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

Question

I want to read an int number bit by bit from a list, for example: decimal list: [28,171,3,324,66]

first: 28 binary: 100111 second: 171 binary: 10101011 ...

I was trying to do the bitwise operation but I don't know the length of the binary(e.g. length of 100111 = 6). So I can't do the bitwise operation. Is there any way to read an arbitrary number?

Can you explain `length of 100111 = 3`? You can get the number of bits with `sizeof(the_number) * 8`. — 001, May 28 '20 at 01:40
@JohnnyMopp So can I do this: for (int i=0; i < sizeof(the_number) * 8; i++) { the_number << i >> sizeof(the_number)-i } — SadSalad, May 28 '20 at 01:45
You can simply keep right-shifting the number until it becomes 0, then you've found the highest bit. — Barmar, May 28 '20 at 01:47

score 0 · Answer 1 · answered May 28 '20 at 01:50

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'

score 0 · Answer 2 · answered May 28 '20 at 01:55

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;
}

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

2 Answers2