Grabbing bit 1 and 6, and 2 through 5 in an unsigned char in C

Question

I have an unsigned char at 6 bytes long. The value being stored within the char is:

Hex: 53167DFD95B7
Binary: 010100 110001 011001 111101 111111 011001 010110 110111

What I need to do is capture bit 1 and bit 6. Then convert that to a decimal. Then capture bit 2-5 and convert that to decimal For example, bit 1 here is 0, bit 6 is 0, so binary 00, is decimal 0. Then for bit 2-5, binary 1010, or decimal 10. Then move to the next group of 6 bits.

Bit 1 is 1, Bit 6 is 1, so binary 11, or decimal 3 Bit 2-5 is binary 1000, or decimal 8

Bit 1 is 0, Bit 6 is 1, so binary 01, or decimal 1 Bit 2-5 is binary 1100, or decimal 12

And so on for the remaining groups of 6 bits.

I'm not really sure how I should be masking, shifting for this. Since this is only 6 bits at a time I am having some difficulty. Any help with this would be greatly appreciated! Thank you all in advance.

EDIT

int getBitVal(unsigned char *keyStrBin, int keyIndex) {

    int keyMod = keyIndex % 8;  
    int keyIn = keyIndex / 8;

    return (((keyStrBin[keyIn]) >> (7 - (keyMod))) & 1);

}
void getSValueMajor(char **tableS, unsigned char *f, unsigned char *sValue) {
    int i, bitOne, bitSix;
    int sCol;

    for (i = 0; i < 8; i++) {
        bitOne = getBitVal(f, 0);
        bitSix = getBitVal(f, 5);
        // Do something here to get only bits 2-5. Doesn't matter if its decimal. Just need the 4 bits.

    }

}

Ill shift by 6 bits I guess at the end of the loop to go to the next 6 bits, but not sure how to read those 4 bits into a variable.

Answer 1

I've updated this because a commenter mentioned that it wasn't using real numbers. I know this but the point was to leave something up to the person doing their homework.

The following is not a neat answer to this but it should give anyone coming to this an idea on how to solve it in a neat way. In the second method I used a string as the thing I'm converting and converted this to a binary number and from there I've performed a few bit manipulations.

With everything in bit fiddling there are always ways to speed things up ie you could use Masks for all 8 bytes to avoid some shifts, you could also unroll the loop etc.

#include <assert.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/*
 010100 == 00 , 1010 === 0 , 10 
 110001 == 11 , 1000 === 3 , 8
 011001 == 10 , 1100 === 2 , 12
 111101 == 11 , 1110 === 3 , 14
 111111 == 11 , 1111 === 3 , 15
 011001 == 01 , 1100 === 1 , 12
 010110 == 00 , 1011 === 0 , 11
 110111 == 11 , 1011 === 3 , 11
*/
int main(void) {
  size_t x = 48; 
  size_t v = 91356068156855;

  size_t one66 = 0;
  size_t two55 = 0;
  size_t bit1  = 0;
  size_t bit6  = 0;

  //Masks
  size_t sixty3 = 63; 
  size_t thirty = 30; 

  size_t b[8];
  b[0] = (v & (sixty3 << 42ULL)) >> 42ULL;
  b[1] = (v & (sixty3 << 36ULL)) >> 36ULL;
  b[2] = (v & (sixty3 << 30ULL)) >> 30ULL;
  b[3] = (v & (sixty3 << 24ULL)) >> 24ULL;
  b[4] = (v & (sixty3 << 18ULL)) >> 18ULL;
  b[5] = (v & (sixty3 << 12ULL)) >> 12ULL;
  b[6] = (v & (sixty3 << 6ULL))  >> 6ULL;
  b[7] = (v & (sixty3 << 0ULL))  >> 0ULL;

  for(x = 0; x < 8;x++) {
    one66 = 0;
    two55 = 0;
    bit1 = (b[x] & 1)  > 0;
    bit6 = (b[x] & 32) > 0;
    one66 |= bit1 << 1;
    one66 |= bit6 << 0;
    two55 = (b[x] & thirty) >> 1;
    printf("%zu %zu\n", one66, two55);
  }

 // Method 2 using a string as the input...
 //                    |     |     |     |     |     |     |     |      
  char     pat[]     = "010100110001011001111101111111011001010110110111";
  size_t   patlength = strlen(pat);
  for(x = 0; x < patlength; x += 6) {
    size_t one6 = 0;
    size_t two5 = 0;
    if(pat[x] == '1') {
      one6 |= 1ULL << 0;
    }
    if(pat[x + 5] == '1') {
      one6 |= 1ULL << 1;
    }
    assert(one6 < 4);
    if(pat[x + 1] == '1') {
      two5 |= 1ULL << 3;
    }
    if(pat[x + 2] == '1') {
      two5 |= 1ULL << 2;
    }
    if(pat[x + 3] == '1') {
      two5 |= 1ULL << 1;
    }
    if(pat[x + 4] == '1') {
      two5 |= 1ULL << 0;
    }
    assert(two5 < 16);
    printf("%zu %zu\n", one6, two5);
  }
  return 0;
}