How do I do bitwise AND of two numbers?

Question

For example let's say that I want to calculate bitwise the AND function between 124 and 100, how exactly should I do this? Please explain briefly. Thank you.

Solution 1

int number = 124 & 100

---

Explanation:

Convert 124 to bitwise format, assuming a 32 bit integer. =

0000 0000 0000 0000 0000 0000 0111 1100

Convert 100 to bitwise format, assuming a 32 bit integer. =

0000 0000 0000 0000 0000 0000 0110 0100

ANDING bitwise:

0000 0000 0000 0000 0000 0000 0111 1100
0000 0000 0000 0000 0000 0000 0110 0100
----------------------------------------
0000 0000 0000 0000 0000 0000 0110 0100
---------------------------------------

Which is 100 in decimal! This will be contained in number in the above code!

The AND gate truth table is:

+-------+---------+
| A | B | A AND B |
+-------+---------+
|0  | 0 |    0    |
|0  | 1 |    0    |
|1  | 0 |    0    |
|1  | 1 |    1    |
+-------+---------+

---

#include <stdio.h>

#define IS_SET(a,b) ((((a & (1 << b)) >> b) > 0?1:0))
#define SET(a,b) ((a |= (1 << b)))
#define CLEAR(a,b) ((a &= ~(1 << b)))

int AND(int a, int b)
{
    int i;
    int c = 0;
    for(i = 0; i < sizeof(int); i++){
      if(IS_SET(a,i) && IS_SET(b,i)){
        SET(c,i);
      }else{
        CLEAR(c,i);
      }
    }
    return c;
}

int main(int argc,char *Argv[])
{
   int a = 10;
   int b = 30;
   //Basic UNIT test!
   if((a & b) == AND(a,b)){
      printf("Test pass!");
   }else{
      printf("Test fail!");
   }
}

Solution 2

Use the bitwise and operator in C & like 124 & 100

Comments

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