Bitwise Operation on a byte and an int

Question

I have a byte array represented as

char * bytes = getbytes(object); //some api function

I want to check whether the bit at some position x is set.

I've been trying this

int mask = 1 << x % 8;
y= bytes[x>>3] & mask;

However y returns as all zeros? What am I doing incorrectly and is there an easier way to check if a bit is set?

EDIT:

I did run this as well. It didn't return with the expected result either.

  int k = x >> 3;
   int mask = x % 8;
    unsigned char byte = bytes[k];
   return (byte & mask);

it failed an assert true ctest I ran. Byte and Mask at this time where "0002" and 2 respectively when printed from gdb.

edit 2: This is how I set the bits in the first place. I'm just trying to write a test to verify they are set.

unsigned long x = somehash(void* a);

unsigned int mask = 1 << (x % 8);

unsigned int location = x >> 3;
char* filter = getData(ref);

filter[location] |= mask;

The expression looks fine to me. What is the type of `y`? Does `bytes` point to consecutive memory region of `char`s? — nglee, Jul 04 '17 at 04:58
One possible source of confusion is bit-endianness. As is, your code interprets `x = 0` as the least significant bit of the first byte. — user3386109, Jul 04 '17 at 04:58
Print everything separatly for debugging, especially `bytes[x>>3]`. — Yunnosch, Jul 04 '17 at 05:06
@nglee y is in the official code listed as a bool which is just an int typedef — knowads, Jul 04 '17 at 05:12
Have you given any thought to complications that will arise from `y= bytes[x>>3] & mask;`, e.g. `y = char & int`? (one byte anded against four). Perhaps using `uint8_t` for all would help? — David C. Rankin, Jul 04 '17 at 05:18
@DavidC.Rankin this is how I set the code in the first place: I'm just trying to write something to verify it. I just need something that returns a 1 if that bit is set. unsigned int mask = 1 << (x % 8); unsigned int location = x >> 3; char* filter = getData(ref); filter[location] |= mask; — knowads, Jul 04 '17 at 05:20
Recall if your system is *little-endian* the least significant bits of an `int` are 3-bytes to the left... — David C. Rankin, Jul 04 '17 at 05:21

babon · Accepted Answer · 2017-07-04T06:41:47.480

0

This would be one (crude perhaps) way from the top of my head:

#include "stdio.h"
#include "stdlib.h"

// this function *changes* the byte array
int getBit(char *b, int bit)
{
  int bitToCheck = bit % 8;
  b = b + (bitToCheck ? (bit / 8) : (bit / 8 - 1));

  if (bitToCheck)
    *b = (*b) >> (8 - bitToCheck);

  return (*b) & 1;
}

int main(void)
{
  char *bytes = calloc(2, 1);
  *(bytes + 1)= 5;  // writing to the appropiate bits
  printf("%d\n", getBit(bytes, 16)); // checking the 16th bit from the left
  return 0;
}

Assumptions:

A byte is represented as:

----------------------------------------
| 2^7 | 2^6 | 2^5 | 2^4 | 2^3 |...     |
----------------------------------------

The left most bit is considered bit number 1 and the right most bit is considered the max. numbered bit (16th bit in a 2 byte object).

It's OK to overwrite the actual byte object (if this is not wanted, use memcpy).

edited Jul 04 '17 at 06:41

answered Jul 04 '17 at 05:36

babon

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I think you have to consider the endiannes of the system like [here](https://stackoverflow.com/a/4181991/8051589), because you have multiple bytes in an array. – Andre Kampling Jul 04 '17 at 06:11
You guys know what. I think it might be something to do with the bool typedef and the api I was using to retrieve the byte. I made everything vanilla ints and it seemed to run as expected. I'm going to look into it further and discuss with the original author – knowads Jul 04 '17 at 06:25
@AndreKampling Ahhh... you just had to make this difficult for me ;P – babon Jul 04 '17 at 06:27
@knowads 'much ado about nothing'? – babon Jul 04 '17 at 06:27
@babon I guess so. I'm still green enough to not realize that I can be completely correct but something can go wrong because of an implementation decision in someone's original API not accounting for certain use cases – knowads Jul 04 '17 at 06:29
@knowads no problem, carry on :) – babon Jul 04 '17 at 06:33

Bitwise Operation on a byte and an int

1 Answers1