Hey folks,

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 x*3.5 by adding 2*x, x and x/2. To calculate 2*x, 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:

`// C++ program for above approach`

`#include <iostream>`

`using`

`namespace`

`std;`

`// 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