes * number of bits in a byte
| this with the final output
For each subsequent byte:
<< left by the (byte width - index) * bits per byte
| this byte with the final output
| the final byte (not shifted) with the final output
>> the final output by the number of unused bits in the byte on the right
And here is my attempt at coding it, which does not give the correct result:
#include <vector>
#include <iostream>
#include <cstdint>
#include <bitset>
template<class byte_type = char>
class BitValues {
private:
std::vector<byte_type> bytes;
public:
static const auto bits_per_byte = 8;
BitValues(std::vector<byte_type> bytes) : bytes(bytes) {
}
template<class return_type>
return_type get_bits(int start, int end) {
auto byte_start = (start - (start % bits_per_byte)) / bits_per_byte;
auto byte_end = (end - (end % bits_per_byte)) / bits_per_byte;
auto byte_width = byte_end - byte_start;
return_type value = 0;
unsigned char first = bytes[byte_start];
first &= (0xff >> start % 8);
return_type first_wide = first;
first_wide <<= byte_width;
value |= first_wide;
for(auto byte_i = byte_start + 1; byte_i <= byte_end; byte_i++) {
auto byte_offset = (byte_width - byte_i) * bits_per_byte;
unsigned char next_thin = bytes[byte_i];
return_type next_byte = next_thin;
next_byte <<= byte_offset;
value |= next_byte;
}
value >>= (((byte_end + 1) * bits_per_byte) - end) % bits_per_byte;
return value;
}
};
int main() {
BitValues<char> bits(std::vector<char>({'\x78', '\xDA', '\x05', '\x5F', '\x8A', '\xF1', '\x0F', '\xA0'}));
std::cout << bits.get_bits<unsigned>(15, 29) << "\n";
return 0;
}
(In action: http://coliru.stacked-crooked.com/a/261d32875fcf2dc0)
I just can't seem to wrap my head around these bit manipulations, and I find debugging very difficult! If anyone can correct the above code, or help me in any way, it would be much appreciated!
Edit:
My bytes are 8 bits long
The integer to return could be 8,16,32 or 64 bits wside
The integer is stored in big endian
You made two primary mistakes. The first is here:
first_wide <<= byte_width;
You should be shifting by a bit count, not a byte count. Corrected code is:
first_wide <<= byte_width * bits_per_byte;
The second mistake is here:
auto byte_offset = (byte_width - byte_i) * bits_per_byte;
It should be
auto byte_offset = (byte_end - byte_i) * bits_per_byte;
The value in parenthesis needs to be the number of bytes to shift right by, which is also the number of bytes byte_i is away from the end. The value byte_width - byte_i has no semantic meaning (one is a delta, the other is an index)
The rest of the code is fine. Though, this algorithm has two issues with it.
First, when using your result type to accumulate bits, you assume you have room on the left to spare. This isn't the case if there are set bits near the right boundry and the choice of range causes the bits to be shifted out. For example, try running
bits.get_bits<uint16_t>(11, 27);
You'll get the result 42 which corresponds to the bit string 00000000 00101010 The correct result is 53290 with the bit string 11010000 00101010. Notice how the rightmost 4 bits got zeroed out. This is because you start off by overshifting your value variable, causing those four bits to be shifted out of the variable. When shifting back at the end, this results in the bits being zeroed out.
The second problem has to do with the right shift at the end. If the rightmost bit of the value variable happens to be a 1 before the right shift at the end, and the template parameter is a signed type, then the right shift that is done is an 'arithmetic' right shift, which causes bits on the right to be 1-filled, leaving you with an incorrect negative value.
Example, try running:
bits.get_bits<int16_t>(5, 21);
The expected result should be 6976 with the bit string 00011011 01000000, but the current implementation returns -1216 with the bit string 11111011 01000000.
I've put my implementation of this below which builds the bit string from the right to the left, placing bits in their correct positions to start with so that the above two problems are avoided:
template<class ReturnType>
ReturnType get_bits(int start, int end) {
int max_bits = kBitsPerByte * sizeof(ReturnType);
if (end - start > max_bits) {
start = end - max_bits;
}
int inclusive_end = end - 1;
int byte_start = start / kBitsPerByte;
int byte_end = inclusive_end / kBitsPerByte;
// Put in the partial-byte on the right
uint8_t first = bytes_[byte_end];
int bit_offset = (inclusive_end % kBitsPerByte);
first >>= 7 - bit_offset;
bit_offset += 1;
ReturnType ret = 0 | first;
// Add the rest of the bytes
for (int i = byte_end - 1; i >= byte_start; i--) {
ReturnType tmp = (uint8_t) bytes_[i];
tmp <<= bit_offset;
ret |= tmp;
bit_offset += kBitsPerByte;
}
// Mask out the partial byte on the left
int shift_amt = (end - start);
if (shift_amt < max_bits) {
ReturnType mask = (1 << shift_amt) - 1;
ret &= mask;
}
}
There is one thing you certainly missed I think: the way you index the bits in the vector is different from what you have been given in the problem. I.e. with algorithm you outlined, the order of the bits will be like 7 6 5 4 3 2 1 0 | 15 14 13 12 11 10 9 8 | 23 22 21 .... Frankly, I didn't read through your whole algorithm, but this one was missed in the very first step.
Interesting problem. I've done similar, for some systems work.
Your char is 8 bits wide? Or 16? How big is your integer? 32 or 64?
Ignore the vector complexity for a minute.
Think about it as just an array of bits.
How many bits do you have? You have 8*number of chars
You need to calculate a starting char, number of bits to extract, ending char, number of bits there, and number of chars in the middle.
You will need bitwise-and & for the first partial char
you will need bitwise-and & for the last partial char
you will need left-shift << (or right-shift >>), depending upon which order you start from
what is the endian-ness of your Integer?
At some point you will calculate an index into your array that is bitindex/char_bit_width, you gave the value 171 as your bitindex, and 8 as your char_bit_width, so you will end up with these useful values calculated:
171/8 = 23 //location of first byte
171%8 = 3 //bits in first char/byte
8 - 171%8 = 5 //bits in last char/byte
sizeof(integer) = 4
sizeof(integer) + ( (171%8)>0?1:0 ) // how many array positions to examine
Some assembly required...
I'm having a little trouble grabbing n bits from a byte.
I have an unsigned integer. Let's say our number in hex is 0x2A, which is 42 in decimal. In binary it looks like this: 0010 1010. How would I grab the first 5 bits which are 00101 and the next 3 bits which are 010, and place them into separate integers?
If anyone could help me that would be great! I know how to extract from one byte which is to simply do
int x = (number >> (8*n)) & 0xff // n being the # byte
which I saw on another post on stack overflow, but I wasn't sure on how to get separate bits out of the byte. If anyone could help me out, that'd be great! Thanks!
Integers are represented inside a machine as a sequence of bits; fortunately for us humans, programming languages provide a mechanism to show us these numbers in decimal (or hexadecimal), but that does not alter their internal representation.
You should review the bitwise operators &, |, ^ and ~ as well as the shift operators << and >>, which will help you understand how to solve problems like this.
The last 3 bits of the integer are:
x & 0x7
The five bits starting from the eight-last bit are:
x >> 3 // all but the last three bits
& 0x1F // the last five bits.
"grabbing" parts of an integer type in C works like this:
You shift the bits you want to the lowest position.
You use & to mask the bits you want - ones means "copy this bit", zeros mean "ignore"
So, in you example. Let's say we have a number int x = 42;
first 5 bits:
(x >> 3) & ((1 << 5)-1);
or
(x >> 3) & 31;
To fetch the lower three bits:
(x >> 0) & ((1 << 3)-1)
or:
x & 7;
Say you want hi bits from the top, and lo bits from the bottom. (5 and 3 in your example)
top = (n >> lo) & ((1 << hi) - 1)
bottom = n & ((1 << lo) - 1)
Explanation:
For the top, first get rid of the lower bits (shift right), then mask the remaining with an "all ones" mask (if you have a binary number like 0010000, subtracting one results 0001111 - the same number of 1s as you had 0-s in the original number).
For the bottom it's the same, just don't have to care with the initial shifting.
top = (42 >> 3) & ((1 << 5) - 1) = 5 & (32 - 1) = 5 = 00101b
bottom = 42 & ((1 << 3) - 1) = 42 & (8 - 1) = 2 = 010b
You could use bitfields for this. Bitfields are special structs where you can specify variables in bits.
typedef struct {
unsigned char a:5;
unsigned char b:3;
} my_bit_t;
unsigned char c = 0x42;
my_bit_t * n = &c;
int first = n->a;
int sec = n->b;
Bit fields are described in more detail at http://www.cs.cf.ac.uk/Dave/C/node13.html#SECTION001320000000000000000
The charm of bit fields is, that you do not have to deal with shift operators etc. The notation is quite easy. As always with manipulating bits there is a portability issue.
int x = (number >> 3) & 0x1f;
will give you an integer where the last 5 bits are the 8-4 bits of number and zeros in the other bits.
Similarly,
int y = number & 0x7;
will give you an integer with the last 3 bits set the last 3 bits of number and the zeros in the rest.
just get rid of the 8* in your code.
int input = 42;
int high3 = input >> 5;
int low5 = input & (32 - 1); // 32 = 2^5
bool isBit3On = input & 4; // 4 = 2^(3-1)
Assuming I have a byte b with the binary value of 11111111
How do I for example read a 3 bit integer value starting at the second bit or write a four bit integer value starting at the fifth bit?
Some 2+ years after I asked this question I'd like to explain it the way I'd want it explained back when I was still a complete newb and would be most beneficial to people who want to understand the process.
First of all, forget the "11111111" example value, which is not really all that suited for the visual explanation of the process. So let the initial value be 10111011 (187 decimal) which will be a little more illustrative of the process.
1 - how to read a 3 bit value starting from the second bit:
___ <- those 3 bits
10111011
The value is 101, or 5 in decimal, there are 2 possible ways to get it:
mask and shift
In this approach, the needed bits are first masked with the value 00001110 (14 decimal) after which it is shifted in place:
___
10111011 AND
00001110 =
00001010 >> 1 =
___
00000101
The expression for this would be: (value & 14) >> 1
shift and mask
This approach is similar, but the order of operations is reversed, meaning the original value is shifted and then masked with 00000111 (7) to only leave the last 3 bits:
___
10111011 >> 1
___
01011101 AND
00000111
00000101
The expression for this would be: (value >> 1) & 7
Both approaches involve the same amount of complexity, and therefore will not differ in performance.
2 - how to write a 3 bit value starting from the second bit:
In this case, the initial value is known, and when this is the case in code, you may be able to come up with a way to set the known value to another known value which uses less operations, but in reality this is rarely the case, most of the time the code will know neither the initial value, nor the one which is to be written.
This means that in order for the new value to be successfully "spliced" into byte, the target bits must be set to zero, after which the shifted value is "spliced" in place, which is the first step:
___
10111011 AND
11110001 (241) =
10110001 (masked original value)
The second step is to shift the value we want to write in the 3 bits, say we want to change that from 101 (5) to 110 (6)
___
00000110 << 1 =
___
00001100 (shifted "splice" value)
The third and final step is to splice the masked original value with the shifted "splice" value:
10110001 OR
00001100 =
___
10111101
The expression for the whole process would be: (value & 241) | (6 << 1)
Bonus - how to generate the read and write masks:
Naturally, using a binary to decimal converter is far from elegant, especially in the case of 32 and 64 bit containers - decimal values get crazy big. It is possible to easily generate the masks with expressions, which the compiler can efficiently resolve during compilation:
read mask for "mask and shift": ((1 << fieldLength) - 1) << (fieldIndex - 1), assuming that the index at the first bit is 1 (not zero)
read mask for "shift and mask": (1 << fieldLength) - 1 (index does not play a role here since it is always shifted to the first bit
write mask : just invert the "mask and shift" mask expression with the ~ operator
How does it work (with the 3bit field beginning at the second bit from the examples above)?
00000001 << 3
00001000 - 1
00000111 << 1
00001110 ~ (read mask)
11110001 (write mask)
The same examples apply to wider integers and arbitrary bit width and position of the fields, with the shift and mask values varying accordingly.
Also note that the examples assume unsigned integer, which is what you want to use in order to use integers as portable bit-field alternative (regular bit-fields are in no way guaranteed by the standard to be portable), both left and right shift insert a padding 0, which is not the case with right shifting a signed integer.
Even easier:
Using this set of macros (but only in C++ since it relies on the generation of member functions):
#define GETMASK(index, size) ((((size_t)1 << (size)) - 1) << (index))
#define READFROM(data, index, size) (((data) & GETMASK((index), (size))) >> (index))
#define WRITETO(data, index, size, value) ((data) = (((data) & (~GETMASK((index), (size)))) | (((value) << (index)) & (GETMASK((index), (size))))))
#define FIELD(data, name, index, size) \
inline decltype(data) name() const { return READFROM(data, index, size); } \
inline void set_##name(decltype(data) value) { WRITETO(data, index, size, value); }
You could go for something as simple as:
struct A {
uint bitData;
FIELD(bitData, one, 0, 1)
FIELD(bitData, two, 1, 2)
};
And have the bit fields implemented as properties you can easily access:
A a;
a.set_two(3);
cout << a.two();
Replace decltype with gcc's typeof pre-C++11.
You need to shift and mask the value, so for example...
If you want to read the first two bits, you just need to mask them off like so:
int value = input & 0x3;
If you want to offset it you need to shift right N bits and then mask off the bits you want:
int value = (intput >> 1) & 0x3;
To read three bits like you asked in your question.
int value = (input >> 1) & 0x7;
just use this and feelfree:
#define BitVal(data,y) ( (data>>y) & 1) /** Return Data.Y value **/
#define SetBit(data,y) data |= (1 << y) /** Set Data.Y to 1 **/
#define ClearBit(data,y) data &= ~(1 << y) /** Clear Data.Y to 0 **/
#define TogleBit(data,y) (data ^=BitVal(y)) /** Togle Data.Y value **/
#define Togle(data) (data =~data ) /** Togle Data value **/
for example:
uint8_t number = 0x05; //0b00000101
uint8_t bit_2 = BitVal(number,2); // bit_2 = 1
uint8_t bit_1 = BitVal(number,1); // bit_1 = 0
SetBit(number,1); // number = 0x07 => 0b00000111
ClearBit(number,2); // number =0x03 => 0b0000011
You have to do a shift and mask (AND) operation.
Let b be any byte and p be the index (>= 0) of the bit from which you want to take n bits (>= 1).
First you have to shift right b by p times:
x = b >> p;
Second you have to mask the result with n ones:
mask = (1 << n) - 1;
y = x & mask;
You can put everything in a macro:
#define TAKE_N_BITS_FROM(b, p, n) ((b) >> (p)) & ((1 << (n)) - 1)
"How do I for example read a 3 bit integer value starting at the second bit?"
int number = // whatever;
uint8_t val; // uint8_t is the smallest data type capable of holding 3 bits
val = (number & (1 << 2 | 1 << 3 | 1 << 4)) >> 2;
(I assumed that "second bit" is bit #2, i. e. the third bit really.)
To read bytes use std::bitset
const int bits_in_byte = 8;
char myChar = 's';
cout << bitset<sizeof(myChar) * bits_in_byte>(myChar);
To write you need to use bit-wise operators such as & ^ | & << >>. make sure to learn what they do.
For example to have 00100100 you need to set the first bit to 1, and shift it with the << >> operators 5 times. if you want to continue writing you just continue to set the first bit and shift it. it's very much like an old typewriter: you write, and shift the paper.
For 00100100: set the first bit to 1, shift 5 times, set the first bit to 1, and shift 2 times:
const int bits_in_byte = 8;
char myChar = 0;
myChar = myChar | (0x1 << 5 | 0x1 << 2);
cout << bitset<sizeof(myChar) * bits_in_byte>(myChar);
int x = 0xFF; //your number - 11111111
How do I for example read a 3 bit integer value starting at the second bit
int y = x & ( 0x7 << 2 ) // 0x7 is 111
// and you shift it 2 to the left
If you keep grabbing bits from your data, you might want to use a bitfield. You'll just have to set up a struct and load it with only ones and zeroes:
struct bitfield{
unsigned int bit : 1
}
struct bitfield *bitstream;
then later on load it like this (replacing char with int or whatever data you are loading):
long int i;
int j, k;
unsigned char c, d;
bitstream=malloc(sizeof(struct bitfield)*charstreamlength*sizeof(char));
for (i=0; i<charstreamlength; i++){
c=charstream[i];
for(j=0; j < sizeof(char)*8; j++){
d=c;
d=d>>(sizeof(char)*8-j-1);
d=d<<(sizeof(char)*8-1);
k=d;
if(k==0){
bitstream[sizeof(char)*8*i + j].bit=0;
}else{
bitstream[sizeof(char)*8*i + j].bit=1;
}
}
}
Then access elements:
bitstream[bitpointer].bit=...
or
...=bitstream[bitpointer].bit
All of this is assuming are working on i86/64, not arm, since arm can be big or little endian.