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