Bitwise operation on char array

Question

So I have a binary representation of a number as a character array. What I need to do is shift this representation to the right by 11 bits.

For example,

I have a char array which is currently storing this string: 11000000111001 After performing a bitwise shift, I will get 110 with some zeros before it.

I tried using this function but it gave me strange output:

char *shift_right(unsigned char *ar, int size, int shift)
{
int carry = 0;                              // Clear the initial carry bit.
while (shift--) {                           // For each bit to shift ...
    for (int i = size - 1; i >= 0; --i) {   // For each element of the array   from high to low ...
        int next = (ar[i] & 1) ? 0x80 : 0;  // ... if the low bit is set, set the carry bit.
        ar[i] = carry | (ar[i] >> 1);       // Shift the element one bit left and addthe old carry.
        carry = next;                       // Remember the old carry for next time.
    }
}
return ar;
}

Any help on this would be very much appreciated; let me know if I'm not being clear.

I have no idea what you think this bitwise-OR with 0x80 is going to do .. we are just operating on characters 1 and 0 here. If you do 0x80 | '1' you will end up with a character that will print as some sort of symbol — M.M
– M.M, Commented May 7, 2014 at 5:46
Your right, I was trying to use the answer provided here (stackoverflow.com/questions/10367616/…) as a guide but that didn't work, — user3610554
– user3610554, Commented May 7, 2014 at 6:25

Mahonri Moriancumer · Accepted Answer · 2014-05-07 06:33:33Z

2

They are just characters...

char *shift_right(unsigned char *ar, int size, int shift)
   {

   memmove(&ar[shift], ar, size-shift);
   memset(ar, '0', shift);

   return(ar);
   };

Or, convert the string to a long-long, shift it, then back to a string:

char *shift_right(char *ar, int size, int shift)
   {
   unsigned long long x;
   char *cp;

   x=strtoull(ar, &cp, 2);  // As suggested by 'Don't You Worry Child'
   x = x >> shift;
   while(cp > ar)
      {
      --cp;
      *cp = (1 & x) ? '1' : '0';
      x = x >> 1;
      }

   return(ar);
   };

edited May 7, 2014 at 6:33

answered May 7, 2014 at 5:46

Mahonri Moriancumer

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5 Comments

M.M Over a year ago

of course, check shift < size and perhaps also shift > 0 before doing this

Lundin Over a year ago

Why you would need long long to store 14 bits of data...?

user3610554 Over a year ago

The second function worked well. What would I need to add to this if I wanted to store the 11 bits that I shifted out?

Lundin Over a year ago

@user3610554 Simply save them before calling the shift function.

Mahonri Moriancumer Over a year ago

@user3610554, keep in mind that 'size' is ignored by the second function. Perhaps it could be used to ensure that size is less than, or equal to, 64 bits? (just an afterthought).

0xF1 · Accepted Answer · 2014-05-07 06:22:55Z

0

If you really want to use bitwise shifting, then you can't do it on a string. Simply not Possible!!

You have to convert it to integer (use strtol for that) then do bitwise shifting. After that, convert it back to string (no standard library function for that, use for loop).

answered May 7, 2014 at 6:22

0xF1

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Comments

Lundin · Accepted Answer · 2014-05-07 06:55:20Z

0

I would advise to keep the code simple and readable.

#include <stdio.h>
#include <stdlib.h>

void shift_right (char* dest, const char* source, int shift_n)
{
  uint16_t val = strtoul(source, NULL, 2);
  val >>= shift_n;

  for(uint8_t i=0; i<16; i++)
  {
    if(val & 0x8000) // first item of the string is the MSB
    {
      dest[i] = '1';
    }
    else
    {
      dest[i] = '0';
    }
    val <<= 1;    // keep evaluating the number from MSB and down
  }

  dest[16] = '\0';
}


int main()
{
  const char str [16+1] = "0011000000111001";
  char str_shifted [16+1];

  puts(str);
  shift_right(str_shifted, str, 11);
  puts(str_shifted);

  return 0;
}

answered May 7, 2014 at 6:55

Lundin

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2 Comments

user3610554 Over a year ago

This works for my example but it doesn't work if my representation has more bits. How could I do it for a representation of 'x' many bits?

Lundin Over a year ago

@user3610554 "x" bits doesn't make any sense, nor does anything which isn't a multiple of 8. There must be a specified maximum limit. When you have that, you can easily modify the above to suit any size.

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