shifting arrays

So what you want is a sort of 'shift with carry.' Looks easy enough:

array[24] >>= 1;
array[24] |= (array[23] << 7);
array[23] >>= 1;
array[23] |= (array[22] << 7);

and so on. Feel free to make a for loop out of it.
If you care about speed or code size, you should implement this in assembly using the RRC instruction.

Regards,
- mike

" want to shift the all array right/left by 1 bit."

Do you mean you want to shift each element independently?

Or do you want to shift bits out of one element and into its adjacent element?

#include <stdio.h>

int main() {
    unsigned char array_1[24] = {
        0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,
        0x09,0x10,0x11,0x12,0x13,0x14,0x15,0x16,
        0x17,0x18,0x19,0x20,0x21,0x22,0x23,0x24};
    int i;

    for (i = 0; i < sizeof(array_1); i++) {
        printf("%02x ",array_1[i]);
    }
    printf("\n");

    // Shift all data right one bit, with carry from lower index.
    for (i = 23; i >= 1; --i) {
        array_1[i] = (array_1[i]>>1) | ((array_1[i-1] << 7) & 0x80);
    }
    array_1[0] >>= 1; // possibly take care of rotate from last entry too.
    for (i = 0; i < sizeof(array_1); i++) {
        printf("%02x ",array_1[i]);
    }
    printf("\n");

    // Shift all data left one bit, with carry from higher index.
    for (i = 0; i < 23; i++) {
        array_1[i] = (array_1[i] << 1) | ((array_1[i+1] >> 7) & 0x01);
    }
    array_1[23] <<= 1; // possibly take care of rotate from first entry too.
    for (i = 0; i < sizeof(array_1); i++) {
        printf("%02x ",array_1[i]);
    }
    printf("\n");
    return 0;
}

Gives result:

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
00 81 01 82 02 83 03 84 04 88 08 89 09 8a 0a 8b 0b 8c 0c 90 10 91 11 92
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

i want to shift the bits out of one element and into the start element like a loop.
in this case i want to rotate all the bits in the array so no bit will lost. the last bit(msb) will enter the place of the first bit(lsb).

many thank for Per Westermark
very elegant, do exactly what i mean.

i added this to Per Westermark code so after 192 loops all the array is rotating to its original form.

unsigned char array_1[25] = {
0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x10,0x11,0x12,0x13,0x14,0x15,0x16,0x17,0x18,0x19,0x20,0x21,0x22,0x23,0x24,0x00};

for (i = 0; i < sizeof(array_1); i++) {
  printf("%02bx ",array_1[i]);
  }
printf("\n");

for (j = 0; j < 192; j++) {
  // Shift all data right one bit,
  // with carry from lower index.
  if (array_1[23] & 0x01) {
         array_1[24] = 0x80;  // buffer for exit '1'
         }
     for (i = 23; i >= 1; --i) {
       array_1[i] =
          (array_1[i]>>1) | ((array_1[i-1]<<7) & 0x80);
         }
  //for (i = 0; i < sizeof(array_1); i++) {
         printf("%02bx ",array_1[i]);
         }
  // possibly take care of rotate from last entry too.
  array_1[0] >>= 1;
  // insert the '1' to entery point
  array_1[0] |= array_1[24];
  for (i = 0; i < sizeof(array_1); i++) {
       printf("%02bx ",array_1[i]); }
  printf("\n");
  array_1[24] = 0x00;  // reset buffer
}

thanks again