how can I get bit-by-bit of an arbitrary decimal number

Here's a version that right shifts the number until zero and builds the buffer from right to left:

#include <stdio.h>

const char *
bitprt(unsigned int val)
{
    char *bp;
    char *rtn;
    static char buf[128 + 1];

    bp = &buf[sizeof(buf) - 1];
    *bp-- = 0;

    *bp = '0';
    rtn = bp;

    for (;  val != 0;  val >>= 1, --bp) {
        if (val & 1)
            *bp = '1';
        else
            *bp = '0';
        rtn = bp;
    }

    return rtn;
}

void
dotest(unsigned int val)
{
    const char *buf;

    buf = bitprt(val);
    printf("%20u: '%s'\n",val,buf);
}

int
main(void)
{

    dotest(28);
    dotest(171);
    dotest(3);
    dotest(324);
    dotest(66);

    return 0;
}

Here's the output of the program:

                  28: '11100'
                 171: '10101011'
                   3: '11'
                 324: '101000100'
                  66: '1000010'

If you are trying to get bit representation from integer value, here is an example of how to do that. You have to check each bit manually, there is not built in utility to iterate through "bits".

#include <stdio.h>

int main(int argc, char const *argv[])
{
  int numbers[] = { 28, 171, 3, 324, 66 };
  int numberslength = (int) (sizeof(numbers) / sizeof(int));

  // each number
  for (int i = 0; i < numberslength; ++i)
  {
    int number = numbers[i];  // get the number
    int mask = 1;             // start at the beginning
    for (int j = 0; j < sizeof(int) * 8; ++j)
    {
      // if the number has a bitwise and in that bit, 1
      printf("%c", number & mask ? '1': '0');

      // move the mask over to the next bit
      mask <<= 1;
    }

    // separate outputs by newline
    printf("\n");
  }

  return 0;
}