Iterate through bits in C

Question

I have a big char * str, where the first 8 characters (which is 64 bits, if I'm not mistaken) represents a bitmap. Is there any way to iterate over these 8 characters and see which bits are 0? I have many problems understanding the concept of bits, since you cannot “see” them in the code, so I cannot think of any way to do this.

Answer 1

Imagine you only have one byte, one char my_char. You can check for individual bits using bitwise operators and bit offsets.

unsigned char my_char = 0xAA;
int what_bit_i_am_testing = 0;

while (what_bit_i_am_testing < 8) {
  if (my_char & 0x01) {
     printf("bit %d is 1\n", what_bit_i_am_testing);
  }
  else {
     printf("bit %d is 0\n", what_bit_i_am_testing);
  }

  what_bit_i_am_testing++;
  my_char = my_char >> 1;
}

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Answer 2

little-endian:

const int cBitmapSize = 8;
const int cBitsCount = cBitmapSize * 8;
const unsigned char cBitmap[cBitmapSize] = /* some data */;

for(int n = 0; n < cBitsCount; n++)
{
  unsigned char Mask = 1 << (n % 8);
  if(cBitmap[n / 8] & Mask)
  {
    // if n'th bit is 1...
  }
}

Answer 3

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int isBitSet = bitmap & (1 << bit_position);

isBitSet 0, . , .

Answer 4

char b :

for (int i=0; i<8; i++) {
  printf("This is the %d-th bit : %d\n",i,(b>>i)&1);
}

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Answer 5

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:

#define LSBIT(X)                    ((X) & (-(X)))
#define CLEARLSBIT(X)               ((X) & ((X) - 1))

, LSbit:

unsigned temp_bits;
unsigned one_bit;

temp_bits = some_value;
for ( ; temp_bits; temp_bits = CLEARLSBIT(temp_bits) ) {
    one_bit = LSBIT(temp_bits);
    /* Do something with one_bit */
}

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