Question:
How can you divide a number by 3 without using *, /, +, – % operator?
Input: 48
Output: 16
The method below implements the method using bit-wise operators.
//a function to add 2 numbers without using +
int add(int x, int y)
{
int a, b;
do
{
a = x & y;
b = x ^ y;
x = a << 1;
y = b;
} while (a);
return b;
}
int divideby3(int num)
{
int sum = 0;
while (num > 3)
{
sum = add(num >> 2, sum);
num = add(num >> 2, num & 3);
}
if (num == 3)
sum = add(sum, 1);
return sum;
}
//the main function
int main(void)
{
printf("Enter a number:- ");
int num;
scanf("%d",&num);
printf("The number divided by 3 is:- %d",divideby3(num));
return 0;
}
For example if your number is 10 then convert to binary
10 =>00001010
if 10 > 3 then shift the binary number 2 bits
Now num will be 00000010 ie, 2
Now sum will be 2
num = (shift the number by 2 bits)+(number BITWISE AND 3)
num = 2+2
Now the number will be 4 then, 4>3 => true so loop will be repeated
4=>00000100 then shift the binary number 2 bits
Now sum=2+1 ie, sum=3 (this value is returned)
num=(shift the number(00000100) by 2 bits)+(number BITWISE AND 3)
num=1+0
ie remainder=1
