Bitand function in C - c

Question

I want to make a bitwise AND computation over integers, but without converting them to binary numbers. For example, I have a integer "10111" (it is integer, not binary) and another integer "01001". I want bitwise AND of these numbers without converting them to binary and then making bitwise AND. I know it is not bitwise what I ask, but I want something similar to this. I know it can be interpreted initially as binary, converted to decimal and then do bitwise AND, but I do not want that. I want something like this:
int a;
int b;
int temp;
double result;
temp = a & b;
while (result != 0) {
if (result % 10 == 1)
count++;
result /= 10;
}
int length = floor(log10(abs(a))) + 1;
result = count / length;
return result;
I want this to check similarity of the Bag of Words(from natural language processing, string of 0s and 1s). I am importing Bag of Words in Monetdb, Column type should be Integer (Not string). If I have for example "10111" and "01001" in the Integer type cells, I want to get "00001" and fraction 1/5, because only 1 positions matches.
Thanks in advance

Answer 1

Might be a bit bulky, but it kind of works) You can optimize it yourself. I hope that I get you correctly.
IDEOne demo
#include <stdio.h>
unsigned int weirdAnd(unsigned int a, unsigned int b) {
unsigned int result = 0;
unsigned int coef = 1;
while (a && b) {
result += ((a % 10) && (b % 10)) * coef;
coef *= 10;
a /= 10;
b /= 10;
}
return result;
}
unsigned int weirdOr(unsigned int a, unsigned int b) {
unsigned int result = 0;
unsigned int coef = 1;
while (a || b) {
result += ((a % 10) || (b % 10)) * coef;
coef *= 10;
a /= 10;
b /= 10;
}
return result;
}
int main(void) {
// your code goes here
unsigned int a = 10110;
unsigned int b = 10011;
printf("%u and \n%u = \n%u\n\n", a, b, weirdAnd(a, b));
printf("%u or \n%u = \n%u\n\n", a, b, weirdOr(a, b));
return 0;
}
Output:
10110 and 10011 = 10010
10110 or 10011 = 10111

Answer 2

The problem is that and works on bits only, and it does not care if the input numbers are given in decimals, octal, hexadecimal, or any other way. To force and to work correctly, you must give it an input in 'binary', that is, only ones and zeroes. To do so you need to grab each digit of the input numbers as if they are binary digits.
The following works as expected:
#include <stdio.h>
int cheap_pow (int base, int exponent)
{
int result = base;
while (exponent-- > 0)
result *= base;
return result;
}
int main (void)
{
int a = 10111, b = 1001 ;
int result, factor;
printf ("BAD: %05d AND %05d = %05d\n", a, b, a & b);
printf ("GOOD: %05d AND %05d = ");
result = 0;
factor = 1;
while (a | b)
{
result += factor * ((a & 1) and (b & 1));
factor *= 10;
a /= 10;
b /= 10;
}
printf ("%05d\n", result);
}
but you must be careful when defining your inputs. When written directly into your source code, a value b = 01001 will be interpreted by the C compiler as octal (base 8), rather than decimal; and it will have the (decimal) value 513. It's just a Rule of C.
If your input comes from elsewhere, you don't need to take this into account – except when you use a standard function such as strtol, where you should carefully read the documentation because it does the same thing:
If base is zero or 16, the string may then include a "0x"
prefix, and the number will be read in base 16; otherwise, a zero
base is taken as 10 (decimal) unless the next character is '0', in
which case it is taken as 8 (octal).
An additional note is this program only will work in the range for signed int, that is (currently) up to 10 "binary" digits. If you need some more, you could switch to a larger type; but beyond that, you are better off with a not-numerical solution and use a character strings throughout.

Answer 3

int a, b;
int tmp = a & b;
int res = 0;
while ((tmp != 0 &&) (tmp / 10 != 0)){
int dig = tmp % 10;
res = (dig == 1)? ++res: res;
tmp /= 10;
}
Try this.