comparing an array of bits

[byenary]

Hello
I have got 2 variabeles in an application the type;
typedef guint32 data[8]; -> so 256 bits
Now i have a load of data of this type derrived from fileheaders.
I want to compare these and get a number of how many bits are set to 1 on the same place.
So for Ex with 8 bits
data1 : 00110011
data2 : 00100010
This should give me 2 as an answer. I don't really know how to do this i am not that expririenced. I guess ORRING the whole array with the other or something like bitshifting is what i should do but i don't know how.

Thank you very much for any help!

[Banfa]

This should give me 2 as an answer. I don't really know how to do this i am not that expririenced. I guess ORRING the whole array with the other or something like bitshifting is what i should do but i don't know how.

XORing (^) the 2 values will give a result that has bits set where the bits in the 2 original values do not have the same value.

Having got this result count the number of bits set in it by, testing bit 0 shifting right 1 place and continuing while the result != 0.

[byenary]

made a little test ap wich does the job

#include <stdio.h>
 
void show_binary(unsigned short u);
void show_lastbit(unsigned short u);
unsigned short count_ones(unsigned short test);
 
int main(void)
{
  unsigned short getal1;
  unsigned short getal2;
  getal1 = 14531;
  getal2 = 14045;
  show_binary(getal1);
  show_binary(getal2);
  unsigned short test = getal1 & getal2;
  show_binary(test);
  printf("aantal = %d",count_ones(test));
  /*
  0 0 1 1 1 0 0 0 1 1 0 0 0 0 1 1
  0 0 1 1 0 1 1 0 1 1 0 1 1 1 0 1
  0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 1
  1 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 aantal = 5
  */
  return 0;
}
 
unsigned short count_ones(unsigned short test){
  int aantal = 0;   
  int i=0;
  for(i=0;i<sizeof(test)*8;i++){  
      if(test & 1){ 
          printf("1 ");
          aantal++;
          }
      else 
          printf("0 ");
  test = test >> 1;      
  }     
    return aantal;
}
 
void show_binary(unsigned short u)
{
  int n;
 
  for( n = 32768; n > 0; n = n / 2 )
    if(u & n) 
        printf("1 ");
    else 
        printf("0 ");
 
  printf("\n");
}
 
void show_lastbit(unsigned short u)
{
  int n;
    if(u & 1) 
      printf("1 ");
  else 
      printf("0 ");
 
  printf("\n");
}