hi,
i'm using uVision3.33 , uC-80c31x2. my question is: I'v initilize buffer array which is been loaded with 24 Byte each interval time. now i want to shift the all array right/left by 1 bit. maybe an example will clarify bettet:
unsigned char array_1[24]; 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};
the array's values in the example are just for example. probably the operand '>>' must be use. the result i wish for is: (shifting all the array right only one time)
unsigned char array_1[24] = { 0x00,0x81,0x01,0x82,0x02,.....,0x92};
thanks in advance.
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
many thank for Per Westermark very elegant, do exactly what i mean.
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).
#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
" 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?
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
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