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
[code=c]
#include <stdio.h>

void show_binary(uns igned short u);
void show_lastbit(un signed short u);
unsigned short count_ones(unsi gned short test);

int main(void)
{
unsigned short getal1;
unsigned short getal2;
getal1 = 14531;
getal2 = 14045;
show_binary(get al1);
show_binary(get al2);
unsigned short test = getal1 & getal2;
show_binary(tes t);
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(unsi gned short test){
int aantal = 0;
int i=0;
for(i=0;i<sizeo f(test)*8;i++){
if(test & 1){
printf("1 ");
aantal++;
}
else
printf("0 ");
test = test >> 1;
}
return aantal;
}

void show_binary(uns igned 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(un signed short u)
{
int n;
if(u & 1)
printf("1 ");
else
printf("0 ");

printf("\n");
}[/code]