C code for binary multiplication by 7 then divide by 2

Given an integer x, write a function that multiplies x with 3.5 and returns the integer result. You are not allowed to use %, /, *.

Examples : Input: 2 Output: 7 Input: 5 Output: 17 (Ignore the digits after decimal point)

Solution:
1. We can get x3.5 by adding 2x, x and x/2. To calculate 2x, left shift x by 1 and to calculate x/2, right shift x by 2. The function should not use any of the arithmetic operators (+, ++, –, -, … etc). Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. The idea is to use subtractor logic. Write a program to subtract one from a given number. The idea is to use bitwise operators. Like addition, Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, … etc). Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. The idea is to use subtractor logic. The use of operators like ‘+’, ‘-‘, ‘’, ‘/’, ‘++’, ‘–‘ …etc. are not allowed.

Another way of doing this could be by doing a binary multiplication by 7 then divide by 2 using only <<, ^, &, and >>.

But here we have to mention that only positive numbers can be passed to this method.

Below is the implementation of the above approach:

// Function to multiple number

// with 3.5

int multiplyWith3Point5( int x){

int r = 0;

// The 3.5 is 7/2, so multiply

// by 7 (x * 7) then

// divide the result by 2

// (result/2) x * 7 -> 7 is

// 0111 so by doing mutiply

// by 7 it means we do 2

// shifting for the number

// but since we doing

// multiply we need to take

// care of carry one.

int x1Shift = x << 1;

int x2Shifts = x << 2;

r = (x ^ x1Shift) ^ x2Shifts;

int c = (x & x1Shift) | (x & x2Shifts)

| (x1Shift & x2Shifts);

while (c > 0) {

c <<= 1;

int t = r;

r ^= c;

c &= t;

}

// Then divide by 2

// r / 2

r = r >> 1;

return r;

}

// Driver Code

int main() {

cout<<(multiplyWith3Point5(5));

return 0;

}

Output

17