Bitwise shifting array of char's

Question

I have got an array of chars that I\'m trying to bitwise shift right >>, then & with another array. I think I have got the wrong idea of

Answer 1

If you want to perform operations such as shifting / OR / XOR / AND / etc.. on arrays, you should perform it in a loop, you cannot perform it directly on the array.

Answer 2

I know this is old topic but i was not satisfied with the answers available, here is something i wrote recently which allows you to specify the amount of bits you can shift by and also there is simple XOR encryption in it.

//https://github.com/ashvin-bhuttoo/CryptoTest/blob/master/CryptoTest/Crypto.cpp
//CRYPTO CONFIGURATION PARAMETERS
#define BIT_SHIFT 3
#define XOR_KEY 0x3C
#define ENABLE_XOR_VARIANCE true
////////////////////////////////

int get_rs_mask(int shift)
{
    switch (shift)
    {
    case 0:
        return 0x00;
    case 1:
        return 0x01;
    case 2:
        return 0x03;
    case 3:
        return 0x07;
    case 4:
        return 0x0F;
    case 5:
        return 0x1F;
    case 6:
        return 0x3F;
    case 7:
        return 0x7F;
    default:
        throw "get_rs_mask -> Error, shift argument outside legal range 0-7";
    }
}

void shift_right(char* buf, int msg_len, int shift)
{
    unsigned char tmp = 0x00, tmp2 = 0x00;
    for (int k = 0; k <= msg_len; k++)
    {
        if (k == 0)
        {
            tmp = buf[k];
            buf[k] >>= shift;
        }
        else
        {
            tmp2 = buf[k];
            buf[k] >>= shift;
            buf[k] |= ((tmp & get_rs_mask(shift)) << (8 - shift));

            if (k != msg_len)
                tmp = tmp2;
        }
    }
}

int get_ls_mask(int shift)
{
    switch (shift)
    {
    case 0:
        return 0x00;
    case 1:
        return 0x80;
    case 2:
        return 0xC0;
    case 3:
        return 0xE0;
    case 4:
        return 0xF0;
    case 5:
        return 0xF8;
    case 6:
        return 0xFC;
    case 7:
        return 0xFE;
    default:
        throw "get_ls_mask -> Error, shift argument outside legal range 0-7";
    }
}

void shift_left(char* buf, int msg_len, int shift)
{
    char tmp = 0x00, tmp2 = 0x00;
    for (int k = msg_len; k >= 0; k--)
    {
        if (k == msg_len)
        {
            tmp = buf[k];
            buf[k] <<= shift;
        }
        else
        {
            tmp2 = buf[k];
            buf[k] <<= shift;
            buf[k] |= ((tmp & get_ls_mask(shift)) >> (8 - shift));

            tmp = tmp2;
        }
    }
}

void crypt(char* buf, int msg_len, bool decrypt = false)
{
    if (!decrypt)
    {
        shift_right(buf, msg_len, BIT_SHIFT);
        for (int k = 0; k < msg_len; k++)
        {
            buf[k] = buf[k] ^ XOR_KEY ^ k * (ENABLE_XOR_VARIANCE ? 2 : 0);
        }
        buf[msg_len] = '\0';
    }
    else
    {
        for (int k = 0; k < msg_len; k++)
        {
            buf[k] = buf[k] ^ XOR_KEY ^ k * (ENABLE_XOR_VARIANCE ? 2 : 0);
        }
        shift_left(buf, (msg_len)-1, BIT_SHIFT);
    }
}

Answer 3

/** Shift an array right.
 * @param ar The array to shift.
 * @param size The number of array elements.
 * @param shift The number of bits to shift.
 */
void shift_right(unsigned char *ar, int size, int shift)
{
    int carry = 0;                              // Clear the initial carry bit.
    while (shift--) {                           // For each bit to shift ...
        for (int i = size - 1; i >= 0; --i) {   // For each element of the array from high to low ...
            int next = (ar[i] & 1) ? 0x80 : 0;  // ... if the low bit is set, set the carry bit.
            ar[i] = carry | (ar[i] >> 1);       // Shift the element one bit left and addthe old carry.
            carry = next;                       // Remember the old carry for next time.
        }   
    }
}   

Answer 4

You can shift only members of that arrays, a char (or an int). You can't shift an entire array. Shifting my_array tries to perform a shift operation on an array type (or a pointer to char) which is impossible. Do this instead:

for (i = 0; i < size; i++) {
  my_array[i] >>= 1;
}

Also you must be careful with chars because they are usually signed, and a char containing a negative value will bring '1' from the left instead of zeros. So you better use unsigned chars.

EDIT: The code above is simplistic. If you intended to shift right the array as a whole, not just each byte on its own, then you need to "manually" copy each LSB to the MSB of the byte to its right. Take a loop at the answer of Richard Pennington.

Answer 5

You have to shift and compare elementwise.

for(i = 0; i < len; ++i)
    array[i] >>= 3;

for example. If you want to move the bits shifted out of one element to the next, it's more complicated, say you're shifting right, then

unsigned char bits1 = 0, bits2 = 0;
for(i = len-1; i >= 0; --i) {
    bits2 = array[i] & 0x07;
    array[i] >>= 3;
    array[i] |= bits1 << 5;
    bits1 = bits2;
}

traversing the array in the other direction because you need the bits from the next higher slot.

Answer 6

/**
 * shift a number of bits to the right
 *
 * @param   SRC         the array to shift
 * @param   len         the length of the array
 * @param   shift       the number of consecutive bits to shift
 *
*/
static void shift_bits_right(uint8_t SRC[], uint16_t len, uint32_t shift) {
    uint32_t i = 0;

    uint8_t start = shift / 8;
    uint8_t rest = shift % 8;
    uint8_t previous = 0;

    for(i = 0; i < len; i++) {
        if(start <= i) {
            previous = SRC[i - start];
        }
        uint8_t value = (previous << (8 - rest)) | SRC[i + start] >> rest;
        SRC[i + start] = value;
    }
}