CODEWITHSUNDEEP
cbit-manipulationbytelookup-tablesbitmask
unconditionalcoder
unconditionalcoder

Reputation: 753

Bit rearrangement/manipulation in C

I want to arrange bits in a byte, to result in a certain order. For example, if the starting byte is as follows 0 1 1 0 1 0 1 0 with bits labeled as 1 2 3 4 5 6 7 8, I want to arrange it so it matches the following positioning: 2 4 3 5 7 1 8 6 this results to: 1 0 1 1 1 0 0 0. What would be the most efficient way of doing so? I read about "look-up" tables but I am not sure how this works. Can someone give an example and an explanation of an efficient way of doing this bit rearrangement in C.

Upvotes: 3

Views: 473

Answers (3)

eXtranium
eXtranium

Reputation: 1

Here is one way to change bit-positions. With & (and-operator) we select certain bits from the char and then shift them to new bit-positions. Finally all the shifted bits will happily join together by | (or-operator). The left shift << will move bits left and right shift >> to the right. I took the freedom to renumber bit-positions. 7 means most-significant bit on the left and 0 is least-significant bit, so the left and right shift operations descripts shifting direction correctly.

And why there is the shift operations first and then AND-operation for the last two rows? – Because char-type can be unsigned and if we do the right shift for the negative value, eg 11111000 (-8), the most-significant bit will be copied; 11111000 >> 2 will result (1 filled from this end -->) 11111110 (-2). (See Right shifting negative numbers in C.)

But back into function:

char changebitpositions (char ch) {

                                  // bit locations (- = don't care)
                                  // before      after
    return (ch & 0x20)            // --5----- => --5----- (no change)
        | ((ch & 0x49) << 1)      // -6--3--0 => 6--3--0-
        | ((ch & 0x12) << 2)      // ---4--1- => -4--1---
        | ((ch >> 5) & 0x04)      // 7------- => -----7--
        | ((ch >> 2) & 0x01);     // -----2-- => -------2

        // original and result:      76543210 => 64531702
}

Upvotes: 0

TonyB
TonyB

Reputation: 947

You could create an array of "unsigned char" with 256 entries. The index into that array would be the current value of the byte to be converted, and the value at that entry would be the "converted" value.

Alternatively, you could use bit masking, and "if" statements... but it would less efficient.

Here's a snippet of the "array" method... with only a few values defined... ... and no output of the output in "binary-text" format.

#include<stdio.h>

unsigned char lkup[256] =
    { 0x00,            /* idx: 0  (0x00) */
      0x02,            /* idx: 1  (0x01) (0b00000001) */
      0x08,            /* idx: 2  (0x02) (0b00000010) */
      0x0a,            /* idx: 3  (0x03) (0b00000011) */
      0x01             /* idx: 4  (0x04) (0b00000100) */
      };

int main(int argc, char **argv)
{
    unsigned char wk = 3;

    printf("Input: %u  output: >%u\n", wk, lkup[wk]);
}

Upvotes: 3

ckruczek
ckruczek

Reputation: 2410

I think I understood what he wants to achieve. This code may help you:

#include <stdio.h>
#include <stdint.h>

int main(void) {
  uint8_t original = 0b01101010;
  uint8_t positions[8] = {1,3,2,4,6,0,7,5};
  uint8_t result = 0;

  for(int i = 0; i < 8; i++)
  {   
     if(original & (1 << (7 - positions[i])))
        result |= (1 << (7-i));
  }
 return 0;
}

The first thing I have done is to create a byte that represents the original value as well as a array of the positions you want to change. Next step ist to look the original byte at the xth. position is zero or one and then shift the value in the result if so. The last for-loop is just for printing the result. I adjusted your indices to be zero-based.

Upvotes: 2

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