How to read/write arbitrary bits in C/C++

Question

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?

Answer 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.)

Answer 2

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

Answer 3

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

Answer 4

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)

Answer 5

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.

Answer 6

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);